7.E: Exercises
- Page ID
- 93820
If \(A\) is an invertible \(n\times n\) matrix, compare the eigenvalues of \(A\) and \(A^{−1}\). More generally, for \(m\) an arbitrary integer, compare the eigenvalues of \(A\) and \(A^m\).
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\(A^mX = λ^mX\) for any integer. In the case of \(−1,\: A^{−1}λX = AA^{−1}X = X\) so \(A^{−1}X = λ^{−1}X\). Thus the eigenvalues of \(A^{−1}\) are just \(λ^{-1}\) where \(λ\) is an eigenvalue of \(A\).
If \(A\) is an \(n\times n\) matrix and \(c\) is a nonzero constant, compare the eigenvalues of \(A\) and \(cA\).
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Say \(AX = λX\). Then \(cAX = cλX\) and so the eigenvalues of \(cA\) are just \(cλ\) where \(λ\) is an eigenvalue of \(A\).
Let \(A,\: B\) be invertible \(n\times n\) matrices which commute. That is, \(AB = BA\). Suppose \(X\) is an eigenvector of \(B\). Show that then \(AX\) must also be an eigenvector for \(B\).
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\(BAX = ABX = AλX = λAX\). Here it is assumed that \(BX = λX\).
Suppose \(A\) is an \(n\times n\) matrix and it satisfies \(A^m = A\) for some \(m\) a positive integer larger than \(1\). Show that if \(λ\) is an eigenvalue of \(A\) then \(|λ|\) equals either \(0\) or \(1\).
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Let \(X\) be the eigenvector. Then \(A^mX = λ^mX,\: A^mX = AX = λX\) and so \[\lambda^m=\lambda\nonumber\] Hence if \(\lambda\neq 0\), then \[\lambda^{m-1}=1\nonumber\] and so \(|\lambda|=1\).
Show that if \(AX = λX\) and \(AY = λY\), then whenever \(k,\: p\) are scalars, \[A(kX+pY)=\lambda (kX+pY)\nonumber\] Does this imply that \(kX+pY\) is an eigenvector? Explain.
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The formula follows from properties of matrix multiplications. However, this vector might not be an eigenvector because it might equal \(0\) and eigenvectors cannot equal \(0\).
Suppose \(A\) is a \(3\times 3\) matrix and the following information is available. \[\begin{aligned}A\left[\begin{array}{r}0\\-1\\-1\end{array}\right]&=0\left[\begin{array}{r}0\\-1\\-1\end{array}\right] \\ A\left[\begin{array}{c}1\\1\\1\end{array}\right]&=-2\left[\begin{array}{c}1\\1\\1\end{array}\right] \\ A\left[\begin{array}{r}-2\\-3\\-2\end{array}\right]&=-2\left[\begin{array}{r}-2\\-3\\-2\end{array}\right]\end{aligned}\] Find \(A\left[\begin{array}{r}1\\-4\\3\end{array}\right]\).
Suppose \(A\) is a \(3\times 3\) matrix and the following information is available. \[\begin{aligned}A\left[\begin{array}{r}-1\\-2\\-2\end{array}\right]&=1\left[\begin{array}{r}-1\\-2\\-2\end{array}\right] \\ A\left[\begin{array}{c}1\\1\\1\end{array}\right]&=0\left[\begin{array}{c}1\\1\\1\end{array}\right] \\ A\left[\begin{array}{r}-1\\-4\\-3\end{array}\right]&=2\left[\begin{array}{r}-1\\-4\\-3\end{array}\right]\end{aligned}\] Find \(A\left[\begin{array}{r}3\\-4\\3\end{array}\right]\).
Suppose \(A\) is a \(3\times 3\) matrix and the following information is available. \[\begin{aligned}A\left[\begin{array}{r}0\\-1\\-1\end{array}\right]&=2\left[\begin{array}{r}0\\-1\\-1\end{array}\right] \\ A\left[\begin{array}{c}1\\1\\1\end{array}\right]&=1\left[\begin{array}{c}1\\1\\1\end{array}\right] \\ A\left[\begin{array}{r}-3\\-5\\-4\end{array}\right]&=-3\left[\begin{array}{r}-3\\-5\\-4\end{array}\right]\end{aligned}\] Find \(A\left[\begin{array}{r}2\\-3\\3\end{array}\right]\).
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}-6&-92&12 \\ 0&0&0\\-2&-31&4\end{array}\right]\nonumber\] One eigenvalue is \(-2\).
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}-2&-17&-6 \\ 0&0&0\\1&9&3\end{array}\right]\nonumber\] One eigenvalue is \(1\).
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}9&2&8 \\ 2&-6&-2 \\ -8&2&-5\end{array}\right]\nonumber\] One eigenvalue is \(-3\).
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}6&76&16 \\ -2&-21&-4 \\ 2&64&17\end{array}\right]\nonumber\] One eigenvalue is \(-2\).
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}3&5&2 \\ -8&-11&-4 \\ 10&11&3\end{array}\right]\nonumber\] One eigenvalue is \(-3\).
Is it possible for a nonzero matrix to have only \(0\) as an eigenvalue?
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Yes. \(\left[\begin{array}{cc}0&1\\0&0\end{array}\right]\) works.
If \(A\) is the matrix of a linear transformation which rotates all vectors in \(\mathbb{R}^2\) through \(60^{\circ}\), explain why \(A\) cannot have any real eigenvalues. Is there an angle such that rotation through this angle would have a real eigenvalue? What eigenvalues would be obtainable in this way?
Let \(A\) be the \(2\times 2\) matrix of the linear transformation which rotates all vectors in \(\mathbb{R}^2\) through an angle of \(θ\). For which values of \(θ\) does \(A\) have a real eigenvalue?
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When you think of this geometrically, it is clear that the only two values of \(θ\) are \(0\) and \(π\) or these added to integer multiples of \(2π\).
Let \(T\) be the linear transformation which reflects vectors about the \(x\) axis. Find a matrix for \(T\) and then find its eigenvalues and eigenvectors.
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The matrix of \(T\) is \(\left[\begin{array}{rr}1&0\\0&-1\end{array}\right]\). The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}0\\1\end{array}\right]\right\}↔ -1,\:\left\{\left[\begin{array}{c}1\\0\end{array}\right]\right\}↔1\nonumber\]
Let \(T\) be the linear transformation which rotates all vectors in \(\mathbb{R}^2\) counterclockwise through an angle of \(π/2\). Find a matrix of \(T\) and then find eigenvalues and eigenvectors.
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The matrix of \(T\) is \(\left[\begin{array}{rr}0&-1\\1&0\end{array}\right]\). The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-i\\1\end{array}\right]\right\}↔ -i,\:\left\{\left[\begin{array}{c}i\\1\end{array}\right]\right\}↔i\nonumber\]
Let \(T\) be the linear transformation which reflects all vectors in \(\mathbb{R}^3\) through the \(xy\) plane. Find a matrix for \(T\) and then obtain its eigenvalues and eigenvectors.
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The matrix of \(T\) is \(\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&-1\end{array}\right]\). The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}0\\0\\1\end{array}\right]\right\}↔-1,\:\left\{\left[\begin{array}{c}1\\0\\0\end{array}\right],\:\left[\begin{array}{c}0\\1\\0\end{array}\right]\right\}↔1\nonumber\]
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}5&-18&-32\\0&5&4\\2&-5&-11\end{array}\right]\nonumber\] One eigenvalue is \(1\). Diagonalize if possible.
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The eigenvalues are \(−1,−1, 1\). The eigenvectors corresponding to the eigenvalues are: \[\left\{\left[\begin{array}{r}10&-2&3\end{array}\right]\right\}↔-1,\:\left\{\left[\begin{array}{r}7\\-2\\2\end{array}\right]\right\}↔1\nonumber\] Therefore this matrix is not diagonalizable.
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}-12&-28&28\\4&9&-8\\-4&-8&9\end{array}\right]\nonumber\] One eigenvalue is \(3\). Diagonalize if possible.
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}2\\0\\1\end{array}\right]\right\}↔1,\:\left\{\left[\begin{array}{r}-2\\1\\0\end{array}\right]\right\}↔1,\:\left\{\left[\begin{array}{r}7\\-2\\2\end{array}\right]\right\}↔3\nonumber\] The matrix \(P\) needed to diagonalize the above matrix is \[\left[\begin{array}{rrr}2&-2&7\\0&1&-2\\1&0&2\end{array}\right]\nonumber\] and the diagonal matrix \(D\) is \[\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&3\end{array}\right]\nonumber\]
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}89&38&268\\14&2&40\\-30&-12&-90\end{array}\right]\nonumber\] One eigenvalue is \(-3\). Diagonalize if possible.
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-6\\-1\\-2\end{array}\right]\right\}↔6,\:\left\{\left[\begin{array}{r}-5\\-2\\2\end{array}\right]\right\}↔-3,\:\left\{\left[\begin{array}{r}-8\\-2\\3\end{array}\right]\right\}↔2\nonumber\] The matrix \(P\) needed to diagonalize the above matrix is \[\left[\begin{array}{rrr}-6&-5&-8\\-1&-2&-2\\2&2&3\end{array}\right]\nonumber\] and the diagonal matrix \(D\) is \[\left[\begin{array}{rrr}6&0&0\\0&-3&0\\0&0&-2\end{array}\right]\nonumber\]
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}1&90&0\\0&-2&0\\3&89&-2\end{array}\right]\nonumber\] One eigenvalue is \(1\). Diagonalize if possible.
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}11&45&30\\10&26&20\\-20&-60&-44\end{array}\right]\nonumber\] One eigenvalue is \(1\). Diagonalize if possible.
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}95&25&24\\-196&-53&-48\\-164&-42&-43\end{array}\right]\nonumber\] One eigenvalue is \(5\). Diagonalize if possible.
Suppose \(A\) is an \(n\times n\) matrix and let \(V\) be an eigenvector such that \(AV = λV\). Also suppose the characteristic polynomial of \(A\) is \[\det (xI-A)=x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0\nonumber\] Explain why \[(A^n+a_{n-1}A^{n-1}+\cdots +a_1A+a_0I)V=0\nonumber\] If \(A\) is diagonalizable, give a proof of the Cayley Hamilton theorem based on this. This theorem says \(A\) satisfies its characteristic equation \[A^n+a_{n-1}A^{n-1}+\cdots +a_1A+a_0I=0\nonumber\]
Suppose the characteristic polynomial of an \(n\times n\) matrix \(A\) is \(1−X^n\). Find \(A^{mn}\) where \(m\) is an integer.
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The eigenvalues are distinct because they are the \(n\)th roots of \(1\). Hence if \(X\) is a given vector with \[X=\sum\limits_{j=1}^na_jV_j\nonumber\] then \[A^{nm}X=A^{nm}\sum\limits_{j=1}^na_jV_j=\sum\limits_{j=1}^na_jA^{nm}V_j=\sum\limits_{j=1}^na_jV_j=X\nonumber\] so \(A^{nm}=I\).
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}15&-24&7\\-6&5&-1\\-58&76&-20\end{array}\right]\nonumber\] One eigenvalue is \(−2\). Diagonalize if possible. Hint: This one has some complex eigenvalues.
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}15&-25&6\\-13&23&-4\\-91&155&-30\end{array}\right]\nonumber\] One eigenvalue is \(2\). Diagonalize if possible. Hint: This one has some complex eigenvalues.
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}-11&-12&4\\8&17&-4\\-4&28&-3\end{array}\right]\nonumber\] One eigenvalue is \(1\). Diagonalize if possible. Hint: This one has some complex eigenvalues.
Find the eigenvalues and eigenvectors of the matrix \[\left[\begin{array}{rrr}14&-12&5\\-6&2&-1\\-69&51&-21\end{array}\right]\nonumber\] One eigenvalue is \(−3\). Diagonalize if possible. Hint: This one has some complex eigenvalues.
Suppose \(A\) is an \(n\times n\) matrix consisting entirely of real entries but \(a + ib\) is a complex eigenvalue having the eigenvector, \(X +iY\) Here \(X\) and \(Y\) are real vectors. Show that then \(a−ib\) is also an eigenvalue with the eigenvector, \(X − iY\). Hint: You should remember that the conjugate of a product of complex numbers equals the product of the conjugates. Here \(a+ib\) is a complex number whose conjugate equals \(a−ib\).
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\(AX = (a+ib)X\). Now take conjugates of both sides. Since \(A\) is real, \[A\overline{X}=(a-ib)\overline{X}\nonumber\]
Let \(A=\left[\begin{array}{cc}1&2\\2&1\end{array}\right]\). Diagonalize \(A\) to find \(A^{10}\).
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First we write \(A=PDP^{-1}\). \[\left[\begin{array}{cc}1&2\\2&1\end{array}\right]=\left[\begin{array}{rr}-1&1\\1&1\end{array}\right]\left[\begin{array}{rr}-1&0\\0&3\end{array}\right]=\left[\begin{array}{rr}-\frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{1}{2}\end{array}\right]\nonumber\] Therefore \(A^{10}=PD^{10}P^{-1}\). \[\begin{aligned}\left[\begin{array}{cc}1&2\\2&1\end{array}\right]^{10}&=\left[\begin{array}{rr}-1&1\\1&1\end{array}\right]\left[\begin{array}{rr}-1&0\\0&3\end{array}\right]^{10}\left[\begin{array}{rr}-\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\end{array}\right] \\ &=\left[\begin{array}{rr}-1&1\\1&1\end{array}\right]\left[\begin{array}{rr}(-1)^{10}&0\\0&3^{10}\end{array}\right]\left[\begin{array}{rr}-\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\end{array}\right] \\ &=\left[\begin{array}{rr}29525&29524 \\ 29524&29525\end{array}\right]\end{aligned}\]
Let \(A=\left[\begin{array}{ccc}1&4&1\\0&2&5\\0&0&5\end{array}\right]\). Diagonalize \(A\) to find \(A^{50}\).
Let \(A=\left[\begin{array}{rrr}1&-2&-1\\2&-1&1\\-2&3&1\end{array}\right]\). Diagonalize \(A\) to find \(A^{100}\).
The following is a Markov (migration) matrix for three locations \[\left[\begin{array}{ccc}\frac{7}{10}&\frac{1}{9}&\frac{1}{5} \\ \frac{1}{10}&\frac{7}{9}&\frac{2}{5} \\ \frac{1}{5}&\frac{1}{9}&\frac{2}{5}\end{array}\right]\nonumber\]
- Initially, there are \(90\) people in location \(1,\: 81\) in location \(2\), and \(85\) in location \(3\). How many are in each location after one time period?
- The total number of individuals in the migration process is \(256\). After a long time, how many are in each location?
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- Multiply the given matrix by the initial state vector given by \(\left[\begin{array}{c}90\\81\\85\end{array}\right]\). After one time period there are \(89\) people in location \(1\), \(106\) in location \(2\), and \(61\) in location \(3\).
- Solve the system given by \((I − A)X_s = 0\) where \(A\) is the migration matrix and \(X_s=\left[\begin{array}{c}x_{1s} \\ x_{2s} \\ x_{3s}\end{array}\right]\) is the steady state vector. The solution to this system is given by \[\begin{aligned}x_{1s}&=\frac{8}{5}x_{3s} \\ x_{2s}&=\frac{63}{25}x_{3s}\end{aligned}\] Letting \(x_{3s} = t\) and using the fact that there are a total of \(256\) individuals, we must solve \[\frac{8}{5}t+\frac{63}{25}t+t=256\nonumber\] We find that \(t = 50\). Therefore after a long time, there are \(80\) people in location \(1,\: 126\) in location \(2\), and \(50\) in location \(3\).
The following is a Markov (migration) matrix for three locations \[\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5} \\ \frac{2}{5}&\frac{2}{5}&\frac{1}{5} \\ \frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\nonumber\]
- Initially, there are \(130\) individuals in location \(1,\: 300\) in location \(2\), and \(70\) in location \(3\). How many are in each location after two time periods?
- The total number of individuals in the migration process is \(500\). After a long time, how many are in each location?
The following is a Markov (migration) matrix for three locations \[\left[\begin{array}{ccc}\frac{3}{10}&\frac{3}{8}&\frac{1}{3} \\ \frac{1}{10}&\frac{3}{8}&\frac{1}{3} \\ \frac{3}{5}&\frac{1}{4}&\frac{1}{3}\end{array}\right]\nonumber\] The total number of individuals in the migration process is \(480\). After a long time, how many are in each location?
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We solve \((I −A)X_s = 0\) to find the steady state vector \(X_s=\left[\begin{array}{c}x_{1s} \\ x_{2s} \\ x_{3s}\end{array}\right]\). The solution to the system is given by \[\begin{aligned}x_{1s}&=\frac{5}{6}x_{3s} \\ x_{2s}&=\frac{2}{3}x_{3s}\end{aligned}\] Letting \(x_{3s} = t\) and using the fact that there are a total of \(480\) individuals, we must solve \[\frac{5}{6}t+\frac{2}{3}t+t=480\nonumber\] We find that \(t = 192\). Therefore after a long time, there are \(160\) people in location \(1,\: 128\) in location \(2\), and \(192\) in location \(3\).
The following is a Markov (migration) matrix for three locations \[\left[\begin{array}{ccc}\frac{3}{10}&\frac{1}{3}&\frac{1}{5} \\ \frac{3}{10}&\frac{1}{3}&\frac{7}{10} \\ \frac{2}{5}&\frac{1}{3}&\frac{1}{10}\end{array}\right]\nonumber\] The total number of individuals in the migration process is \(1155\). After a long time, how many are in each location?
The following is a Markov (migration) matrix for three locations \[\left[\begin{array}{ccc}\frac{2}{5}&\frac{1}{10}&\frac{1}{8} \\ \frac{3}{10}&\frac{2}{5}&\frac{5}{8} \\ \frac{3}{10}&\frac{1}{2}&\frac{1}{4}\end{array}\right]\nonumber\] The total number of individuals in the migration process is \(704\). After a long time, how many are in each location?
A person sets off on a random walk with three possible locations. The Markov matrix of probabilities \(A = [a_{ij}]\) is given by \[\left[\begin{array}{ccc}0.1&0.3&0.7 \\ 0.1&0.3&0.2\\0.8&0.4&0.1\end{array}\right]\nonumber\] If the walker starts in location \(2\), what is the probability of ending back in location \(2\) at time \(n = 3\)?
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\[X_3=\left[\begin{array}{c}0.38\\0.18\\0.44\end{array}\right]\nonumber\] Therefore the probability of ending up back in location \(2\) is \(0.18\).
A person sets off on a random walk with three possible locations. The Markov matrix of probabilities \(A = [a_{ij}]\) is given by \[\left[\begin{array}{ccc}0.5&0.1&0.6\\0.2&0.9&0.2\\0.3&0&0.2\end{array}\right]\nonumber\] It is unknown where the walker starts, but the probability of starting in each location is given by \[X_0=\left[\begin{array}{r}0.2\\0.25\\0.55\end{array}\right]\nonumber\] What is the probability of the walker being in location \(1\) at time \(n = 2\)?
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\[X_2=\left[\begin{array}{r}0.367\\0.4625\\0.1705\end{array}\right]\nonumber\] Therefore the probability of ending up in location \(1\) is \(0.367\).
You own a trailer rental company in a large city and you have four locations, one in the South East, one in the North East, one in the North West, and one in the South West. Denote these locations by SE, NE, NW, and SW respectively. Suppose that the following table is observed to take place.
| SE | NE | NW | SW | |
|---|---|---|---|---|
| SE | \(\frac{1}{3}\) | \(\frac{1}{10}\) | \(\frac{1}{10}\) | \(\frac{1}{5}\) |
| NE | \(\frac{1}{3}\) | \(\frac{7}{10}\) | \(\frac{1}{5}\) | \(\frac{1}{10}\) |
| NW | \(\frac{2}{9}\) | \(\frac{1}{10}\) | \(\frac{3}{5}\) | \(\frac{1}{5}\) |
| SW | \(\frac{1}{9}\) | \(\frac{1}{10}\) | \(\frac{1}{10}\) | \(\frac{1}{2}\) |
In this table, the probability that a trailer starting at NE ends in NW is \(1/10\), the probability that a trailer starting at SW ends in NW is \(1/5\), and so forth. Approximately how many will you have in each location after a long time if the total number of trailers is \(413\)?
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The migration matrix is \[A=\left[\begin{array}{cccc}\frac{1}{3}&\frac{1}{10}&\frac{1}{10}&\frac{1}{5} \\ \frac{1}{3}&\frac{7}{10}&\frac{1}{5}&\frac{1}{10} \\ \frac{2}{9}&\frac{1}{10}&\frac{3}{5}&\frac{1}{5} \\ \frac{1}{9}&\frac{1}{10}&\frac{1}{10}&\frac{1}{2}\end{array}\right]\nonumber\] To find the number of trailers in each location after a long time we solve system \((I − A)X_s = 0\) for the steady state vector \(X_s=\left[\begin{array}{c}x_{1s} \\ x_{2s} \\ x_{3s} \\ x_{4s}\end{array}\right]\). The solution to the system is \[\begin{aligned} x_{1s}&=\frac{9}{10}x_{4s} \\ x_{2s}&=\frac{12}{5}x_{4s} \\ x_{3s}&=\frac{8}{5}x_{4s}\end{aligned}\] Letting \(x_{4s} = t\) and using the fact that there are a total of \(413\) trailers we must solve \[\frac{9}{10}t+\frac{12}{5}t+\frac{8}{5}t+t=413\nonumber\] We find that \(t = 70\). Therefore after a long time, there are \(63\) trailers in the SE, \(168\) in the NE, \(112\) in the NW and \(70\) in the SW.
You own a trailer rental company in a large city and you have four locations, one in the South East, one in the North East, one in the North West, and one in the South West. Denote these locations by SE, NE, NW, and SW respectively. Suppose that the following table is observed to take place.
| SE | NE | NW | SW | |
|---|---|---|---|---|
| SE | \(\frac{1}{7}\) | \(\frac{1}{4}\) | \(\frac{1}{10}\) | \(\frac{1}{5}\) |
| NE | \(\frac{2}{7}\) | \(\frac{1}{4}\) | \(\frac{1}{5}\) | \(\frac{1}{10}\) |
| NW | \(\frac{1}{7}\) | \(\frac{1}{4}\) | \(\frac{3}{5}\) | \(\frac{1}{5}\) |
| SW | \(\frac{3}{7}\) | \(\frac{1}{4}\) | \(\frac{1}{10}\) | \(\frac{1}{2}\) |
In this table, the probability that a trailer starting at NE ends in NW is 1/10, the probability that a trailer starting at SW ends in NW is 1/5, and so forth. Approximately how many will you have in each location after a long time if the total number of trailers is 1469.
The following table describes the transition probabilities between the states rainy, partly cloudy and sunny. The symbol p.c. indicates partly cloudy. Thus if it starts off p.c. it ends up sunny the next day with probability \(\frac{1}{5}\). If it starts off sunny, it ends up sunny the next day with probability \(\frac{2}{5}\) and so forth.
| rains | sunny | p.c. | |
|---|---|---|---|
| rains | \(\frac{1}{5}\) | \(\frac{1}{5}\) | \(\frac{1}{3}\) |
| sunny | \(\frac{1}{5}\) | \(\frac{2}{5}\) | \(\frac{1}{3}\) |
| p.c. | \(\frac{3}{5}\) | \(\frac{2}{5}\) | \(\frac{1}{3}\) |
Given this information, what are the probabilities that a given day is rainy, sunny, or partly cloudy?
The following table describes the transition probabilities between the states rainy, partly cloudy and sunny. The symbol p.c. indicates partly cloudy. Thus if it starts off p.c. it ends up sunny the next day with probability \(\frac{1}{10}\). If it starts off sunny, it ends up sunny the next day with probability \(\frac{2}{5}\) and so forth.
| rains | sunny | p.c. | |
|---|---|---|---|
| rains | \(\frac{1}{5}\) | \(\frac{1}{5}\) | \(\frac{1}{3}\) |
| sunny | \(\frac{1}{10}\) | \(\frac{2}{5}\) | \(\frac{4}{9}\) |
| p.c. | \(\frac{7}{10}\) | \(\frac{2}{5}\) | \(\frac{2}{9}\) |
Given this information, what are the probabilities that a given day is rainy, sunny, or partly cloudy?
You own a trailer rental company in a large city and you have four locations, one in the South East, one in the North East, one in the North West, and one in the South West. Denote these locations by SE, NE, NW, and SW respectively. Suppose that the following table is observed to take place.
| SE | NE | NW | SW | |
|---|---|---|---|---|
| SE | \(\frac{5}{11}\) | \(\frac{1}{10}\) | \(\frac{1}{10}\) | \(\frac{1}{5}\) |
| NE | \(\frac{1}{11}\) | \(\frac{7}{10}\) | \(\frac{1}{5}\) | \(\frac{1}{10}\) |
| NW | \(\frac{2}{11}\) | \(\frac{1}{10}\) | \(\frac{3}{5}\) | \(\frac{1}{5}\) |
| SW | \(\frac{3}{11}\) | \(\frac{1}{10}\) | \(\frac{1}{10}\) | \(\frac{1}{2}\) |
In this table, the probability that a trailer starting at NE ends in NW is 1/10, the probability that a trailer starting at SW ends in NW is 1/5, and so forth. Approximately how many will you have in each location after a long time if the total number of trailers is 407?
The University of Poohbah offers three degree programs, scouting education (SE), dance appreciation (DA), and engineering (E). It has been determined that the probabilities of transferring from one program to another are as in the following table.
| SE | DA | E | |
|---|---|---|---|
| SE | \(.8\) | \(.1\) | \(.3\) |
| DA | \(.1\) | \(.7\) | \(.5\) |
| E | \(.1\) | \(.2\) | \(.2\) |
where the number indicates the probability of transferring from the top program to the program on the left. Thus the probability of going from DA to E is \(.2\). Find the probability that a student is enrolled in the various programs.
In the city of Nabal, there are three political persuasions, republicans (R), democrats (D), and neither one (N). The following table shows the transition probabilities between the political parties, the top row being the initial political party and the side row being the political affiliation the following year.
| R | D | N | |
|---|---|---|---|
| R | \(\frac{1}{5}\) | \(\frac{1}{6}\) | \(\frac{2}{7}\) |
| D | \(\frac{1}{5}\) | \(\frac{1}{3}\) | \(\frac{4}{7}\) |
| N | \(\frac{3}{5}\) | \(\frac{1}{2}\) | \(\frac{1}{7}\) |
Find the probabilities that a person will be identified with the various political persuasions. Which party will end up being most important?
The following table describes the transition probabilities between the states rainy, partly cloudy and sunny. The symbol p.c. indicates partly cloudy. Thus if it starts off p.c. it ends up sunny the next day with probability \(\frac{1}{5}\). If it starts off sunny, it ends up sunny the next day with probability \(\frac{2}{7}\) and so forth.
| rains | sunny | p.c. | |
|---|---|---|---|
| rains | \(\frac{1}{5}\) | \(\frac{2}{7}\) | \(\frac{5}{9}\) |
| sunny | \(\frac{1}{5}\) | \(\frac{2}{7}\) | \(\frac{1}{3}\) |
| p.c. | \(\frac{3}{5}\) | \(\frac{3}{7}\) | \(\frac{1}{9}\) |
Given this information, what are the probabilities that a given day is rainy, sunny, or partly cloudy?
Find the solution to the initial value problem \[\begin{aligned}\left[\begin{array}{c}x\\y\end{array}\right]'&=\left[\begin{array}{rr}0&-1\\6&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\\ \left[\begin{array}{c}x(0) \\ y(0)\end{array}\right]&=\left[\begin{array}{c}2\\2\end{array}\right]\end{aligned}\] Hint: form the matrix exponential \(e^{At}\) and then the solution is \(e^{At}C\) where \(C\) is the initial vector,
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The solution is \[e^{At}C=\left[\begin{array}{c}8e^{2t}-6e^{3t} \\ 18e^{3t}-16e^{2t}\end{array}\right]\nonumber\]
Find the solution to the initial value problem \[\begin{aligned}\left[\begin{array}{c}x\\y\end{array}\right]'&=\left[\begin{array}{rr}-4&-3\\6&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\\ \left[\begin{array}{c}x(0) \\ y(0)\end{array}\right]&=\left[\begin{array}{c}3\\4\end{array}\right]\end{aligned}\] Hint: form the matrix exponential \(e^{At}\) and then the solution is \(e^{At}C\) where \(C\) is the initial vector.
Find the solution to the initial value problem \[\begin{aligned}\left[\begin{array}{c}x\\y\end{array}\right]'&=\left[\begin{array}{rr}-1&2\\-4&5\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right]\\ \left[\begin{array}{c}x(0) \\ y(0)\end{array}\right]&=\left[\begin{array}{c}2\\2\end{array}\right]\end{aligned}\] Hint: form the matrix exponential \(e^{At}\) and then the solution is \(e^{At}C\) where \(C\) is the initial vector.
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). \[A=\left[\begin{array}{rrr}11&-1&-4 \\ -1&11&-4\\-4&-4&14\end{array}\right]\nonumber\] Hint: Two eigenvalues are \(12\) and \(18\).
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The eigenvectors and eigenvalues are: \[\left\{\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\1\\1\end{array}\right]\right\}\leftrightarrow 6,\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}-1\\1\\0\end{array}\right]\right\}\leftrightarrow 12,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}-1\\-1\\2\end{array}\right]\right\}\leftrightarrow 18\nonumber\]
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). \[A=\left[\begin{array}{rrr}4&1&-2\\1&4&-2\\-2&-2&7\end{array}\right]\nonumber\] Hint: One eigenvalue is \(3\).
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The eigenvectors and eigenvalues are: \[\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}-1\\1\\0\end{array}\right],\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\1\\1\end{array}\right]\right\}\leftrightarrow 3,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}-1\\-1\\2\end{array}\right]\right\}\leftrightarrow 9\nonumber\]
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{rrr}-1&1&1\\1&-1&1\\1&1&-1\end{array}\right]\nonumber\] Hint: One eigenvalue is \(-2\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2} \\ 0\end{array}\right],\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ -\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow -2\nonumber\] \[\left[\begin{array}{c}\sqrt{3}/3&-\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&0&\frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]^T\left[\begin{array}{rrr}-1&1&1\\1&-1&1\\1&1&-1\end{array}\right]\nonumber\] \[\left[\begin{array}{c}\sqrt{3}/3&-\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&0&\frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\nonumber\] \[=\left[\begin{array}{rrr}1&0&0\\0&-2&0\\0&0&-2\end{array}\right]\nonumber\]
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{rrr}17&-7&-4 \\ -7&17&-4 \\ -4&-4&14\end{array}\right]\nonumber\] Hint: Two eigenvalues are \(18\) and \(24\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 6,\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ -\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 18,\left\{\left[\begin{array}{c}-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2} \\ 0\end{array}\right]\right\}\leftrightarrow 24\nonumber\] The matrix \(U\) has these as its columns.
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{rrr}13&1&4\\1&13&4\\4&4&10\end{array}\right]\nonumber\] Hint: Two eigenvalues are \(12\) and \(18\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ -\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 6,\left\{\left[\begin{array}{c}-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2} \\ 0\end{array}\right]\right\}\leftrightarrow 12, \left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 18.\nonumber\] The matrix \(U\) has these as its columns.
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}-\frac{5}{3}&\frac{1}{15}\sqrt{6}\sqrt{5}&\frac{8}{15}\sqrt{5} \\ \frac{1}{15}\sqrt{6}\sqrt{5}&-\frac{14}{5}&-\frac{1}{15}\sqrt{6} \\ \frac{8}{15}\sqrt{5}&-\frac{1}{15}\sqrt{6}&\frac{7}{15} \end{array}\right]\nonumber\] Hint: The eigenvalues are \(-3,-2,1\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ 0\\ \frac{1}{6}\sqrt{5}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}-\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{5}\sqrt{5} \\ \frac{1}{15}\sqrt{2}\sqrt{15}\end{array}\right]\right\} -2,\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ \frac{2}{5}\sqrt{5} \\ \frac{1}{30}\sqrt{30}\end{array}\right]\right\}\leftrightarrow -3\nonumber\] These vectors are the columns of \(U\).
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}3&0&0 \\ 0&\frac{3}{2}&\frac{1}{2} \\ 0&\frac{1}{2}&\frac{3}{2}\end{array}\right]\nonumber\]
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}0\\-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}0\\ \frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 2,\left\{\left[\begin{array}{c}1\\0\\0\end{array}\right]\right\}\leftrightarrow 3.\nonumber\] These vectors are the columns of the matrix \(U\).
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}2&0&0\\0&5&1\\0&1&5\end{array}\right]\nonumber\]
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}1\\0\\0\end{array}\right]\right\}\leftrightarrow 2,\left\{\left[\begin{array}{c}0\\ -\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 4, \left\{\left[\begin{array}{c}0\\ \frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 6.\nonumber\] These vectors are the columns of \(U\).
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}\frac{4}{3}&\frac{1}{3}\sqrt{3}\sqrt{2}&\frac{1}{3}\sqrt{2} \\ \frac{1}{3}\sqrt{3}\sqrt{2}&1&-\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{2}&-\frac{1}{3}\sqrt{3}&\frac{5}{3}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(0,2,2\) where \(2\) is listed twice because it is a root of multiplicity \(2\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{5}\sqrt{2}\sqrt{5} \\ \frac{1}{5}\sqrt{3}\sqrt{5} \\ \frac{1}{5}\sqrt{5}\end{array}\right]\right\}\leftrightarrow 0,\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ 0\\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right],\left[\begin{array}{c}\frac{1}{5}\sqrt{2}\sqrt{5} \\ \frac{1}{5}\sqrt{3}\sqrt{5} \\ -\frac{1}{5}\sqrt{5}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] The columns are these vectors.
Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}1&\frac{1}{6}\sqrt{3}\sqrt{2}&\frac{1}{6}\sqrt{3}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{2}&\frac{3}{2}&\frac{1}{12}\sqrt{2}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{6}&\frac{1}{12}\sqrt{2}\sqrt{6}&\frac{1}{2}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(2,1,0\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ 0\\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 0,\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ -\frac{1}{2}\sqrt{2} \\ \frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{2}\sqrt{2} \\ \frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] The columns are these vectors.
Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix \[A=\left[\begin{array}{ccc}\frac{1}{3}&\frac{1}{6}\sqrt{3}\sqrt{2}&-\frac{7}{18}\sqrt{3}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{2}&\frac{3}{2}&-\frac{1}{12}\sqrt{2}\sqrt{6} \\ -\frac{7}{18}\sqrt{3}\sqrt{6}&-\frac{1}{12}\sqrt{2}\sqrt{6}&-\frac{5}{6}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(1,2,-2\).
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ \frac{1}{2}\sqrt{2} \\ \frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 1, \left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ 0\\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow -2, \left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{2}\sqrt{2} \\ -\frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] Then the columns of \(U\) are these vectors.
Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix \[A=\left[\begin{array}{ccc}-\frac{1}{2}&-\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{1}{10}\sqrt{5} \\ -\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{7}{5}&-\frac{1}{5}\sqrt{6} \\ \frac{1}{10}&\sqrt{5}&-\frac{1}{5}\sqrt{6}&-\frac{9}{10}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(-1,2,-1\) where \(-1\) is listed twice because it has multiplicity \(2\) as a zero of the characteristic equation.
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The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ 0 \\ \frac{1}{6}\sqrt{5}\sqrt{6}\end{array}\right],\left[\begin{array}{c}\frac{1}{3}\sqrt{2}\sqrt{3} \\ \frac{1}{5}\sqrt{5} \\ \frac{1}{15}\sqrt{2}\sqrt{15}\end{array}\right]\right\}\leftrightarrow -1,\left\{\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ -\frac{2}{5}\sqrt{5} \\ \frac{1}{30}\sqrt{30}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] The columns of \(U\) are these vectors. \[\left[\begin{array}{ccc}-\frac{1}{6}\sqrt{6}&\frac{1}{3}\sqrt{2}\sqrt{3}&\frac{1}{6}\sqrt{6} \\ 0&\frac{1}{5}\sqrt{5}&-\frac{2}{5}\sqrt{5} \\ \frac{1}{6}\sqrt{5}\sqrt{6} &\frac{1}{15}\sqrt{2}\sqrt{15}&\frac{1}{30}\sqrt{30}\end{array}\right]^T\left[\begin{array}{ccc}-\frac{1}{2}&-\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{1}{10}\sqrt{5} \\ -\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{7}{5}&-\frac{1}{5}\sqrt{6} \\ \frac{1}{10}\sqrt{5}&-\frac{1}{5}\sqrt{6}&-\frac{9}{10}\end{array}\right].\nonumber\] \[\left[\begin{array}{ccc}-\frac{1}{6}\sqrt{6}&\frac{1}{3}\sqrt{2}\sqrt{3}&\frac{1}{6}\sqrt{6} \\ 0&\frac{1}{5}\sqrt{5}&-\frac{2}{5}\sqrt{5} \\ \frac{1}{6}\sqrt{5}\sqrt{6}&\frac{1}{15}\sqrt{2}\sqrt{15}&\frac{1}{30}\sqrt{30}\end{array}\right]=\left[\begin{array}{rrr}-1&0&0\\0&-1&0\\0&0&2\end{array}\right]\nonumber\]
Explain why a matrix \(A\) is symmetric if and only if there exists an orthogonal matrix \(U\) such that \(A = U^TDU\) for \(D\) a diagonal matrix.
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If \(A\) is given by the formula, then \[A^T=U^TD^TU=U^TDU=A\nonumber\] Next suppose \(A = A^T\). Then by the theorems on symmetric matrices, there exists an orthogonal matrix \(U\) such that \[UAU^T=D\nonumber\] for \(D\) diagonal. Hence \[A=U^TDU\nonumber\]
Show that if \(A\) is a real symmetric matrix and \(λ\) and \(µ\) are two different eigenvalues, then if \(X\) is an eigenvector for \(λ\) and \(Y\) is an eigenvector for \(µ\), then \(X •Y = 0\). Also all eigenvalues are real. Supply reasons for each step in the following argument. First \[\lambda X^T\overline{X}=(AX)^T\overline{X}=X^TA\overline{X}=X^T\overline{AX}=X^T\overline{\lambda X}=\overline{\lambda}X^T\overline{X}\nonumber\] and so \(\lambda=\overline{\lambda}\). This shows that all eigenvalues are real. It follows all the eigenvectors are real. Why? Now let \(X,\: Y,\:µ\) and \(λ\) be given as above. \[\lambda (X\bullet Y)=\lambda X\bullet Y=AX\bullet Y=X\bullet AY=X\bullet\mu Y=\mu (X\bullet Y)=\mu (X\bullet Y)\nonumber\] and so \[(\lambda -\mu )X\bullet Y=0\nonumber\] Why does it follow that \(X\bullet Y=0\)?
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Since \(\lambda\neq\mu\), it follows \(X\bullet Y=0\).
Find the Cholesky factorization for the matrix \[\left[\begin{array}{ccc}1&2&0 \\ 2&6&4\\0&4&10\end{array}\right]\nonumber\]
Find the Cholesky factorization for the matrix \[\left[\begin{array}{rrr}4&8&0\\8&17&2\\0&2&13\end{array}\right]\nonumber\]
Find the Cholesky factorization for the matrix \[\left[\begin{array}{rrr}4&8&0\\8&20&8\\0&8&20\end{array}\right]\nonumber\]
Find the Cholesky factorization for the matrix \[\left[\begin{array}{rrr}1&2&1\\2&8&10\\1&10&18\end{array}\right]\nonumber\]
Find the Cholesky factorization for the matrix \[\left[\begin{array}{rrr}1&2&1\\2&8&10\\1&10&26\end{array}\right]\nonumber\]
Suppose you have a lower triangular matrix \(L\) and it is invertible. Show that \(LL^T\) must be positive definite.
Using the Gram Schmidt process or the \(QR\) factorization, find an orthonormal basis for the following span: \[span\left\{\left[\begin{array}{c}1\\2\\1\end{array}\right],\left[\begin{array}{r}2\\-1\\3\end{array}\right],\left[\begin{array}{c}1\\0\\0\end{array}\right]\right\}\nonumber\]
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Using the \(QR\) factorization, we have: \[\left[\begin{array}{rrr}1&2&1\\2&-1&0\\1&3&0\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{6}\sqrt{6}&\frac{3}{10}\sqrt{2}&\frac{7}{15}\sqrt{3} \\ \frac{1}{3}\sqrt{6}&-\frac{2}{5}\sqrt{2}&-\frac{1}{15}\sqrt{3} \\ \frac{1}{6}\sqrt{6}&\frac{1}{2}\sqrt{2}&-\frac{1}{3}\sqrt{3}\end{array}\right]\left[\begin{array}{ccc}\sqrt{6}&\frac{1}{2}\sqrt{6}&\frac{1}{6}\sqrt{6} \\ 0&\frac{5}{2}\sqrt{2}&\frac{3}{10}\sqrt{2} \\ 0&0&\frac{7}{15}\sqrt{3}\end{array}\right]\nonumber\] A solution is then \[\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{6} \\ \frac{1}{6}\sqrt{6}\end{array}\right],\left[\begin{array}{c}\frac{3}{10}\sqrt{2} \\ -\frac{2}{5}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right],\left[\begin{array}{c}\frac{7}{15}\sqrt{3} \\ -\frac{1}{15}\sqrt{3} \\ -\frac{1}{3}\sqrt{3}\end{array}\right]\nonumber\]
Using the Gram Schmidt process or the \(QR\) factorization, find an orthonormal basis for the following span: \[span\left\{\left[\begin{array}{c}1\\2\\1\\0\end{array}\right],\left[\begin{array}{r}2\\-1\\3\\1\end{array}\right],\left[\begin{array}{c}1\\0\\0\\1\end{array}\right]\right\}\nonumber\]
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\[\left[\begin{array}{rrr}1&2&1\\2&-1&0\\1&3&0\\0&1&1\end{array}\right]=\left[\begin{array}{cccc}\frac{1}{6}\sqrt{6}&\frac{1}{6}\sqrt{2}\sqrt{3}&\frac{5}{111}\sqrt{3}\sqrt{37}&\frac{7}{111}\sqrt{111} \\ \frac{1}{3}\sqrt{6}&-\frac{2}{9}\sqrt{2}\sqrt{3}&\frac{1}{333}\sqrt{3}\sqrt{37}&-\frac{2}{111}\sqrt{111} \\ \frac{1}{6}\sqrt{6}&\frac{5}{18}\sqrt{2}\sqrt{3}&-\frac{17}{333}\sqrt{3}\sqrt{37}&-\frac{1}{37}\sqrt{111} \\ 0&\frac{1}{9}\sqrt{2}\sqrt{3}&\frac{22}{333}\sqrt{3}\sqrt{37}&-\frac{7}{111}\sqrt{111}\end{array}\right]\nonumber\] \[\left[\begin{array}{ccc}\sqrt{6}&\frac{1}{2}\sqrt{6}&\frac{1}{6}\sqrt{6} \\ 0&\frac{3}{2}\sqrt{2}\sqrt{3}&\frac{5}{18}\sqrt{2}\sqrt{3} \\ 0&0&\frac{1}{9}\sqrt{3}\sqrt{37} \\ 0&0&0\end{array}\right]\nonumber\] Then a solution is \[\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{6} \\ \frac{1}{6}\sqrt{6}&0\end{array}\right],\left[\begin{array}{c}\frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{2}{9}\sqrt{2}\sqrt{3} \\ \frac{5}{18}\sqrt{2}\sqrt{3} \\ \frac{1}{9}\sqrt{2}\sqrt{3}\end{array}\right],\left[\begin{array}{c}\frac{5}{111}\sqrt{3}\sqrt{37} \\ \frac{1}{333}\sqrt{3}\sqrt{37} \\ -\frac{17}{333}\sqrt{3}\sqrt{37} \\ \frac{22}{333}\sqrt{3}\sqrt{37}\end{array}\right]\nonumber\]
- \(\left[\begin{array}{ccc}1&2&3\\0&3&4\\0&0&1\end{array}\right]\)
- \(\left[\begin{array}{cc}2&1\\2&1\end{array}\right]\)
- \(\left[\begin{array}{rr}1&2\\-1&2\end{array}\right]\)
- \(\left[\begin{array}{cc}1&1\\2&3\end{array}\right]\)
- \(\left[\begin{array}{rrr}\sqrt{11}&1&3\sqrt{6} \\ \sqrt{11}&7&-\sqrt{6} \\ 2\sqrt{11}&-4&-\sqrt{6}\end{array}\right]\) Hint: Notice that the columns are orthogonal.
Using a computer algebra system, find a QR factorization for the following matrices.
- \(\left[\begin{array}{rrr}1&1&2\\3&-2&3\\2&1&1\end{array}\right]\)
- \(\left[\begin{array}{rrrr}1&2&1&3\\4&5&-4&3\\2&1&2&1\end{array}\right]\)
- \(\left[\begin{array}{rr}1&2\\3&2\\1&-4\end{array}\right]\) Find the thin QR factorization of this one.
A quadratic form in three variables is an expression of the form \(a_1x^2 + a_2y^2 + a_3z^2 + a_4xy+a_5xz+a_6yz\). Show that every such quadratic form may be written as \[\left[\begin{array}{ccc}x&y&z\end{array}\right]A\left[\begin{array}{c}x\\y\\z\end{array}\right]\nonumber\] where \(A\) is a symmetric matrix.
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\[\left[\begin{array}{ccc}x&y&z\end{array}\right]\left[\begin{array}{ccc}a_1&a_4/2&a_5/2 \\ a_4/2&a_2&a_6/2 \\ a_5/2&a_6/2&a_3\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]\nonumber\]
Given a quadratic form in three variables, \(x, y,\) and \(z\), show there exists an orthogonal matrix \(U\) and variables \(x′ , y ′ ,z ′\) such that \[\left[\begin{array}{c}x\\y\\z\end{array}\right]=U\left[\begin{array}{c}x'\\y'\\z'\end{array}\right]\nonumber\] with the property that in terms of the new varaibles, the quadratic form is \[\lambda_1(x')^2+\lambda_2(y')^2+\lambda_3(z')^2\nonumber\] where the numbers, \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\) are the eigenvalues of the matrix \(A\) in Exercise \(\PageIndex{78}\).
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The quadratic form may be written as \[\vec{x}^TA\vec{x}\nonumber\] where \(A = A^T\). By the theorem about diagonalizing a symmetric matrix, there exists an orthogonal matrix \(U\) such that \[U^TAU=D,\:A=UDU^T\nonumber\] Then the quadratic form is \[\vec{x}^TUDU^T\vec{x}=(U^T\vec{x})^TD(U^T\vec{x})\nonumber\] where \(D\) is a diagonal matrix having the real eigenvalues of \(A\) down the main diagonal. Now simply let \[\vec{x}'=U^T\vec{x}\nonumber\]
Consider the quadratic form \(q\) given by \(q = 3x_1^2 −12x_1x_2 −2x_2^2\).
- Write \(q\) in the form \(\vec{x}^TA\vec{x}\) for an appropriate symmetric matrix \(A\).
- Use a change of variables to rewrite \(q\) to eliminate the \(x_1x_2\) term.
Consider the quadratic form \(q\) given by \(q = −2x_1^2 +2x_1x_2 −2x_2^2\).
- Write \(q\) in the form \(\vec{x}^TA\vec{x}\) for an appropriate symmetric matrix \(A\).
- Use a change of variables to rewrite \(q\) to eliminate the \(x_1x_2\) term.
Consider the quadratic form \(q\) given by \(q = 7x_1^2 +6x_1x_2 −x_2^2\).
- Write \(q\) in the form \(\vec{x}^TA\vec{x}\) for an appropriate symmetric matrix \(A\).
- Use a change of variables to rewrite \(q\) to eliminate the \(x_1x_2\) term.