5.1: Review of Linear Algebra
- Page ID
- 409
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this discussion, we expect some familiarity with matrices. We will rely heavily on calculators and computers to work out the problems. Consider some examples.
Solve the system of equations
\[\begin{align} &4x &+y &&+3z &=2 \\ &x &-2y &&-5z &=3 \\ &5x & &&+2z &=1. \end{align}\]
Solution
We write this system as the matrix equation
\[ Ax = b \]
where
\[A = \begin{pmatrix} 4 &1 &3 \\ 1 &-2 &-5 \\ 5 &0 &2 \end{pmatrix} \;\;\; b=\begin{pmatrix} 2 \\ 3 \\1 \end{pmatrix}. \]
To solve this matrix equation we take the inverse of both sides (which is possible since the determinant of \(A\) is -13 which is not equal to zero. We have
\[x=A^{-1}b. \]
Using a calculator we find that
\[A^{-1} = \dfrac{1}{13} \begin{pmatrix} 4 & 27 &-10 \\ 2 &7 &-5 \\ -1 & -23 & 9 \end{pmatrix} . \]
Multiplying by \(b\) gives
\[x=\begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} . \]
What we mean by "\(x\)" is the vector \(<x,y,z>\). The solution is
\[ x = 1 \;\;\; y = 4 \;\;\; z = -2. \]
Find the solution of
\[\begin{align} &3x &+2y &&-z &=5 \\ &2x &+y &&-z &=2 \\ &5x &+4y &&-z &=11 \end{align}.\]
Solution
A quick check shows that we cannot solve this problem in the same way, since the determinant of \(A\) is 0. Instead, we rref the augmented matrix
\[\left(\begin{array}{ccc|c} 3 &2 &-1 &5 \\ 2 &1 &-1 &2 \\ 5 &4 &-1 & 11 \end{array} \right) \]
to get
\[\left(\begin{array}{ccc|c} 1 &0 &-1 &-1 \\ 0 &1 &1 &4 \\ 0 &0 &0 &0 \end{array}\right). \]
Putting this back into equation form, we get
\[x-z=-1 \;\;\; \text{and} \;\;\; y+z =4. \]
We write this as
\[x=-1+z \;\;\; y=4-z \;\;\; z=z. \]
Letting \(z= t\) be the parameter we get parametric equations for the solution set
\[x=-1+t \;\;\; y=4-t \;\;\; z=t. \]
Recall that vectors \(v_1,...,v_n \) are called linearly independent if
\[c_1v_1+...+c_nv_n=0 \]
implies that all of the constants \(c_i\) are zero. A theorem from linear algebra tell us that if we have \(n\) vectors in \(\mathbb{R}^n\) then they will be linearly independent if and only if the determinant of the matrix whose columns are these vectors has nonzero determinant.
Show that the vectors
\[ u = <1,4,-2> \;\;\; v = <0,3,5> \;\;\; \text{and} \;\;\; w = <1,2,3> \]
are linearly independent.
Solution
We find the determinant
\[det\begin{pmatrix} 1 &0 &1 \\ 4 &3 &2 \\ -2 &5 &3 \end{pmatrix} =25. \]
Since the determinant is nonzero, the vectors are linearly independent.
For systems of differential equations, eigenvalues and eigenvectors play a crucial role. We recall their definitions below.
Definition: Eigenvalues and Eigenvectors
Let \(A\) be an \(n \times n\) matrix. Then \( \lambda \) is an eigenvalue for \(A\) with eigenvector \(v\) if
\[Av = \lambda v\]
Find the eigenvalues and eigenvectors for
\[A=\begin{pmatrix} 6 &4 \\ -3 & -1 \end{pmatrix}. \]
Solution
If
\[ Av = \lambda v \]
then
\[A -\lambda = 0\]
Taking determinants of both sides, we get
\[\begin{align}(6 - \lambda)(-1 - \lambda) + 12 &= 0 \\ \lambda ^2 - 5\lambda + 6 &= 0 \\ (\lambda - 2)(\lambda - 3) &= 0 \end{align}\]
The eigenvalues are
\[\lambda=2 \;\;\; \text{and} \;\;\; \lambda=3 \]
To find the eigenvectors, we plug the eigenvalues into the equation
\[ A -\lambda = 0 \]
and find the null space of the left hand side. For the eigenvalue \(\lambda = 2\), we have
\[A-\lambda I = \begin{pmatrix} 4 &4 \\ -3 &-3 \end{pmatrix} \]
The first row gives
\[y = -x\]
so that an eigenvector corresponding to the eigenvalue \(\lambda = 2\) is
\[v_2 = \begin{pmatrix}1\\-1 \end{pmatrix}. \]
For the eigenvalue \(\lambda = 3\), we have
\[A-\lambda I = \begin{pmatrix} 3&4 \\ -3 &-4 \end{pmatrix} . \]
The first row gives
\[ 3y = -4x\]
so that an eigenvector corresponding to the eigenvalue \(\lambda = 3\) is
\[v_3 = \begin{pmatrix}3\\-4\end{pmatrix} .\]
Typically, we want to normalize the eigenvectors, that is find unit eigenvectors. We get
\[ u_2 =\begin{pmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix} \;\;\; \text{and} \;\;\; u_3 =\begin{pmatrix}\frac{3}{5}\\-\frac{-4}{5}\end{pmatrix}. \]
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)