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Mathematics LibreTexts

5.1: Review of Linear Algebra

  • Page ID
    409
  • [ "article:topic", "eigenvalues", "eigenvectors" ]

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    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.

    Example \(\PageIndex{1}\)

    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 

    \[ x = 1 \;\;\; y = 4 \;\;\; z = -2. \]

    Example \(\PageIndex{2}\)

    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.

    Example \(\PageIndex{3}\)

    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 

    \[Av = \lambda v\]

    Example \(\PageIndex{4}\)

    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

    \[ u_2 =\begin{pmatrix}\frac{1}{\sqrt{2}}\\-\frac{1}{\sqrt{2}}\end{pmatrix} \;\;\; \text{and} \;\;\; u_3 =\begin{pmatrix}\frac{3}{5}\\-\frac{-4}{5}\end{pmatrix}. \]

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