Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 23

MATH152 April 2015

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[hide] Question A 23
The matrix $A={\begin{bmatrix}7&0&-10\\5&-3&-5\\5&0&-8\end{bmatrix}}$ has the eigenvalue $-3$. Find all eigenvectors corresponding to this eigenvalue.

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[show] Hint
The eigenvectors $\mathbf {v}$ of eigenvalue $-3$ satisfy the equation $A\mathbf {v} =-3\mathbf {v}$; in other words, $(A+3I)\mathbf {v} =\mathbf {0} .$ Find all solutions of this linear system.

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[show] Solution

We would like to find the eigenvectors of the matrix $A={\begin{bmatrix}7&0&-10\\5&-3&-5\\5&0&-8\end{bmatrix}}$ corresponding to the eigenvalue $\lambda =-3$. Recall (from the hint or otherwise) that the eigenvectors of $A$ corresponding to the eigenvalue $\lambda$ are those vectors $\mathbf {v}$ such that $(A-\lambda I)\mathbf {v} =\mathbf {0}$. When $\lambda =-3$, this equation reads ${\begin{bmatrix}10&0&-10\\5&0&-5\\5&0&-5\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}0\\0\\0\end{bmatrix}}.$ We solve this system using row reduction. Beginning with the augmented system matrix $\left[{\begin{array}{ccc|c}10&0&-10&0\\5&0&-5&0\\5&0&-5&0\end{array}}\right],$ we subtract half the first row from the second and third rows, and then divide the first row by 10, giving $\left[{\begin{array}{ccc|c}1&0&-1&0\\0&0&0&0\\0&0&0&0\end{array}}\right].$ Therefore, any vector satisfying $x-z=0$ ($y$ is free) will be an eigenvector, i.e., the desired eigenvectors are $\color {blue}{\begin{bmatrix}x\\y\\x\end{bmatrix}},\ x,y\in \mathbb {R} .$