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

MATH152 April 2015

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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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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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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 $\color {blue}{\begin{bmatrix}x\\y\\x\end{bmatrix}},\ x,y\in \mathbb {R} .$