Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 09

MATH152 April 2015

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Question A 09
Find an eigenvector corresponding to eigenvalue 2 for the matrix below: ${\begin{bmatrix}5&-3\\6&-4\end{bmatrix}}$

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
Recall that an eigenvector of a matrix $\mathbf {A}$ with corresponsing eigenvalue $\mathbf {\lambda }$ is a non-zero vector $\mathbf {v}$ that satisfies $\mathbf {A} \mathbf {v} =\mathbf {\lambda } \mathbf {v} .$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Denote the matrix in question by $A.$ We therefore seek a vector $\mathbf {v}$ for which $A\mathbf {v} =2\mathbf {v} ,$ which amounts to solving the homogeneous system $(A-2I)\mathbf {v} =\mathbf {0} .$ In augmented matrix form, we have $\left[{\begin{array}{cc|c}5-2&-3&0\\6&-4-2&0\end{array}}\right]=\left[{\begin{array}{cc|c}3&-3&0\\6&-6&0\end{array}}\right].$ We see immediately (or perhaps after Gaussian elimination) that $\mathbf {v} =\color {blue}{\begin{bmatrix}1\\1\end{bmatrix}}$ is an eigenvector of eigenvalue 2. (Note that any vector of the form ${\begin{bmatrix}t\\t\end{bmatrix}},\,t\in \mathbb {R}$ would thus also be an eigenvector of eigenvalue 2.)