Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 05 (a)

MATH152 April 2011

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Question B 05 (a)
Consider the matrix below: $A={\begin{bmatrix}3&1&-3\\-1&0&2\\1&0&0\end{bmatrix}}$. It is known that $\lambda _{1}=1$ is an eigenvalue of A. Find an eigenvector associated with the eigenvalue $\lambda _{1}=1$.

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?

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Hint
How do we solve $A\mathbf {x} =\lambda \mathbf {x}$ for x with a given $\lambda$?

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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. We want to solve something of the form ${\begin{aligned}A\mathbf {x} &=\lambda \mathbf {x} (A-\lambda {}I)\mathbf {x} &=\mathbf {0} \end{aligned}}$ where I is the identity matrix. In this case we have $\lambda _{1}=1$ so we have that $A-\lambda _{1}I=A-I={\begin{bmatrix}2&1&-3\\-1&-1&2\\1&0&-1\end{bmatrix}}$. To solve (A-I)x we write x=[a,b,c] and solve ${\begin{bmatrix}2&1&-3\\-1&-1&2\\1&0&-1\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}={\begin{bmatrix}0\\0\\0\end{bmatrix}}$ which is equivalent to solving ${\begin{aligned}2a+b-3c&=0,\\-a-b+2c&=0,\\a-c&=0.\end{aligned}}$ This has solution b=c=a for any arbitrary a. Therefore we take a=1, without loss of generality, and our eigenvetor is $\mathbf {x} ={\begin{bmatrix}1\\1\\1\end{bmatrix}}$.