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I am given a 4x4 matrix such and a basis for $R^4$, $\{u_1,u_2,u_3,u_4\}$. Suppose the following conditions: $$Bu_1 = 2u_1 \\Bu_2 = 0\\Bu_3 = u_4\\Bu_4 = u_3$$

First, I am asked to find the eigenvalues. It's fairly obvious that $1$ and $2$ are eigenvalues. Then I manipulate the lower 2 equations as such:

$$Bu_3 = u_4\\B(Bu_3) = Bu_4\\B^2u_3 = u_3$$

Hence an eigenvalue of $B^2$ is supposed to be 1. Square rooting this, I obtain eigenvalues of $1$ and $-1$ for $B$.

Thereafter, I need to find the associated eigenvectors. Again, for $\lambda = 1$ is $u_1$ and $\lambda = 0$ is $u_2$. But for the other two, I cannot figure it out. The answer key says that that of $\lambda = 1$ is $u_3+u_4$, and that of $\lambda = -1$ is $u_3-u_4$. I cannot comprehend why. Am I missing out on come manipulation?

Any advice is appreciated! Thank you!

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$Bu_3=u_4$ and $Bu_4=u_3$ $\Rightarrow$ $Bu_3+Bu_4=B(u_3+u_4)=u_4+u_3$ hence 1 is an eigenvalue for $u_3+u_4$

$Bu_3=u_4$ $\Rightarrow$ $-Bu_3=B(-u_3)=-u_4$

$B(-u_3)=-u_4$ and $Bu_4=u_3$ $\Rightarrow$ $B(u_3-u_4)=-(u_3-u_4)$ hence -1 is an eigenvalue

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  • $\begingroup$ Thank you! Can't believe I failed to see something to obvious haha. Your help is appreciated! $\endgroup$
    – a9302c
    Nov 21, 2020 at 15:17

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