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I have this problem on a homework that I must do two ways to receive full credit on. I attempted the problem one way and was wondering if it was right, and if anybody could give me any hints or insights on how to approach the problem another way.

$\textbf{Question}:$ A certain $4$x$4$ matrix transforms $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\rightarrow\begin{bmatrix}b\\a\\d\\c\end{bmatrix}$. Find all of its eigenvectors. You must do this problem with two different approaches to receive full credit.

$\textbf{My Attempt}:$ If we represent the matrix like this $$\begin{bmatrix}x_1&x_2&x_3&x_4\\y_1&y_2&y_3&y_4\\z_1&z_2&z_3&z_4\\f_1&f_2&f_3&f_4\end{bmatrix}$$ and multiply is against the the original vector, we get $$ax_1+bx_2+cx_3+dx_4=b\\ay_1+by_2+cy_3+dy_4=a\\az_1+bz_2+cz_3+dc_4=d\\af_1+bf_2+cf_3+df_4=c$$ which tells me which tells me that every row will have one $1$ in it and the rest of the entries will be zero. So the transformation matrix would be $$\begin{bmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}$$ Now that I have this matrix, I think it will be easy enough to find its eigenvectors using $det(A-\lambda I)=0$, etc.

Is this solution correct? Can anyone think of any other ways to solve this question so that I can get full credit? I was thinking maybe there could be two ways to find the eigenvectors now that I have found the matrix, which would count as two different approaches to the problem, but I can't think of how I would do that. Or is there another way to find the transform? Thanks so much for any help you guys can give!

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    $\begingroup$ Work directly with the definition of the transformation and that of an eigenvector: $A\mathbf v=\lambda\mathbf v$. $\endgroup$ – amd Nov 17 '16 at 18:46
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You have correctlty identified one way.

Now consider what eigenvector is. It is the vector that the matrix transforms to a multiple of the eigenvector.

So we're talking about $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\rightarrow\begin{bmatrix}ka\\kb\\kc\\kd\end{bmatrix}$

But you have $\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\rightarrow\begin{bmatrix}b\\a\\d\\c\end{bmatrix}$

Therefore an eigenvector will satisfy $\begin{bmatrix}ka\\kb\\kc\\kd\end{bmatrix} = \begin{bmatrix}b\\a\\d\\c\end{bmatrix}$

Taken together $ka=b$ and $kb=a$ yields $k^2a=a \Rightarrow k=1$ or $k=-1$ with eigenvectors $\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$ and $\begin{bmatrix}1\\-1\\0\\0\end{bmatrix}$

Can you see what the other two eigenvectors must be?

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