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!
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?