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Let $\alpha = \{\vec{a_1},\vec{a_2},\vec{a_3},\vec{a_4}\}$ be an orthonormal basis for $\mathbf{R}^4$. Let $\mathbf{A}:\mathbf{R}^4 \rightarrow \mathbf{R}^4$ be a linear orthogonal map with the following properties: $\mathbf{A}^2=\mathbf{I}$, $\mathbf{A}\vec{a_1}=\vec{a_3}$, and $\mathbf{A}\vec{a_2}=\vec{a_4}$. Find the eigenvalues and an orthonormal basis of eigenvectors of $\mathbf{A}$. I know that the eigenvalues are either 1, -1 or both because the length is invariant in an orthogonal mapping, which means the real roots of the characteristic polynomial are either 1 or -1. But I don't know how I know which eigenvalue is the correct one, and how I would go about finding the eigenvectors.

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You know that $A.\vec{a_3}$ has norm $1$ and it is orthogonal to both $\vec{a_3}\left(=A.\vec{a_1}\right)$ and to $\vec{a_4}\left(=A.\vec{a_2}\right)$. Therefore, it is of the form $\gamma\vec{a_1}+\beta\vec{a_2}$, with $\gamma^2+\beta^2=1$. And you know that $A.\vec{a_4}$ has norm $1$ and it is orthogonal to $\vec{a_3}\left(=A.\vec{a_1}\right)$, to $\vec{a_4}\left(=A.\vec{a_2}\right)$, and to $\alpha\vec{a_1}+\beta\vec{a_2}\left(=A.\vec{a_3}\right)$. Therefore, it is equal to $\mp\beta\vec{a_1}\pm\gamma\vec{a_2}$. So, the matrix of $A$ with respect to $\alpha$ is$$\begin{bmatrix}0&0&\gamma&\mp\beta\\0&0&\beta&\pm\gamma\\1&0&0&0\\0&1&0&0\end{bmatrix}.$$But$$\operatorname{Id}=\begin{bmatrix}0&0&\gamma&\mp\beta\\0&0&\beta&\pm\gamma\\1&0&0&0\\0&1&0&0\end{bmatrix}^2=\begin{bmatrix}\gamma&\mp\beta&0&0\\\beta&\pm\gamma&0&0\\0&0&\gamma&\mp\beta\\0&0&\beta&\pm\gamma\end{bmatrix},$$amd therefore $\gamma=1$, $\beta=0$, and the matrix of $A$ with respect to $\alpha$ is, in fact$$\begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}.$$Can you take it from here?

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  • $\begingroup$ Yes, thank you. $\endgroup$ – Garnel Mar 7 at 17:01

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