Finding the eigenvalues and an orthornormal basis of eigenvectors of an orthogonal mapping

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.

1 Answer

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?

• Yes, thank you. – Garnel Mar 7 at 17:01