Eigenvalues-eigenvectors

Question

Let $A \in \mathbb{R}^{3\times3}$ be a matrix with the eigenvalues $\lambda_1 = -1, \lambda_2 = 1$ and $\lambda_3 = 5$. Show that there is a 2-dimensional subspace $U \subset \mathbb{R}^{3}$ such that the following equivalence holds for all $x \in \mathbb{R}^{3}$:

$x \in U \Leftrightarrow A^2 x = x$

Answer 1

I hope this is right. As we are given 3 distinct eigen values, it implies that this matrix is diagonalisable. Using Spectral theory, we can write $A$ as $$VAV^*=D=\begin{bmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5\end{bmatrix}$$ where $V$ is a unitary matrix. Then, clearly $A^2x=(V^*DV)^2x=(V^*DVV^*DV)x=(V^*D^2V)x$, and so $A^2x=x \implies D^2(Vx)=Vx$.

Now, we have $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 25 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$$ where $[a \ \ b \ \ c ]^T=Vx=y$. Then we have $$\begin{bmatrix} a \\ b \\ 25c \end{bmatrix}=\begin{bmatrix} a \\ b \\ c \end{bmatrix}$$

A simple way to get a two dimensional space that satisfies this would be to assume $c=0$. Then for all $y \in U$, where $U:=\{(x,y,0):x,y \in \mathbb{R}\}$, we have that $A^2y=y$.

Answer 2

It seems to me that you can do this. Let $x$ and $y$ be chosen such that $Ax=x$ and $Ay=-y$. Then $U=\langle x,y\rangle$. It is easy to check that $U$ is the desired one.