Find a subspace $U$ of $\mathbb{R^3}$ s.t. $T(U)\subseteq U$ and s.t. $T_{|U}$ on $U$ is diagonalizable in $\mathbb{R^3}$.

Question

I got stuck in this point of this exercise, do you have ideas about? Thank you.

Let $T:\mathbb{R^3}\to \mathbb{R^3}$ be an operator represented (canonical basis) by the matrix $A$ = $ \begin{bmatrix} 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0 \\ 0 & 2 & 3 \\ \end{bmatrix} $

Find a subspace $U$ of $\mathbb{R^3}$, $dim(U)=2$, s.t. $T(U)\subseteq U$ and s.t. $T_{|U}$ on $U$, the restriction of $T$ on $U$, is diagonalizable in $\mathbb{R^3}$.

Answer 1

First, eigenvalues:

$$|xI-A|=\begin{vmatrix}x&-1&\frac12\\ 0&\,\,\;x&0\\ 0&-2&x-3\end{vmatrix}=x^2(x-3)$$

Well, you have two different eigenvalues for your operator, and we know that any= operator on an $\;n\,-$ dimensional vector space with $\;n\;$ different eigenvalues is diagonalizable, so...what do you think your $\;U\;$ is going to be?

Answer 2

Since $(1,0,0)$ is an eigenvector of $A$ (with eigenvalue $0$), and $(-1,0,6)$ is an eigenvector of $A$ (with eigenvalue $3$)…