This is a continuation of this question: Find real number $a$ such that matrix $A$ is NOT diagonalisable
For the matrix $A =$ \begin{bmatrix}2&5&-1\\0&2&1\\-1&8&-1\end{bmatrix}
Give W, a subspace of $R^3$ such that the following 2 properties hold:
i. $f_{A}(W) \subset W$
ii. The representation matrix of $f'_{A}: R^3 /W \rightarrow R^3 /W$ induced by $f_{A}$ is diagonalisable.
I believe that to satisfy property i, I just need W to be a subspace spanned by one of the eigenvectors of A:
\begin{equation} v_{1} = \begin{bmatrix}4\\1\\1\end{bmatrix} \ \text{or} \ v_{2} = \begin{bmatrix}2\\-1\\5\end{bmatrix} \end{equation}
I am unfortunately at a loss as to which I should choose in order to satisfy property 2. I understand the idea of quotient spaces of course, but am unsure what this induced map $f'_{A}: R^3 /W \rightarrow R^3 /W$ is. Could someone please enlighten me on how to find it?
We note that the characteristic polynomial of $A$ is $\chi_A(x) = -(x-3)^2 (x+3).$ Likewise, this is the minimal polynomial of $A.$ Considering that the minimal polynomial of $A$ does not split into distinct linear factors in $\mathbb{R}[x],$ it follows that $A$ is not diagonalizable. By the Primary Decomposition Theorem, we have that $\mathbb{R}^3 = \ker(T-3I)^2 \oplus \ker(T+3I),$ and $W = \ker(T-3I)^2$ is a $T$-invariant subspace. Furthermore, the minimal polynomial $\mu_T(x)$ (or $\mu_A(x)$) is the least common multiple of minimal polynomial of $T$ restricted to $W$ and the minimal polynomial of $T$ restricted to $\ker(T+3I).$ Evidently, we have that $\mathbb{R}^3 / W \cong \ker(T+3I)$ so that the minimal polynomial of the induced operator $T'$ on $\mathbb{R}^3 / W$ is $\mu_{T'}(x) = x+3$ so that $T'$ is diagonalizable.