Let $V$, a vector space above $\mathbb{Q}$ such that $\dim(V) = 3$ and let $B$, a basis to $V$ such that $$A:=[T]_B = \begin{bmatrix} 2 & 0 & 0 \\[0.3em] 1 & 2 & 0 \\[0.3em] 0 & 0 & 3 \end{bmatrix}$$
Let $W = \ker(T-2I)$. Show there's no $U\subseteq V$ such that $V = W \oplus U$.
Solution:
- Let's assume by contradiction there's such $U$.
- Since $T(T-2I) = (T-2I)T$ then $W$ is $T$ invariant. Therefore, the decomposition of $V$ is a direct sum of invariant subspaces (Why?)
- Easy to evaluate that the minimal polynomial, $m_T = (x-2)^2(x-3)$. Let $C_1 = \{w\}$, a base of $W$ and $C_2 = \{ u_1, u_2 \}$ a basis of $U$. Then $C = \{w, u_1, u_2\}$ is a basis of $V$.
- Therefore, $$[T]_C = \begin{bmatrix} 2 & 0 & 0 \\[0.3em] 0 & \alpha & \beta \\[0.3em] 0 & \gamma & \lambda \end{bmatrix}$$
I know that there's a theorem claims that if $V = W_1 \oplus \ldots \oplus W_k$ and all of $W_i$ are $T$-invariant then $[T]_B = \text{diag}(A_1,\ldots,A_k)$ and $[T|_{W_i}]_{B_i} = A_i$, but still don't fully understand why it has to be this form.
From this point, the solution is clear:
$$(x-2)^2 m_C(X) = m_T = \text{lcm} \{ (x-2), m_C(X) \}$$
Which implies $$m_C(X) = (x-2)^2(x-3)$$
Which contradicts the fact that $\deg([T|_U]_{B_2}) = 2$ but it's minimal polynomial has the degree $3$.
I'd be glad to get a clarification for the two question marked with bold.
Thanks
Answer to the second question:
We can write any vector $x \in V$ as a block-vector $$ x = (x_1,x_2,\dots,x_k) := \pmatrix{x_1 \\ x_2 \\ \vdots \\ x_k} $$ where for each $i$, $x_i \in W_i$ (and is written with respect to the basis $B_i$).
If $[T]_B$ denotes the matrix of $T$ with respect to the matrix described, then by the $T$-invariance of each subspace, we have $$ [T]_B\pmatrix{x_1 \\ x_2 \\ \vdots \\ x_k} = \pmatrix{A_1x_1 \\ A_2x_2 \\ \vdots \\ A_kx_k} $$ This gives us precisely that $[T]_B = \text{diag}(A_1,\dots,A_k)$.