Matrix of transformation for which a subspace is invariant

Question

I have the following problem:

Let $V$ be a finite dimensional vector space, $T: V\to V$ a linear transformation and $W$ a proper subspace of $V$, dimension $n$, invariant under $T$. Prove that there exists a base $\beta$ of $V$ such that:

$$ T = \begin{pmatrix} A & B\\ 0 & C\\ \end{pmatrix} $$

is the matrix associated to $T$ in the base $\beta$ and $A$ is $n\times n$.

My questions are:

• What does it mean that $W$ is invariant under $T$? That each element of $W$ is fixed under $T$? Or that $T(W) = W$?
• Can you provide a proof? I've been struggling with this for quite a while.

Thank you!

Answer 1

The subspace $W$ is invariant under $T$ if $T(W) \subseteq W$. It doens't mean that for each $w \in W$ we have $Tw = w$ nor $T(W) = W$ (for example, $T$ might be the zero transformation and then any subspace is $T$-invariant but $T(W) = \{ 0 \} \subsetneq W$ for non-trival $W$).

Regarding the proof, choose some basis $(v_1,\dots,v_n)$ of $W$ and complete it arbitrary to a basis $\beta = (v_1,\dots,v_n,v_{n+1},\dots,v_{n+m})$ of $V$. Denote by $[Tv_i]_{\beta}$ the column vector whose entries are the coefficients of $Tv_i$ with respect to the basis $\beta$. Since $Tv_i \in W$ for $1 \leq i \leq n$, the $n + 1, \dots, n + m$ entries of $[Tv_i]_{\beta}$ must be zero (make sure under you understand why!). But $[Tv_i]_{\beta}$ is the $i$-th column of the matrix $[T]_{\beta}$ which represents $T$ with respect to the basis $\beta$. Hence, you have proved the result.