Existence of a block upper triangular form matrix representation for a linear operator

Question

Let $T:V \rightarrow V$ be a linear operator on a finite dimensional vector space over $F$. Let $W \subset V$ be a subspace which is $T$-invariant. Show that there exists an ordered basis $\mathcal{B}$ for $V$ such that $$[T]_{\mathcal{B}}=\begin{pmatrix}A & B \\ 0 & C \end{pmatrix}$$ where $A$ is a matrix representation of ${T|}_W$.

I know that if $V$ is the direct sum of two invariant subspaces $W_1,W_2$ then we can write $[T]$ as a diagonal block form. But I have no clue how to prove the claim above. Any ideas?

Answer 1

Long comment:

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$[T]_{\mathcal B}$ means both the elements of $V$ and $T(V)$ are expressed as linear combinations of the vectors from the basis $\mathcal B$. This is always possible because $T:V\to V\implies T(V)\subseteq V$.

Let $B_1=\left\{b_1,\ldots,b_k\right\}$ and let $X$ be some direct complement of $W$ in $V$ with a basis $B_2=\left\{b_{k+1},\ldots,b_n\right\}$.

Now, the columns of $[T]_{\mathcal B}$ are exactly the vectors $T\left(b_1\right), T\left(b_2\right),\ldots,T\left(b_k\right)$ written as linear combinations of $b_1,b_2,\ldots,b_k,b_{k+1},\ldots,b_n$ (in the same order).

@user7530 has already noted $X$ doesn't have to be invariant under $T$.

That means, for some $v_1\in X$, it might be possible that $T(v_1)\in X$.

So, for some $j\in\{k+1,\ldots,n\},\quad T\left(b_j\right)=\sum_{i=1}^n\alpha_ib_i$, where $\sum_{i=1}^k\alpha_ib_i$ isn't necessarily $0$, however, we cannot express $T\left(b_1\right),\ldots,T\left(b_k\right)$ over $b_{k+1},\ldots,b_n$.

Whether or not $X$ is invariant under $T$ will only affect the block $B$.

Answer 2

Pick an orthogonal basis $B_1$ for $W$, then extend it to an orthogonal basis $\mathcal{B} = B_1 \oplus B_2$ of $V$.

Now examine how $T$ acts on the elements of $\mathcal{B}$ that come from $B_1$, and from $B_2$.

Notice that $B_1$ being $T$-invariant by no means implies that $B_2$ is $T$-invariant. Consider for instance $$T(x,y) = (x+y,0).$$

The linear space spanned by $(1,0)$ is $T$-invariant, but the complement spanned by $(0,1)$ is not.