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

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?

• Start with a basis for $W$ and extend it to a basis for $V$. Then use the fact that $W$ is $T$-invariant. Feb 8, 2016 at 5:07

Long comment:

$$[T]_{\mathcal B}$$ 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 $$B$$.

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.

• If you extend to an orthonormal basis for $V$ then you get block diagonal form, not block upper triangular form Feb 8, 2016 at 5:15
• @OliverJones Why do you think so? Consider how $T$ acts on an arbitrary element $v$ not in $W$. Why do you think $Tv$ remains orthogonal to $W$? Feb 8, 2016 at 5:17
• Won't $B_2$ be $T$-invariant? Feb 8, 2016 at 5:20
• @OliverJones Why do you think so? Feb 8, 2016 at 5:21
• Because of orthogonality. Feb 8, 2016 at 5:21