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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?

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  • $\begingroup$ Start with a basis for $W$ and extend it to a basis for $V$. Then use the fact that $W$ is $T$-invariant. $\endgroup$ – Oliver Jones Feb 8 '16 at 5:07
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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.

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  • $\begingroup$ If you extend to an orthonormal basis for $V$ then you get block diagonal form, not block upper triangular form $\endgroup$ – Oliver Jones Feb 8 '16 at 5:15
  • $\begingroup$ @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 $W$? $\endgroup$ – user7530 Feb 8 '16 at 5:17
  • $\begingroup$ Won't $B_2$ be $T$-invariant? $\endgroup$ – Oliver Jones Feb 8 '16 at 5:20
  • $\begingroup$ @OliverJones Why do you think so? $\endgroup$ – user7530 Feb 8 '16 at 5:21
  • $\begingroup$ Because of orthogonality. $\endgroup$ – Oliver Jones Feb 8 '16 at 5:21

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