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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.