0
$\begingroup$

Let $f:V\to V$ be a linear operator on a finite-dimensional vector space $V$, and let $W\subset V$ be an $f$-invariant subspace. Let $B_W$ be a basis for $W$ and extend it to a basis $B$ for $V$.

Show that $$\mathcal{M}_{BB}(f)=\begin{pmatrix}\mathcal{M_{B_WB_W}}(f) & * \\ 0 & \mathcal{M_{AA}}(\bar{f}) \end{pmatrix}$$

where $\bar f$ is the induced linear operator on $V/W$ and $\mathcal{A}=\{v+W:v\in B \setminus B_W\}$

It is clear to me why the LHS of the matrix is as shown since $W$ is $f$-invariant. However, the RHS is not clear to me.

$\endgroup$
1
$\begingroup$

The induced linear operator $\bar{f}:V/W \to V/W$ is defined as follows: for any $v+W \in V/W$, $$\bar{f}(v+W) := q_{V/W}(f(v+w)),$$ where $w$ is any element of $W$, and $P_W:V \to V/W$ is the quotient map. This is well-defined (i.e. the choice of $w$ does not matter), since $W$-invariance implies $$q_{V/W}(f(v+w))=q_{V/W}(f(v)) + q_{V/W}(f(w)) = f(v)+W.$$


We now turn to representing this in the basis $B$.

The quotient space $V/W$ can be represented as vectors in $\mathbb{R}^{\dim(V)-\dim(W)}$, i.e., vectors with length equal to the height of the bottom block in your above block matrix. The quotient map $q_{V/W}:V \to V/W$ in basis $B$ simply takes a vector $v \in \mathbb{R}^{\dim(V)}$ and deletes the top block.

So, if $v+W$ is represented by some vector in $\mathbb{R}^{\dim(V)-\dim(W)}$, then $\bar{f}(v+W)=q_{V/W}(f(v+w))$ is computed as follows.

  • extend the small vector $v+W$ into a vector in $\mathbb{R}^{\dim(V)}$, filling the new components with arbitrary entries. this yields $v+w$ for some $w \in W$
  • compute $f(v+w)$ by multiplying your matrix $\mathcal{M}_{BB}(f)$ and this vector $v+w$
  • compute $q_{V/W}(f(v+w))$ by deleting the top block of $f(v+w)$

You can see that this is precisely the same as multiplying $\mathcal{M}_{\mathcal{AA}}(\bar{f})$ with your original small vector $v+W$.

$\endgroup$

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.