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.


1 Answer 1


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 $v+W$.


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