matrix representation of T|w where W is invariant space

Question

$V$ is a vector space above field $F$ and $T:V→V$ is a linear operator. $W$ is an invariant subspace. $T|_w:W→W$ is the reduced operator.

How do I prove that there is such base $B$ that $[T]_B$ is $$\begin{pmatrix}X&Y\\ 0&Z\end{pmatrix}$$ where X,Y,Z are block matrices.

And if $V=W_1⊕W_2$ where $W_1,W_2$ are invariant sub-spaces of $V$ then there is such $B$ so: $$[T]_B=\begin{pmatrix}X&0\\ 0&Z\end{pmatrix}$$

Answer 1

Say $dim(V)=n \quad dim(W)=k$

Start by taking a base $B_W=(v_1,v_2,...,v_k)$ for the sub-space W, where $v_i\in W$

You can then complete $B_W$ to a base for $V$ by adding $n-k$ independent vectors from V.

Call this base $B=(v_1,...,v_k,v_{k+1},..,v_n)$.

Now try and write the matrix $[T]_B$. Notice it will be a block matrix.

Use the fact that W is T-invarient.

Also notice: $\forall [1\le i \le n] : v_i \in V \;$ because W is a sub-space of V.

I guess you will be able to figure out the rest. Good luck!

Answer 2

The columns of $[T]_B$ are the coordinates of the images of the basis vectors. Suppose $\dim W=m$ and $\dim V=n$. Recall that the coordinates of a vector relative to some ordered basis $B$ are the coefficients of the unique linear combination of those basis vectors that produces the vector. Take any ordered basis $B_W = (w_1,\dots,w_m)$ of $W$ and complete it to a basis $B=(w_1,\dots,w_m,v_{m+1},\dots,v_n)$ of $V$. The last $n-m$ coordinates of any element of $W$ relative to this basis are zero (why?). Since $W$ is $T$-invariant, we have in particular $T[w_i]\in W$, and as the coordinates of these images under $T$ form the first $m$ columns of $[T]_B$, it will have an $(n-m)\times m$ block of zeros in the lower left.

For the second part, choose any pair of bases for $W_1$ and $W_2$ and concatenate them. By the above, you have the required lower-left block of zeros, and a symmetric argument shows that you will also have a $m\times(n-m)$ block of zeros in the upper right.