Can someone help me with this problem?

Let $V$ be an $n$-dimensional vector space, and let $T: V \rightarrow V$ be a linear transformation. Suppose that $W$ is a $T$-invariant subspace of $V$ having dimension $k$. Show that there is a basis $\beta$ for $V$ such that $[T]_{\beta}$ has the form $\begin{pmatrix} A & B \\ O & C \end{pmatrix}$, where $A$ is a $k \times k$-matrix and $O$ is the $(n-k) \times k$ zero matrix.

Attempt at a solution: Let $(v_1, v_2, \ldots, v_k)$ be a basis for $W$ and extend it to a basis $\beta = (v_1, v_2, \ldots, v_n)$ for $V$. Since $W$ is $T$-invariant, we have $\forall v_j \in W$, $T(v_j) \in W$ for $j= 1, 2, \ldots, k$. Hence $T(v_j) = \sum_{i=1}^{k} a_{ij} v_i$.

Don't know where to go from here. How to find A, B, C and $O$?


Let $T: V \rightarrow V$ be a linear transformation and $\beta = \{ v_1, \dots, v_n \}$ be a basis constructed as to place $\{ v_1, \dots , v_k \}$ in $W$. Define $$ [T]_{\beta} = [[T(v_1)]_{\beta}|\cdots | [T(v_k)]_{\beta}|[T(v_{k+1})]_{\beta}|\cdots | [T(v_n)]_{\beta}] $$ Here $[c_1v_1+ \cdots +c_nv_n]_{\beta} = [c_1, \cdots , c_n]^T$. This is the $\beta$-coordinate map. By the $T$-invariance of $W$ we have the existence of $a_{ij}$ for $1 \leq i,j \leq k$ such that $$ T(v_j) = \sum_{i=1}^k a_{ij}v_i $$ now, just to emphasize what's happening, I'll focus on $j=1$, $$ T(v_1) = \sum_{i=1}^k a_{i1}v_i = a_{11}v_1+ \cdots +a_{k1}v_k+0v_{k+1}+ \cdots +0v_n $$ of course we can easily read off the coordinates of $T(v_1)$ given the expression above: $$ [T(v_1)]_{\beta} = [a_{11}, \dots, a_{k1}, 0, \dots , 0]^T$$ Likewise, for $j=2, \dots k$ we find $k$-possibly nonzero coefficients in the image paired with $n-k$ necessary $0$'s. In contrast, we get to say nothing about $T(v_{k+1})$ to $T(v_n)$. So, what does that mean?

  • $\begingroup$ Thanks for the answer, I think I'm starting to understand. But why can we deduce that $T(v_j) = \sum_{i=1}^{k} a_{ij} v_i$ on the basis of $W$ being $T$-invariant? $\endgroup$ – Kamil Jan 2 '15 at 15:38
  • $\begingroup$ Because $T(W) \subseteq W$ implies $T(v_j) \in W$ for each $j=1, \dots k$. If it were otherwise then the image of your $W$-basis would be outside $W$ which would contradict the invariance of $W$ under $T$. Also, the first $k$-elements of $\beta$ span $W$ so we know it is possible to find the $a_{ij}$ in your post. $\endgroup$ – James S. Cook Jan 2 '15 at 15:40
  • $\begingroup$ I see. And the $c_i$'s are just the coordinate vectors of every $v_i$ with respect to the basis $\beta$? $\endgroup$ – Kamil Jan 2 '15 at 15:43
  • $\begingroup$ Yep, I mean, I'm identifying $c_1=a_{11}$ for $[T(v_1)]_{\beta}$ in my answer. $\endgroup$ – James S. Cook Jan 2 '15 at 15:46
  • $\begingroup$ Notice, there is no condition on the $B$ or $C$ matrices, so those columns of the matrix don't really say much except that $v_{k+1}$ to $v_n$ map to a span of $\beta$. $\endgroup$ – James S. Cook Jan 2 '15 at 15:48

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