I've run into the following problem:

If $V$ has dimension $n$ and $U$ is a $T$-invariant subspace of dimension $m$, prove that there is a basis $\mathcal B$ for $V$ so that the matrix of $T$ with respect to $\mathcal B$ has the following shape: $$[T]_{\mathcal B } = \left [ \begin{array}{c c c c c c } a_{1,1} & \dots & a_{1,m} & 0& \dots &0 \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & \dots & a_{m,m} & 0 & \dots & 0 \\ b_{1,1} & \dots & b_{1,m} & c_{1,1} & \dots & c_{1,n-m} \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ b_{n-m,1} & \dots & b_{n-m,m} & c_{n-m,1} & \dots & c_{n-m,n-m} \end{array} \right ]$$ My idea (possibly completely wrong):

We first get $\mathcal{C} = \{u_1,u_2,\dots,u_m\}$ a basis of $U$ and calculate the matrix of $T|_U$ with respect to $\mathcal{C}$. Now this will be matrix $A \in Mat_{m \times m}$ with entries $[a]_{ij}$, where the $i$-th row is the coordinates of $T|_U(u_i)$ with respect to $\mathcal{C}$. Now we complete $\mathcal{C}$ into $\mathcal{B} = \{u_1,\dots, u_m,v_{m+1},\dots , v_n \}$. Finally we make a matrix with respect to $\mathcal{B}$ the first $m$ rows will be just the same as in $A$ followed by $n-m$ zeroes as each $T(u_i)$ can be expressed from vectors in $\mathcal{C}$, so all "added" vectors will have coefficients zero in the linear combination. Now the thing that confuses me is that the las $n-m$ rows should just be coordinates $T(v_i)$ with respect to $\mathcal{B}$, so I don't know why should we separate them into blocks $B$ and $C$.

Any help is appriciated.


Essentially the idea is correct. If we now that $U$ is an invariant subspace of dimension $m$, we now that $T(U) \subseteq U$; If we complete $U$ to basis of $V$, let's say $\mathcal{B} = \left\lbrace u_{1},\cdots u_{m},v_{m+1},\cdots,v_{n}\right \rbrace$ where the first $m$ vectors span a basis of $U$, and we compute the change of basis with respect to this basis $\mathcal{B}$, the condition $T(U) \subseteq U$ translates into : The coordinates of the image of the vectors in $U$ will be express as a linear combination just of the first $m$ vectors of the chosen basis, hence just as a linear combination of $u_{1},\cdots u_{m}$, and this is exactly the appearance of all the $0$'s after the first $m-$th coordinate in the coordinates releted to T(span($U$)).


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