# Matrices and T-invariant subspaces

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($$U$$)).