TITLE: Prove that direct sum of linear transformations is a block matrix
QUESTION [7 upvotes]: In linear algebra, I'm facing a question I cannot quite formalize into a full proof. I want to prove that a direct sum of linear transformations is a block matrix. Here's the formal question:

Assume $V$ is a vector space with $U_1,\ldots,U_n$ as subspaces such that $V = U_1\oplus\cdots\oplus U_n$.
Assume that each $T_i: U_i\to U_i$ is a linear transformation.
Assume that $T = T_1 \oplus T_2 \oplus\cdots\oplus T_n$.

How do I formally prove that $T$ can be represented as follows?
$$\begin{bmatrix}A_1&&&0\\&A_2&&\\&&\dots\\0&&&A_m\end{bmatrix}$$
where $A_1,\ldots,A_m$ are square matrices representing $T_1,\ldots,T_n$ accordingly.
Thanks!

REPLY [10 votes]: What Chris is referring to is forming a basis of $V$ by taking the union of bases for each $U_i$.  Let's explicitly do the case $n=2$, and then you can generalize.
Let $V=U\oplus W$, and let $\{u_1,\ldots,u_n\}$ and $\{w_1,\ldots,w_m\}$ be bases for $U$ and $W$ respectively.  Let $T_U:U\rightarrow U$ be represented by $A_U$ and $T_W:W\rightarrow W$ be represented by $A_W$.
Now form an ordered basis of $V$ by taking the union of the above bases, $\{u_1,\ldots,u_n,w_1,\ldots,w_m\}$, and let $T=T_U\oplus T_W$.  Then notice that $T(u_i)=T_U(u_i)$, so expressing $T(u_i)$ as a linear combination of our basis of $V$, we see that all of the $w_i$'s will have a zero coefficient, and the coefficients of the $u_i$ will be those taken from the matrix $A_U$.  This gives us a matrix of the form: 
$$\begin{pmatrix}
A_U&*\\
0&*
\end{pmatrix}$$
See if you can complete the form of the matrix by noticing that $T(w_i)=T_W(w_i)$.