Problem 1. (20 points) Let $V$ be a 10-dimensional real vector space and $U_{1}, U_{2}$ two linear subspaces
such that $U_{1} \subseteq U_{2}$, $\dim U_{1} = 3$ and $\dim U_{2} = 6$. Let $E$ be the set of all linear maps $T: V \rightarrow V$ which
have $U_{1}$ and $U_{2}$ as invariant subspaces (i.e., $T(U_{1}) \subseteq U_{1}$ and $T(U_{2}) \subseteq U_{2}$). Calculate the dimension of $E$
as a real vector space.
Solution: Link to the solution: https://www.imc-math.org.uk/imc1998/prob_sol1.pdf
My struggle: I get what they mean at the first part of the solution (so the $\textit{choose a basis}$ part). But after that part they determine in one go what the matrix of the linear map looks like and out of the matrix they conclude the dimension. My question is: Why is the matrix given in the answer logical and my follow-up question would be: why do they count the dots for the dimension of the vector space? Thanks in advance.
Well, the matrix of $T$ relative of the basis $\{v_1,\dots,v_{10}\}$ is the matrix with entries $a_{i,j}$ that satisfy $$T(v_j) = a_{1,j}v_1 + a_{2,j}v_2 + \cdots + a_{10,j}v_{10} \textrm{ for each $j \in \{1,\dots,10\}$.}$$ So, as $U_1$ is $T$-invariant we have that $T(v_j) \in U_1$ for $j = 1,2,3$, and then $T(v_j)$ can be written as a linear combination only of the vectors $v_1,v_2$ and $v_3$. This is why $a_{4,j} = \cdots = a_{10,j} = 0$ in this case. The same happens with $j = 4,5,6$.