Rank of a Linear Operator over matrix space

Question

Suppose $Q\in M_{10\times 10}(\mathbb{R})$ is a matrix of rank 5. Let $T:M_{10\times 10}(\mathbb{R})\to M_{10\times 10}(\mathbb{R})$ be the linear transformation defined by $T(P)=QP$. Then the rank of $T$ is?

Answer 1

Let $m$ and $n$ be positive integers with $m\leq n$. Fix a base field $\mathbb{K}$ and write $\mathcal{R}:=\text{Mat}_{n\times n}(\mathbb{K})$. Suppose that $Q\in \mathcal{R}$ is of rank $m$. We shall prove that the linear transformation $T_Q:R\to R$ sending $P\mapsto QP$ for all $P\in \mathcal{R}$ is a linear map of rank $mn$. That is, we shall prove that $\dim_\mathbb{K}\big(\text{im}(T)\big)=mn$.

Let $q_k$ be the $k$-th column vector of $Q$ for $k\in \{1,2,\ldots,n\}=:[n]$. Suppose that $q_{j_1},q_{j_2},\ldots,q_{j_m}$ are linearly independent columns of $Q$ where $j_1,j_2,\ldots,j_m\in[n]$ satisfies $j_1<j_2<\ldots<j_m$. Write $E_{i,j}$ for the matrix with $1$ at the $(i,j)$-entry and $0$ everywhere else (where $i,j\in [n]$). Then, the image $X_{i,j}:=T_Q(E_{i,j})$ of $E_{i,j}$ under $T_Q$ is the matrix where the $j$-th column equals $q_i$, and other columns are $0$. It is easy to prove that the matrices $X_{i,j}$ for $i\in\{j_1,j_2,\ldots,j_m\}$ and $j\in[n]$ are linearly independent elements in $\mathcal{R}$ which span $\text{im}(T)$, and I shall leave this task to the OP.