Rank of linear transformation from $M_{n×p}(\mathbb{R})$ to $M_{m×p}(\mathbb{R})$

Question

Let define $J: M_{n×p}(\mathbb{R})\rightarrow M_{m×p}(\mathbb{R})$ by

$$J(X)=TX$$ for $T\in M_{m×n}(\mathbb{R})$ then what is $rank(J)?$

My attempt:

(I know that, When $T:\mathbb{R^n}\rightarrow \mathbb{R^m}$ is multiplication by $m×n$ matrix $A$ then $rank(T)=rank(A)$)

So according to this, I think $rank(J)= rank(T)$ (but here domain and codomain are not usual Euclidean spaces) so I am confused?

Further I am unable to prove that, $rank(J)=rank(T)$

Please help..

Answer 1

Notice that if you have a matrix $$ X=\begin{pmatrix}x_{11} & \cdots & x_{1p}\\ \vdots & & \vdots\\ x_{n1} & \cdots & x_{np} \end{pmatrix}=\begin{pmatrix}| & & |\\ X_{1} & \cdots & X_{p}\\ | & & | \end{pmatrix}$$ where I'm using column vectors, then $$TX=\begin{pmatrix}| & & |\\ TX_{1} & \cdots & TX_{p}\\ | & & | \end{pmatrix}$$ so if we view $M_{n\times p}(\mathbb{R}) \sim \underbrace{\mathbb{R}^n \times \cdots \times \mathbb{R}^n}_p=(\mathbb{R}^n)^p$ then $J:(\mathbb{R}^n)^p \longrightarrow (\mathbb{R}^m)^p$ functions like $J(X_1,\ldots,X_p)=(TX_1,\ldots,TX_p)$. This made me think of using that $\text{rank}(J)=\dim \text{Im}(J)$. We can let $\{t_1,\ldots,t_r\}$ be a base of $\text{Im}(T)$ ($t_i=Ta_i$ for some $a_i\in \mathbb{R}^n$). Then $r=\text{rank}(T)$ and it's not hard to see that the following collection is a base for $\text{Im}(J)$

$$ (t_1,0,\ldots,0),(t_2,0,\ldots,0),\ldots (t_r,0,\ldots,0) $$ $$ (0,t_1,0,\ldots,0),(0,t_2,0,\ldots,0),\ldots (0,t_r,0,\ldots,0) $$ $$\cdots$$ $$ (0,\ldots,0,t_1),(0,\ldots,0,t_2),\ldots (0,\ldots,0,t_r) $$ which yields $\boxed{\text{rank}(J)=p\cdot r=p\cdot \text{rank}(T)}$.

It shouldn't be hard to generalise this, if you take linear maps $T_i:V_i \longrightarrow W_i$ between finite dimensional vector spacesfor $i=1,\ldots,p$ and then define the map $(T_1\times \cdots \times T_p) :V_1\times \cdots \times V_p \longrightarrow W_1\times \cdots \times W_p$ like $(T_1\times \cdots \times T_p)(v_1,\ldots,v_p)=(T_1v_1,\ldots,T_pv_p)$ then I think It should hold that

$$ \text{rank}(T_1\times \cdots \times T_p)= \text{rank}(T_1)+\cdots +\text{rank}(T_p) $$