Let $T\in Mat_{m\times n}(\mathbb{R}), X\in Mat_{n\times p}(\mathbb{R})$. Given the subspace $S=\{X|TX=0\}$, Find $dimS$

Question

Let $T\in Mat_{m\times n}(\mathbb{R}), X\in Mat_{n\times p}(\mathbb{R})$.

Given the subspace $S =\{X \ |T X=0 \}$, Find the dimension of $S$.

My attempt: Actually, I have tried that for any column $C_i$ of $X$, $C_i \in Null(A)$, and corresponding to every vector $C_i \in Null(A)$, there exists an $ X \in Mat_{n\times p}$ such that $X \in S$. And, the dimension of $S \leq np$, but how to find it ???

Let $T\in Mat_{m\times n}(\mathbb{R}), X\in Mat_{n\times p}(\mathbb{R})$.Given the subspace $S=\{X|TX=0\}$, Find $dimS$

Answer 1

By the rank-nullity theorem, $\dim\ker T=n-r$, where $r=\operatorname{rank}T$. Let $\{v_1,\ldots,v_n\}$ be a basis of $\mathbb R^n$ such that $\{v_{r+1},\ldots,v_n\}$ is a basis of $\ker T$. Then a matrix $X\in\mathbb R^{n\times p}$ belongs to $S$ if and only if, for each $w\in\mathbb R^p$, $X.w\in\operatorname{span}\bigl(\{v_{r+1},\ldots,v_n\}\bigr)$. It is not hard to deduce from this that $\dim S=(n-r)\times p$.