urgent help needed for linear algebra question

Question

Let’s say we have the matrix $A \in\mathbb{R}^{(m×n)}$ associated with the mapping $T : \mathbb{R}^n → \mathbb{R}^m$. How can I show that if there exists a matrix $B\in\mathbb{R}^{(n×m)}$ such that $BA=I_n$ then T is one to one? if T is one to one then how can I show that there exists a matrix B such that $BA=I_n$ \ Here is my approach: the transformation is onto if we are able to solve $Ax=b$ right? so in this case, $x=B$ and $AB=I_m$. Is that right?

Similarly, how can I show that if there exists a matrix $B\in\mathbb{R}^{(n×m)}$ such that $AB=I_m$ then T is onto? if T is onto then how can I show that there exists a matrix B such that $AB=I_m$

Answer 1

No, your approach is not correct. How can you take $x=B$? After all, $x$ is a vector and you're after a matrix.

If there is a matrix $B$ such that $BA=\operatorname{Id}_n$, then, for each vector $x\in\ker T$ you have$$0=T(x)=BA.x=\operatorname{Id}_n.x=x.$$So, $\ker T=\{0\}$.

And if $T$ is on-to-one, then there is a linear map $U\colon\mathbb{R}^m\longrightarrow\mathbb{R}^n$ such that $U\circ T=\operatorname{Id}_n$. So, let $B$ be the matrix of $B$ with respect to the canonical basis.

Answer 2

In the first case there exists a liner map $S:\mathbb R^{m} \to \mathbb R^{n}$ such that $ST=I_n$. [ This is just the linear map whose matrix is $b$]. Now $Tx=Ty$ implies $S(T(x))=S(T(y))$ so $x=y$. Hence $T$ is one to one. Similarly, in teh second case ther eexists a linear map $V$ such that $TV=I_m$. If $y \in \mathbb R^{m}$ then $y=T(x)$ where $x=Vy$. Hence $T$ is onto.