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I was asked to determine whether a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is injective. I proved that $[T]$ doesn't have a left inverse and therefore it is not injective, but I am stuck with showing the logic equivalence between $T$'s invectiveness and the existence of $T$'s left inverse.

What I am trying to prove:

Suppose a $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$, $\forall x,x' \in \mathbb{R}^n$ such that $T(x) = T(x')$ implies $x = x'$ iff $\exists A$ such that $A[T] = I_n$.

It's quite easy to show that if $[T]$ has a left inverse then $T$ is injective:

Let $[S] = [T]^{-1}$ and $T(x) = T(x')$, then $x' = S(T(x')) = S(T(x) = x'$.

But I failed to show that:

If $T$ is injective then there must exist $A$ such that $A[T] = I_n$.

I am trying to construct a linear left inverse of $T$. It's trivial to prove that a left inverse of $T$ is linear for all $y$ such that $\exists x.T(x) = y$, but I don't know, when $m > n$, how to define $S$ for a $y$ that there is no corresponding $x$ in $T$'s domain. Because the set $Y = \{y = T(x): x \in \mathbb{R}^n\}$ is a subspace of $\mathbb{R}^m$, I could not represent $x \notin Y$ in terms of $Y$'s elements.

Any hints? I just started to learn linear algebra this quarter, so I really don't know some high level concepts, though I could try to read Wikipedia.

Minsheng Liu
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Since $Y$ is a subspace of $\Bbb R^m$, there exists other subspaces $W$ such that $Y \cap W = \{0\}$ and $Y + W = \Bbb R^m$. I.e., $\Bbb R^m = Y \oplus W$. Choose one and define $S$ to carry $W$ to $0$.

To see the decomposition is always possible, set $Y_0 = Y$, and as long as $Y_k \ne \Bbb R^m$, inductively pick $y_{k+1} \notin Y_k$, and set $Y_{k+1} = \text{Span}(Y_k \cup \{y_{k+1}\})$. For some $k \le m$, you will have $Y_k = \Bbb R^m$. Then set $W = \text{Span}(\{y_1, y_2, ..., y_k\})$.

Paul Sinclair
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