# Why does $(A^TA)^{-1} = I$ imply that $T_A$ is injective?

Let $$T_A : \mathbb R^n \to \mathbb R^m$$ be given by the matrix $$\mathbf{A}$$. I have been told A has a left inverse if $$T_A$$ is injective. Also, I have been told that $$\mathbf{A}$$ has a left inverse if $$(A^TA)^{-1} = I$$.

Therefore, I was thinking that $$(A^TA)^{-1} = I$$ must imply that $$T_A$$ is injective. If this is true, can anyone give an explanation of this implication?

• A weaker condition: assume that $A^TA$ is invertible, i.e. $(A^TA)^{-1}A^TA=I$. That's all you need to make a conclusion that $A$ has a left inverse. – A.Γ. Aug 16 at 22:16

Suppose $$(A^T A)^{-1} = I$$. Then $$A^T A = I$$. Hence $$A$$ has a left inverse ($$A^T$$). Therefore, $$T_A$$ is injective.