# T injective implies T*T invertible?

Suppose $$V$$, $$W$$ are inner product spaces, and $$V$$ is finite-dimensional. I need to prove that if $$T \in \mathcal{L}(M)$$ is an injective linear map, and $$T^*$$ is the adjoint of $$T$$, then $$T^*T$$ is invertible.

I got as far as to say that if $$T$$ is injective, then $$T^*$$ is surjective. But I don't know how to show that $$T^*T$$ is invertible. Showing that $$T^*T$$ is surjective or injective would imply invertibility, but I'm not sure how to do that either. I was hoping to find a way to show that $$T^*$$ is injective (which would then imply that $$T^*T$$ is injective) but I wasn't able to.

• What is $M$? What is the domain and range of $T$? – irchans Aug 14 at 1:37
• $T^*Tx = 0$ implies $\langle Tx, Tx \rangle = \langle T^* Tx,x \rangle = 0$ implies $Tx = 0$ implies $x = 0$. So, $T^*T$ injective. – mathworker21 Aug 14 at 2:16

## 1 Answer

Hint: Show that $$T^*Tx = 0$$ implies that $$Tx = 0$$, which means that $$T^*T$$ is also injective.