Assume $(V,\langle \ , \ \rangle_V)$ and $(W,\langle \ , \ \rangle_W)$ are finite dimensional inner product spaces and $T : V \rightarrow W$ is an injective linear transformation. Prove that $T^*T : V \rightarrow V$ is injective.
I am not sure how to prove this. I have shown that $T^*$ is surjective, but I do not know how to proceed. Can anyone offer some hints as to where to go?
EDIT: I just thought of a way, maybe someone can tell me if it is correct? Let $v \in V$ such that $T^* T v = 0$. Taking the inner product with $v$ on both sides (is this valid?) we have $$ \langle T^* T v,v \rangle = \langle T v, T v \rangle = 0 $$ This implies that $T v = 0$. Since $T$ is injective this means that $v = 0$.
Hint:
In general it holds $\ker T =\ker T^*T$ so one is injective if and only if the other is.
Namely, clearly $\ker T \subseteq \ker T^*T$. Conversely
$$x \in \ker T^*T \implies T^*Tx = 0 \implies 0 = \langle T^*Tx, x\rangle = \langle Tx, Tx\rangle \implies Tx = 0\implies x \in \ker T$$