Prove $N(T^*T) = N(T)$, and $\dim(R(T^*T))=\dim(R(T))$.

Question

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Let $T: V\to V$ be linear and $V$ a finite dimensional inner product space. Prove $N(T^*T) = N(T)$, and $\dim(R(T^*T))=\dim(R(T))$.

$\textbf{Solution:}$ Consider the relation $(T+U)^*=T^* + U^*$. We wish to prove the left hand side is equal to the right hand side.

Let $x\in N(T^*T)$ then $T^*T(x) = 0$. $$0=\langle T^*T(x), x\rangle = \langle T(x), T(x) \rangle.$$ This implies, $T(x) = 0$ for all $x\in N(T)$. Thus, $N(T^*T) \subseteq N(T). \hspace{35 pt} (1)$

For $x\in N(T)$, $$T^*T(x) = T^*(0) = 0.$$ Therefore, $x\in N(T^*T).$ Therefore, $N(T) \subseteq N(T^*T). \hspace{35 pt} (2)$

From (1) and (2), $N(T^*T) = N(T)$.

Again, since the dimension is finite, $$R(T^*T) = N(T^*T)^\perp = N(T)^\perp = R(T^*).$$ Thus, rank$(T^*T)=$rank$(T^*) = $rank$(T)$. Hence, rank$(T^*T)=$rank$(T)$.

Answer 1

$\langle T^{*}Tx , x \rangle =\langle Tx , Tx \rangle =\|Tx\|^{2}$ so $T^{*}Tx =0$ implies $Tx=0$ The converse is obvious.

The second part follows from rank-nullity theorem.