1. ## Adjoint of linear operators

Question: Let $T$ be a linear operator on a finite-dimensional vector space $V$. Show that:
1. $N(T^*T)=N(T)$
2. $rank(T) = rank(T^*)$
3. $rank(A^*A)=rank(AA^*)=rank(A)$

I figured that if I can prove (1), then it follows via the dimension theorem that $rank(T^*T)=rank(TT^*)=rank(T)$. Then if I just find a basis, I can use the previous fact to solve (3). I just cannot get started on (1), though. And what about (2)? Any tips would be helpful - thanks!

2. Originally Posted by Last_Singularity
Question: Let $T$ be a linear operator on a finite-dimensional vector space $V$. Show that:
1. $N(T^*T)=N(T)$
2. $rank(T) = rank(T^*)$
3. $rank(A^*A)=rank(AA^*)=rank(A)$

I figured that if I can prove (1), then it follows via the dimension theorem that $rank(T^*T)=rank(TT^*)=rank(T)$. Then if I just find a basis, I can use the previous fact to solve (3). I just cannot get started on (1), though. And what about (2)? Any tips would be helpful - thanks!
The space V has to be more than a vector space. It must be an inner-product space in order for the adjoint operation to be defined.

For (1), it's clear that if Tx=0 then T*Tx=0. So N(T) ⊆ N(T*T). For the reverse inclusion, notice that $T^*Tx=0\ \Longrightarrow\ 0 = \langle T^*Tx,x\rangle = \langle Tx,Tx\rangle = \|Tx\|^2 \Longrightarrow\ Tx=0$.

For (2), suppose T has (finite-dimensional) range $W_0$, and let $W_1$ be the orthogonal complement of $W_0$ in V. If $x\in W_1$ and $y\in V$ then $\langle T^*x,y\rangle = \langle x,Ty\rangle = 0$, so that $T^*(V) = T^*(W_0)$. Therefore $\text{rank}(T^*) \leqslant \text{dim}(W_0) = \text{rank}(T).$ The same argument with T and T* interchanged gives the reverse inequality.