Rank of adjoint operator

Question

$ \newcommand{\span}{\operatorname{span}} \newcommand{\rank}{\operatorname{rank}} \newcommand{\ip}[1]{ \left \langle #1 \right \rangle } $ Suppose $T: V \rightarrow W$ is a linear transformation, where $v$ and $W$ are finite-dimensional vector spaces. If $T^*$ is the adjoint of $T$, show that:

$$\rank(T) = \rank(T^*)$$

My attempt:

Suppose $B = \{v_1, ..., v_m\}$ and $C = \{w_1, ..., w_n\}$ are bases of $V$ and $W$ respectively. Applying the Gram-Schmidt process, $B' = \{s_1, ..., s_m\}$ and $C' = \{t_1, ..., t_m\}$ are orthonormal bases of $V$ and $W$ respectively. Suppose $v \in V$ and $w \in W$, then $v \in \span(s_1, ..., s_m)$ and $w \in \span(t_1, ..., t_n)$. So

$$v = a_1s_1 + ... a_ms_m$$ for scalars $a_1, ..., a_m$

, and

$$w = b_1t_1 + ... b_nt_n$$ for scalars $b_1, ..., b_m$.

By definition, $\ip{ Tv, w }$ = $\ip{ v, T^*w }$ for $v \in V$, $w \in W$. Hence,

\begin{align*} &\ip{ T\sum_{i=1}^{m} a_is_i, \sum_{j=1}^{n} b_jt_j } = \ip { \sum_{i=1}^{m} a_is_i, T^*\sum_{j=1}^{n} b_jt_j }\\ &\implies \ip{ \sum_{i=1}^{m} a_iTs_i, \sum_{j=1}^{n} b_jt_j }= \ip{ \sum_{i=1}^{m} a_is_i, \sum_{j=1}^{n} b_jT^*t_j }\\ &\implies \sum_{i=1}^{m} \sum_{j=1}^{n} \ip{a_iTs_i, b_jt_j} = \sum_{i=1}^{m} \sum_{j=1}^{n}\ip{a_is_i, b_jT^*t_j} \end{align*}

From here, I'm stuck. I'm not sure how this would simplify to the point of implying that $\rank(T) = \rank(T^*)$.

Any assistance is much appreciated. I'm also open to other proofs...

Answer 1

Here are some hints that may help you avoid getting down too much in the weeds:

$\tag 1\operatorname{dim} V = \operatorname{dim}N(T) + \operatorname{dim} N(T)^{\perp}$ and

$\tag2 \operatorname{dim} V = \operatorname{dim}N(T) + \operatorname{dim} R(T)$ so

$\tag3 \operatorname{dim} N(T)^{\perp}=\operatorname{dim} R(T)$

which means that it suffices to prove that

$\tag4 \operatorname{dim} R(T^*)=\operatorname{dim} N(T)^{\perp}.$ And for this it is enough to show that

$\tag5 R(T^*)^{\perp}=N(T).$

So, let $x\in R(T^*)^{\perp}.$ Then, $0=\langle x,T^*y\rangle$ for all $y\in W$ so...

and if $x\in N(T)$ then $0=\langle T(x)*y\rangle$ for all $y\in W$ so...