Let be $V$ an inner product space over $\mathbb{C}$ with finit dimension and a linear operator $T:V\rightarrow V$. Prove that $\DeclareMathOperator{\Image}{Image}\DeclareMathOperator{\Ker}{Ker}$$\Image(T^{*})=\Ker(T)^{\perp}$. Extra note: $T^{*}$ is the adjoint operator

I've proved it, but I've a doubt in one step. Here is my proof:

First Part: $\Image(T^{*}) \subseteq \Ker(T)^{\perp}$

\begin{align*} \text{Let be } v \in \Ker(T) \text{ and }w\in \Image(T^{*}), \text{i.e. }w=T^{*}u \ \ \ \ \ \ \ \ \ \text{for some } u \in V \end{align*}

\begin{equation*} \Rightarrow \left \langle T^{*}u,v \right \rangle=\left \langle u,Tv \right \rangle=\left \langle u,0 \right \rangle=0\\ \Rightarrow T^{*}u\in \Ker(T)^{\perp} \\ \therefore \:\Image(T^{*}) \subseteq \Ker(T)^{\perp} \end{equation*}

Second Part: $\Ker(T)^{\perp}\subseteq \Image(T^{*})$

\begin{align*} \text{Let be } w \in \Ker(T)^{\perp} \text{ and }v\in \Ker(T), \end{align*}

\begin{equation*} \Rightarrow \left \langle w,v \right \rangle=0 \ \ \ \text{ and also, is true that } \left \langle u,Tv\right \rangle=0 \ \ \ \ \ \forall u \in V\\\text{Thus, }\left \langle w,v \right \rangle=\left \langle u,Tv\right \rangle=\left \langle T^{*}u,v \right \rangle=0\\\left \langle w,v \right \rangle=\left \langle T^{*}u,v \right \rangle \end{equation*}

And, here is my doubt: If we know that $\left \langle w,v \right \rangle=\left \langle T^{*}u,v \right \rangle$. Then, can I assure that $w=T^{*}u $ $ \ \ \ \text{for some } u\in V$?

If the answer to my question is yes, then $w \in \Image(T^{*})$, and we've finished. I really appreciate your help, maybe it is a trivial question but I prefer to verify that this step is correct. Thank you very much!

  • 1
    $\begingroup$ When you wrote $\langle T^*,v\rangle$, my guess is that you meant $\langle T^*u,v\rangle$. $\endgroup$ – José Carlos Santos Sep 16 '20 at 18:56
  • $\begingroup$ @JoséCarlosSantos Yes, sorry that was my mistake $\endgroup$ – luisegf Sep 16 '20 at 19:08
  • $\begingroup$ @Physor Sorry! $T^{*}$ is the adjoint operator of $T$ $\endgroup$ – luisegf Sep 16 '20 at 19:09
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    $\begingroup$ It might be easier to prove that $\operatorname{Image}(T^*)^\perp =\operatorname{Ker}(T)$, and then use that $V$ is finite dimensional. $\endgroup$ – Mor A. Sep 16 '20 at 19:30
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    $\begingroup$ In a finite dimensional inner product space, $U^{\perp\perp}=U$ for any subspace $U$. So once you prove that $\operatorname{Image}(T^*)^\perp=\operatorname{Ker}(T)$ (fairly easy to do so directly), then taking the orthogonal complement on both sides will give you the desired result. $\endgroup$ – Mor A. Sep 16 '20 at 19:44

Your doubt makes sense. If you had proved that$$(\forall v\in V):\langle w,v\rangle=\langle T^*u,v\rangle,$$then, yes, it would follow that $w=T^*u$. However, you only proved that the equality $\langle w,v\rangle=\langle T^*u,v\rangle$ holds for some $v$'s, and that is not enough.

On the other hand, you proved that $\operatorname{Image}(T^*)\subseteq\operatorname{Ker}(T)^\perp$. So, in order to complete your proof, it is enough that you prove that $\dim\operatorname{Image}(T^*)=\dim\operatorname{Ker}(T)^\perp$. Let $n=\operatorname{rank}T$. Then $\dim\operatorname{Ker}(T)=\dim(V)-n$ and therefore $\dim\operatorname{Ker}(T)^\perp=n$. On the other hand, since, if you fix an orthonormal basis $B$ of $V$, the matrix of $T^*$ with respect to $B$ is equal to the conjugate transpose of the matrix of $T$ with respect to $B$, $\operatorname{rank}T^*=\operatorname{rank}T=n$.


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