# Doubt in the proof that $\operatorname{Image}(T^{*})=\operatorname{Ker}(T)^{\perp}$

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!

• When you wrote $\langle T^*,v\rangle$, my guess is that you meant $\langle T^*u,v\rangle$. – José Carlos Santos Sep 16 '20 at 18:56
• @JoséCarlosSantos Yes, sorry that was my mistake – luisegf Sep 16 '20 at 19:08
• @Physor Sorry! $T^{*}$ is the adjoint operator of $T$ – luisegf Sep 16 '20 at 19:09
• It might be easier to prove that $\operatorname{Image}(T^*)^\perp =\operatorname{Ker}(T)$, and then use that $V$ is finite dimensional. – Mor A. Sep 16 '20 at 19:30
• 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. – 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 $$B$$, $$\operatorname{rank}T^*=\operatorname{rank}T=n$$.