# Revisted: What does $R(T^*)^{\perp}$ mean?

A PLACE WHERE I ADD MY THOUGHTS AS I GO:

$$w\in R(T^H)^{\perp}:=\{v\in V ~:~ \langle v, T^Hv\rangle =0~\forall~w\in R(T^H)\}$$

$$\langle v , T^Hv \rangle = \langle ? , ? \rangle =0$$

QUESTION:

Let $T$ be a linear operator on a finite diensional inner product space. Prove that $R(T^*)^{\perp}=N(T)$ and $R(T^*)=N(T)^{\perp}$. $$\overset{?}{\downarrow}$$ Let $T$ be a linear operator on a finite diensional inner product space. Prove that $R(T^H)^{\perp}=N(T)$ and $R(T^H)=N(T)^{\perp}$. What does $R(T^H)^{\perp}$ mean? I've never seen the "$^{\perp}$" used that way.

$$\circ\circ\circ\circ~Data~\circ\circ\circ\circ$$

$$W^{\perp} := \{v\in V : \langle v,w\rangle=0 ~\forall~w\in W\}$$ $$R(T):=\{Tv~:~v\in V\}$$ $$N(T):=\{v\in V~:~Tv=0\}$$ $$T^H:=\bar{T}^{\top}$$

COMMENTS:

Is $T$ assumed to be left multiplication by $A$? That is, can I say $\bar{T}$ is the linear operator $\bar{T}:V\rightarrow W$, defined by $\bar{T}v=\bar{A}v$, where $\bar{A}=\bar{(a_{ij})}=(\bar{a}_{ij})$ and $A=(a_{ij})$? Is that what this thing is saying? Similarly, can I say $T^{\top}$ is the linear operator $T^{\top}:V\rightarrow W$, defined by $T^{\top}v=A^{\top}v$, where $A^{\top}=(a_{ji})$?

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Orthogonal complement. –  Michael Jun 5 '13 at 22:05
I know, but of the range of $T^*$. What does that mean? –  Tyquan Pesik Jun 5 '13 at 22:06
Whoever deleted their answer. I wanted to at least read it before you deleted it. –  Tyquan Pesik Jun 5 '13 at 22:07
$R$ for range, $T^*$ for the conjugate transpose of $T$ and $\perp$ for the orthogonal complement –  yoyo Jun 5 '13 at 22:28
@yoyo Thank you. –  Tyquan Pesik Jun 5 '13 at 22:37

## 1 Answer

This symbol is for the "orthogonal complement" of a subspace. In particular, you are asked to prove that the orthogonal complement of the range of $T^*$ is equal to the kernel of $T$, and that the range of $T^*$ is equal to the orthogonal complement of the kernel of $T$.

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I see, so this proof will fall from the definitions of each, right? –  Tyquan Pesik Jun 5 '13 at 22:08
@TyquanPesik Yes. And the definition of $T^*$. –  M Turgeon Jun 5 '13 at 22:10