If $T$ injective or $T$ surjective, what is the composition $T^\ast T$? (where $T^\ast$ denotes adjoint of linear map $T$)

Question

Let $T:V \to W$ be a linear transformation between inner product spaces. Then $T^\ast: W \to V$ denotes the linear transformation with the property that for every $v \in V$ and $w \in W$, $$\langle T(v),w \rangle = \langle v, T^\ast(w) \rangle.$$ We call $T^\ast$ the adjoint of $T:V \to W$.

$\cdot$ If $T$ is injective is $T^\ast T$ injective (or possibly a bijection)?

$\cdot$ If $T$ is surjective is $T T^\ast$ surjective (or a bijection)?

How can we prove this? Any pointers in the right direction appreciated.

As also pointed out by others, the adjoint goes from $W^$ to $V^$. How do you propose to compose them? With matrices the distinction can be blurred and you can identify the dual space, albeit non-canonically. Please clarify. As you see the answer depends... — Jyrki Lahtonen, Nov 22 '15 at 07:06

score 2 · Answer 1 · answered Nov 22 '15 at 04:46

2

Consider the left-shift operator on $\ell^2$ $$ T((x_n)) = (x_2,x_3,\ldots) $$ Then $$ T^{\ast}((x_n)) = (0,x_1,x_2,\ldots) $$ Then $T$ is surjective, and $$ T^{\ast}T((x_n)) = (0,x_2,x_3,\ldots) $$ Hence, $T^{\ast}T$ is neither injective nor surjective.

answered Nov 22 '15 at 04:46

Prahlad Vaidyanathan

31,554

What does $T^*$ mean in this question? – Arashium Nov 22 '15 at 04:51
@Arashium: As mentioned by the OP, it is the adjoint of $T$. – Prahlad Vaidyanathan Nov 22 '15 at 04:54
@Shalop: Agreed, but this is standard convention. Furthermore, the compostion $T^{\ast}T$ only makes sense under this identification. Else, the OP's question does not make sense. – Prahlad Vaidyanathan Nov 22 '15 at 05:17

score 1 · Answer 2 · answered Nov 22 '15 at 04:45

1

start with

$$T^*\times T=T\times T^*=\det(T) I_n$$

answered Nov 22 '15 at 04:45

Arashium

2,541

And when the space is infinite-dimensional?? – Jyrki Lahtonen Nov 22 '15 at 07:03
@JyrkiLahtonen, Then I have no idea :D – Arashium Nov 22 '15 at 08:05
Besides, it looks like you thought of the adjugate rather than the adjoint. – Daniel Fischer Nov 24 '15 at 22:24
@DanielFischer How do you define adjoint? – Arashium Nov 25 '15 at 11:57
For inner product spaces, the adjoint of $T\colon V \to W$ is the linear operator $T^{\ast}\colon W \to V$ characterised by the identity $\langle T(v), w\rangle_W = \langle v, T^{\ast}(w)\rangle_V$ for all $v\in V$ and $w\in W$. – Daniel Fischer Nov 25 '15 at 12:01

If $T$ injective or $T$ surjective, what is the composition $T^\ast T$? (where $T^\ast$ denotes adjoint of linear map $T$)

2 Answers2