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I have some misunderstandings about the claim :

$T: V\to V$ is an inner products space.

$\ker(T^*)=(\operatorname{Im}(T))^\perp$

I don't get why $(\operatorname{Im}(T))^\perp$ isn't $\ker (T)$ ? (I know that $\ker(T)=\ker(T)^* \iff TT^*=T^*T$, this is not that case)

I will be grateful for an explanation or example, thanks!

Chickenmancer
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Algo
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    I wrote an explanation of this fact here: https://math.stackexchange.com/a/467256/40119 – littleO Jan 16 '22 at 10:53
  • @erez It is not true that $\ker(T) = \ker(T^) \iff TT^ = T^T$, but it is true that $TT^ = T^T \implies \ker(T) = \ker(T^)$. For example, the operator corresponding to the matrix $$ \pmatrix{0&0&0\0&1&1\0&0&1} $$ does not satisfy $TT^* = T^T$, but it does satisfy $\ker(T) = \ker(T^)$. – Ben Grossmann Jan 17 '22 at 18:14
  • @erez Regarding your question, you might find it helpful to consider the operator corresponding to the matrix $$ \pmatrix{0&1\0&0}, $$ which is given by $T(x,y) = (y,0)$. Note that we actual have $\operatorname{Im}(T) = \ker(T)$ in this case! – Ben Grossmann Jan 17 '22 at 18:15

2 Answers2

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Just compute.
\begin{align} x \in \ker(T^*) &\Longleftrightarrow T^*(x) = 0 \\ &\Longleftrightarrow \forall y\quad \langle T^*(x),y\rangle = 0 \\ &\Longleftrightarrow \forall y\quad \langle x,T(y)\rangle = 0 \\ &\Longleftrightarrow x \perp \operatorname{im}(T) \\ &\Longleftrightarrow x \in \big(\operatorname{im}(T)\big)^\perp \end{align}

GEdgar
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I don't know if a proof is good enough explanation, but this is what works for me.

Lets take $ v \in \ker(T^*)$, we want to show that $ v \in \operatorname{Im}(T)^\perp$ so let's take $ v_2 \in \operatorname{Im}(T)$ and we want to show that $ \langle v,v_2 \rangle = 0$.

$ v_2 \in \operatorname{Im}(T)$ so there is $u \in V$ such that $ Tu = v_2$.

$$ \langle v,v_2 \rangle = \langle v, Tu \rangle = \langle T^*v, u \rangle = (v \in \ker T^*) = \langle 0, u \rangle = 0$$

On the other hand, if we take $ v \in \operatorname{Im}(T)^\perp$ and we want to show that $ v \in \ker T^*$, so we need to show that $T^*v = 0$

one way of showing it is using the norm. $$ \lVert T^*v \rVert = \langle T^*v, T^*v \rangle = \langle v , T(T^*v) \rangle = (T(T^*v) \in \operatorname{Im}(T), v \in \operatorname{Im}(T)^\perp) = 0 $$ $$ \Rightarrow T^*v = 0$$

It wasn't an example but I hope it helps.

Chickenmancer
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Nadav Kalma
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