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Given an inner product space $V$ and a normal operator $T$, prove that $\ker T=\ker TT^*$

The solution I found mentions that using the fact that $T$ is normal we know it is diagonalizable, so we have an orthonormal basis of eigenvectors $\left \{ v_1, \ldots , v_n \right \}$.

Now we can notice that if $v_i$ is an eigenvector for $T$ with eigenvalue $\lambda_i$ then it is also an eigenvector for $T^*$ with eigenvalue $ \bar{\lambda}_i$. I don't understand why this is true, explanation appreciated.

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  • $\begingroup$ Have you tried using the definition of eigenvalues and eigenvectors? Vector $v$ is an eigenvector of the operator $T$ with eigenvalue $\lambda$, if $Tv = \lambda v$. Besides, the definition of a normal operator is $TT^* = T^*T$. $\endgroup$ – Limsup Jul 10 at 22:32
  • $\begingroup$ @Infima Yes I have tried, I still don't get it $\endgroup$ – paxtibimarce Jul 10 at 22:41
  • $\begingroup$ Ah, and the definition, in such space, of operator $T^*$ is as follows. Let $B(\cdot, \cdot)$ be the inner product. Then for every $x,y \in V$: $B(Tx, y) = B(x, T^*y)$. Now it should be all clear. $\endgroup$ – Limsup Jul 10 at 22:44
  • $\begingroup$ I know it dude, I still don't see why it is true... I know all the basic definitions $\endgroup$ – paxtibimarce Jul 10 at 22:46
  • $\begingroup$ I am going to write it as an answer, as in here it does not fit well. $\endgroup$ – Limsup Jul 10 at 22:46
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Hint

Claim 1 If $T$ is normal then $\|Tv\|=\|T^{*}v\|$ (actually it is an iff statement) $$\langle Tv, Tv \rangle=\langle v, T^{*}Tv \rangle=\langle v, TT^{*}v \rangle=\langle T^{*}v, T^{*}v \rangle.$$

Claim 2 If $T$ is normal then $T-\lambda I$ is also normal.

See if you can show $$(T-\lambda I)(T-\lambda I)^{*}= \dotsb=(T-\lambda I)^{*}(T-\lambda I).$$

Now use claim 1 for the normal operator $T-\lambda I$ to get $$\|(T-\lambda_i I)v_i\|=\|(T-\lambda_i I)^{*}v_i\|.$$ So if $v_i$ is the eigenvector for $T$, then the norm on LHS is $0$. This means $\|(T-\lambda_i I)^{*}v_i\|=0 \implies (T-\lambda_i I)^{*}v_i=0 \implies (T^{*}-\bar{\lambda_i}I)v_i=0$

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  • $\begingroup$ why ⟨v,TT∗v⟩=⟨T∗v,T∗v⟩? $\endgroup$ – paxtibimarce Jul 10 at 22:55
  • $\begingroup$ @paxtibimarce This is the property of the inner product $\langle Tx, y \rangle =\langle x, T^{*}y\rangle $ and the fact that $(T^{*})^{*}=T$. Use the property with $T^{*}$. $\endgroup$ – Anurag A Jul 10 at 22:58
  • $\begingroup$ shouldn't it be $\overline{\left \langle T^*v, T^*v \right \rangle}?$ $\endgroup$ – paxtibimarce Jul 10 at 23:03
  • $\begingroup$ @paxtibimarce No. I think you may want to check the properties again. $\endgroup$ – Anurag A Jul 10 at 23:04
  • $\begingroup$ I wonder why my lecturer thinks this is trivial lol $\endgroup$ – paxtibimarce Jul 10 at 23:11
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The original claim is much easier than the route you're taking. First, $\ker T\subset \ker (T^*T) = \ker (TT^*)$.

Next, if $x\in\ker(T^*T)$, then $0=\langle x,T^*Tx\rangle =\langle Tx,Tx\rangle$, so $Tx=0$. Thus, $\ker(TT^*)\subset\ker T$.

We conclude that $\ker T = \ker(T^*T)$.

(If this is what your lecturer was referring to as "trivial," perhaps (s)he's right?)

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  • $\begingroup$ Yes I solved it this way, but there are more following questions which are made easier using what my lecturer used so this is why I asked it. $\endgroup$ – paxtibimarce Jul 11 at 10:09

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