# $v_i$ is an eigenvector for $T$ with eigenvalue $\lambda _i$ then it's eigenvector for $T^*$ with eigenvalue $\bar{\lambda}_i$ given normal $T$

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.

• 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 $TT^* = T^*T$. – Limsup Jul 10 at 22:32
• @Infima Yes I have tried, I still don't get it – paxtibimarce Jul 10 at 22:41
• 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. – Limsup Jul 10 at 22:44
• I know it dude, I still don't see why it is true... I know all the basic definitions – paxtibimarce Jul 10 at 22:46
• I am going to write it as an answer, as in here it does not fit well. – Limsup Jul 10 at 22:46

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 $$\|(T-\lambda_i I)^{*}v_i\|=0 \implies (T-\lambda_i I)^{*}v_i=0 \implies (T^{*}-\bar{\lambda_i}I)v_i=0$$

• why ⟨v,TT∗v⟩=⟨T∗v,T∗v⟩? – paxtibimarce Jul 10 at 22:55
• @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 $T^{*}$. – Anurag A Jul 10 at 22:58
• shouldn't it be $\overline{\left \langle T^*v, T^*v \right \rangle}?$ – paxtibimarce Jul 10 at 23:03
• @paxtibimarce No. I think you may want to check the properties again. – Anurag A Jul 10 at 23:04
• I wonder why my lecturer thinks this is trivial lol – paxtibimarce Jul 10 at 23:11

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?)

• 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. – paxtibimarce Jul 11 at 10:09