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