Show that $ker(Id−T)=ker(Id−T^*)$

Question

On a Hilbert space, assume T is linear bounded and $||T||\leq1$ Show that $ker(Id−T)=ker(Id−T^*)$ The hint is to consider $||I-T^*x||$ for $x\in kerI-T$

So what I did was I took the square of $||I-T^*x||$ and got $<x-Tx-T^*x+TT^*x,x>=<-T^*x,x>+<TT^*x,x>$ using the fact that $Tx=x$. Also we can assumes that $||T||=||T^*||=1$ otherwise $||I-T^*||$ is automatically invertible by a theorem. Now at this point, I'm not sure how to proceed.

Answer 1

Since $\|T^{*}\|=\|T\|$ we get $$\|x-T^{*}x\|^{2}= \langle x-T^{*}x, x-T^{*}x \rangle$$ $$=\|x\|^{2}+\|T^{*}x\|^{2}-\langle x, T^{*}x \rangle-\langle T^{*}x, x \rangle$$ $$\leq \|x\|^{2}+\|T^{*}\|^{2}\|x\|^{2}-\langle Tx, x \rangle-\langle x, Tx \rangle$$ $$=\|T\|^{2}\|x\|^{2}-\|x\|^{2}\leq 0$$ if $Tx=x$.