= Solution
For a <normal operator>, (a) gives $\|Tx\|=\|T^*x\|$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^*}$.
For real $\lambda$, the <adjoint operator>[adjoint] of $T-\lambda I$ is $T^*-\lambda I$. Expanding both products shows that their difference is $T^*T-TT^*=0$. Thus $T-\lambda I$ is normal, and applying the <kernel of a linear map> conclusion to it gives
$$
\boxed{\ker(T-\lambda I)=\ker(T^*-\lambda I)\quad(\lambda\in\mathbb R).}
$$
This proves equality of each real <eigenspace>, hence the same <eigenvalues> and <eigenvectors> on the stated real <vector space>. If neither has real <eigenvalues>, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline\lambda$ for the <adjoint operator>[adjoint]; no claim of the same complex <eigenvalues> is needed here.
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