Linear Algebra Done Right · Problem 7B.9

A normal operator is one where $T$ commutes with its adjoint—a very special structure that forces the operator to be simultaneously diagonalizable by the same unitary basis. The key insight is that normality is equivalent to $T^*$ being expressible as a polynomial in $T$ itself, which happens precisely when $T^*$ lies in the commutative algebra generated by $T$. The forward direction uses the spectral theorem to show that if $T$ is normal, both $T$ and $T^*$ can be written as polynomials in a common spectral measure; the reverse direction shows that if $T^* = p(T)$, then $T$ and $T^*$ automatically commute.

A normal operator is one where $T$ commutes with its adjoint—a very special structure that forces the operator to be simultaneously diagonalizable by the same unitary basis. The key insight is that normality is equivalent to $T^*$ being expressible as a polynomial in $T$ itself, which happens precisely when $T^*$ lies in the commutative algebra generated by $T$. The forward direction uses the spectral theorem to show that if $T$ is normal, both $T$ and $T^*$ can be written as polynomials in a common spectral measure; the reverse direction shows that if $T^* = p(T)$, then $T$ and $T^*$ automatically commute.