If $T$ is normal, then $S=T-I$ is normal and $\sigma(S)=\{0\}$. The <C-star identity> and normality imply $\lVert S^n\rVert=\lVert S\rVert^n$. The <spectral radius formula> consequently gives
$$
\lVert S\rVert=r(S)=0,
$$
so $T=I$.

Normality is essential. On $\ell^2$, let $N$ send $e_2$ to $e_1$ and every other basis vector to zero. Then $N\ne0$ but $N^2=0$, so
$$
\sigma(I+N)=1+\sigma(N)=\{1\},
$$
while $I+N\ne I$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 4 / b / Solution

{c}
{title2=$\lVert a^*a\rVert=\lVert a\rVert^2$}

The C-star identity is the norm axiom $\lVert a^*a\rVert=\lVert a\rVert^2$. It implies that the involution is isometric and that a normal element has norm equal to its spectral radius.


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 Incoming links: C-star identity