The following is exercise 7.A.28 from Linear Algebra Done Right, fourth edition by Sheldon Axler:
Suppose $T \in \mathcal{L}(V)$ is normal. Prove that if $\lambda \in \mathbf{F}$, then the minimal polynomial of $T$ is not a polynomial multiple of $(x - \lambda)^2$.
My attempt:
Let the minimal polynomial of $T$ be denoted by $p$ with degree $m$. Suppose, for sake of contradiction, $p$ is a polynomial multiple of $(x - \lambda)^2$. This implies that there exists a monic polynomial $s$ with degree $m - 2$ such that
$$ p(x) = (x - \lambda)^2 s(x) \implies (T - \lambda I)^2 s(T) = 0 \tag{1} $$
for all $x \in \mathbf{F}$. My goal is to somehow show $s(T) = 0$, which would give us the contradiction. I tried to show this by considering the following cases:
Case 1 ($\lambda$ is not an eigenvalue of $T$): In this case, we must have that $T - \lambda I$ is invertible. Left-multiplying the inverse of $T - \lambda I$ twice in equation $(1)$ implies $s(T) = 0$, as was to be shown for this case.
Case 2 ($\lambda$ is an eigenvalue of $T$): Since $p(x) = (x - \lambda)(x - \lambda) s(x)$, this case implies that $T$ is not diagonalizable.
This is where I got stuck. I could break up Case 2 into two subcases where $V$ is a complex or real vector space. If it's complex, then $T$ being non-diagonalizable will contradict exercise 5.D.15 from the text (I paraphrased it):
$$ \text{T is diagonalizable} \ \iff \text{There does not exist} \ \lambda \in \mathbf{C} \ \text{such that} \ p \ \text{is a polynomial multiple of}\ (z - \lambda)^2 $$
but then how do I treat the real case? Also, I haven't even used anywhere the hypothesis that $T$ is normal, so I must be going in the wrong direction. Anyone have some suggestions?
