Prove that a square upper-triangular matrix that is not diagonal is not normal.
This is a very standard exercise in linear algebra, which serves as a powerful lemma from which the
spectral theorem is derived. We will prove the contrapositive of this statement, that is, that a normal,
square, upper-triangular matrix must be diagonal; this by induction on the dimension.
The base case is trivial – since a matrix with a single element already satisfies almost any predicate – so assume that the property holds for $n$, it shall be proven for $n+1$. Let then $A \in \mathscr M_{n+1}(\CC)$, and consider the following block partition of $A$: $$ A = \left(\begin{array}{c|ccc} a_{11} & a_{12} & \cdots & a_{1,n+1}\\\\ \hline & a_{22} & \cdots & a_{2,n+1}\\\\ & & \ddots & \vdots\\\\ & & & a_{n+1,n+1} \end{array}\right) = \begin{pmatrix} a_{11} & T\\\\ & \tilde A \end{pmatrix}; \tag{$*$} $$ imposing normality on $(*)$ then gives $$ \begin{align*} A \adj{A} = \begin{pmatrix} a_{11} & T\\\\ & \tilde A \end{pmatrix} \begin{pmatrix} \bar a_{11} & \\\\ \adj{T} & \adj{\tilde A} \end{pmatrix} &= \begin{pmatrix} |a_{11}|^2 + T\adj{T} & T \adj{\tilde A}\\\\ \tilde A \adj{T} & \tilde A \adj{\tilde A} \end{pmatrix}\\\\ &= \begin{pmatrix} |a_{11}|^2 & \bar a_{11} T\\\\ a_{11} \adj{T} & \adj{T} T + \adj{\tilde A} \tilde A \end{pmatrix} = \begin{pmatrix} \bar a_{11} & \\\\ \adj{T} & \adj{\tilde A} \end{pmatrix} \begin{pmatrix} a_{11} & T\\\\ & \tilde A \end{pmatrix} = \adj{A} A, \end{align*} $$ and therefore we are dealing with the equations shown below $$ T \adj T = 0, \qquad \tilde A \adj{\tilde A} = \adj{T} T + \adj{\tilde A} \tilde A. $$
Notice that $T$ is a row matrix, and as such $\adj T$ is a vector; by the definition of the standard inner product on $\CC^n$ and its positive-definiteness $$ T \adj T = \langle \adj T, \adj T \rangle = \lVert \adj T \rVert^2 = 0 \implies \adj T = 0 \implies T = 0. $$ Thus, the second may be rewritten into $\tilde A \adj{\tilde A} = \adj{\tilde A} \tilde A$; $\tilde A$ is then normal, and invoking the induction hypothesis it is also diagonal, as it was to be proven.
The base case is trivial – since a matrix with a single element already satisfies almost any predicate – so assume that the property holds for $n$, it shall be proven for $n+1$. Let then $A \in \mathscr M_{n+1}(\CC)$, and consider the following block partition of $A$: $$ A = \left(\begin{array}{c|ccc} a_{11} & a_{12} & \cdots & a_{1,n+1}\\\\ \hline & a_{22} & \cdots & a_{2,n+1}\\\\ & & \ddots & \vdots\\\\ & & & a_{n+1,n+1} \end{array}\right) = \begin{pmatrix} a_{11} & T\\\\ & \tilde A \end{pmatrix}; \tag{$*$} $$ imposing normality on $(*)$ then gives $$ \begin{align*} A \adj{A} = \begin{pmatrix} a_{11} & T\\\\ & \tilde A \end{pmatrix} \begin{pmatrix} \bar a_{11} & \\\\ \adj{T} & \adj{\tilde A} \end{pmatrix} &= \begin{pmatrix} |a_{11}|^2 + T\adj{T} & T \adj{\tilde A}\\\\ \tilde A \adj{T} & \tilde A \adj{\tilde A} \end{pmatrix}\\\\ &= \begin{pmatrix} |a_{11}|^2 & \bar a_{11} T\\\\ a_{11} \adj{T} & \adj{T} T + \adj{\tilde A} \tilde A \end{pmatrix} = \begin{pmatrix} \bar a_{11} & \\\\ \adj{T} & \adj{\tilde A} \end{pmatrix} \begin{pmatrix} a_{11} & T\\\\ & \tilde A \end{pmatrix} = \adj{A} A, \end{align*} $$ and therefore we are dealing with the equations shown below $$ T \adj T = 0, \qquad \tilde A \adj{\tilde A} = \adj{T} T + \adj{\tilde A} \tilde A. $$
Notice that $T$ is a row matrix, and as such $\adj T$ is a vector; by the definition of the standard inner product on $\CC^n$ and its positive-definiteness $$ T \adj T = \langle \adj T, \adj T \rangle = \lVert \adj T \rVert^2 = 0 \implies \adj T = 0 \implies T = 0. $$ Thus, the second may be rewritten into $\tilde A \adj{\tilde A} = \adj{\tilde A} \tilde A$; $\tilde A$ is then normal, and invoking the induction hypothesis it is also diagonal, as it was to be proven.