Upper Triangular Matrices are Normal If and Only if They are Diagonal

Theorem

• With $V$ as a finite dimensional inner product space
• With $T\in \mathcal{L}(V)$
• With $\beta$ as an Orthogonal Basis for $V$
• Suppose that matrix $A=[T{]}_{\beta }$ is Upper Triangular
• Then, $T$ is normal $⟺$ $T$ is a Diagonal Matrix

Proof

1. Since $\beta$ is an Orthonormal Basis, $A^{\ast }=[T^{\ast }{]}_{\beta }$
2. If $A$ is diagonal, then $A{A}^{\ast }=A^{\ast }A$, then this implies: $[T{T}^{\ast }{]}_{\beta }=[T^{\ast }T{]}_{\beta }$ which implies $T{T}^{\ast }=T^{\ast }T$
3. Conversely, suppose $T$ is Normal, and $[T{]}_{\beta }$ is Upper Triangular
4. Then, if $\beta =\{{\alpha }_1,\dots ,{\alpha }_n\}$
   1. $T_{{\alpha }_1}=a_{11}{\alpha }_1+0{\alpha }_2+\cdots +0{\alpha }_n$
   2. By our previous theorem, we can say that $T_{{\alpha }_1}^{\ast }={\overline{a}}_{11}{\alpha }_1+0{\alpha }_2+\cdots +0{\alpha }_n$ as $T_{{\alpha }_1}$ is an Eigenvector with eigenvalue.
   3. This implies that $a_{1i}=0,\forall i\ge 2$
   4. We continue with Induction, to show that ${\alpha }_{jj+1}=0,\forall i\in \{i,\dots ,n-j\}$
5. Thus, $A$ is Diagonal