Question

Show that a triangular matrix is normal if it is diagonal

Answer 1

Answer:

Given:

A triangular matrix is normal if it is diagonal.

For n=2, Let A= $\begin{bmatrix} a & 0 \\ b & c \end{bmatrix}$ such that

A^∗A=AA^∗= $\begin{bmatrix} \overline a & b \\ b & \overline c \end{bmatrix} ​$$\begin{bmatrix} a & 0 \\ b & c \end{bmatrix}$ -$\begin{bmatrix} a & 0 \\ b & c \end{bmatrix}$$\begin{bmatrix} \overline a & \overline b \\ 0 & \overline c \end{bmatrix} ​$

= $\begin{bmatrix} ||a||^2+||b||^2−||a||^2 & \overline bc−a \overline b \\ \overline cb−b \overline a & |c||^2−(||b||^2+||c||^2) \end{bmatrix}$

= $\begin{bmatrix} b||^2 & \overline bc−a \overline b \\ \overline cb−b \overline a & −||b||^2 \end{bmatrix}$

So b=0 and A is diagonal ,if the result is true.For n≥2, Let A= $\begin{bmatrix} T & 0 \\ v & a \end{bmatrix}$ where T is a n x n triangular matrix,v a 1 x n matrix and a is a complex number . Since A^∗A=AA^∗  we have,

$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$= $\begin{bmatrix} T^* & v* \\ 0 & \overline a \end{bmatrix}$$\begin{bmatrix} T & 0 \\ v & a \end{bmatrix}$-$\begin{bmatrix} T & 0 \\ v & a \end{bmatrix}$ $\begin{bmatrix} T^* & v* \\ 0 & \overline a \end{bmatrix}$

=  $\begin{bmatrix} T^*T + vv*& v*a \\ \overline a v & |a|^2 \end{bmatrix}$-  $\begin{bmatrix} TT^* & Tv* \\ vT* & |v|^2 +|a|^2 \end{bmatrix}$

=  $\begin{bmatrix} T^*T- TT*+vv* & v*a-Tv* \\ \overline a v-vT & -|v|^2 \end{bmatrix}$

Hence v=0 and T is normal .since T is lower triangular,then T is normal if A is diagonal.