Proof. If $A$ is a diagonal matrix, then the complex conjugate $A^\ast$ is also a diagonal matrix. Since arbitrary diagonal matrices commute, it follows that $A^\ast A = A A^\ast$ . Thus any diagonal matrix is a normal triangular matrix.

Next, suppose $A=(a_{ij})$ is a normal upper triangular matrix. Thus $a_{ij}=0$ for $i>j$ , so for the diagonal elements in $A^\ast A$ and $AA^\ast$ , we obtain \begin{eqnarray*} (A^\ast A)_{ii} &=& \sum_{k=1}^i |a_{ki}|^2, \\ (AA^\ast)_{ii} &=& \sum_{k=i}^n |a_{ik}|^2. \\ \end{eqnarray*}For $i=1$ , we have $$ |a_{11}|^2 = |a_{11}|^2+|a_{12}|^2+\cdots + |a_{1n}|^2.$$ It follows that the only non-zero entry on the first row of $A$ is $a_{11}$ . Similarly, for $i=2$ , we obtain $$ |a_{12}|^2 + |a_{22}|^2 = |a_{22}|^2+\cdots + |a_{2n}|^2.$$ Since $a_{12}=0$ , it follows that the only non-zero element on the second row is $a_{22}$ . Repeating this argument for all rows, we see that $A$ is a diagonal matrix. Thus any normal upper triangular matrix is a diagonal matrix.

Suppose then that $A$ is a normal lower triangular matrix. Then it is not difficult to see that $A^\ast$ is a normal upper triangular matrix. Thus, by the above, $A^\ast$ is a diagonal matrix, whence also $A$ is a diagonal matrix.

Bibliography

1

V.V. Prasolov, Problems and Theorems in Linear Algebra, American Mathematical Society, 1994.