theorem for normal triangular matrices

Theorem 1

([1], pp. 82)A square matrix is diagonalif and only if it is normal and triangular.

Proof. If $A$ is a diagonal matrix, then the complex conjugate$A^{\ast }$ is also a diagonal matrix. Since arbitrary diagonal matricescommute, it follows that $A^{\ast }A=A{A}^{\ast }$.Thusany 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$A{A}^{\ast }$, we obtain

${(A^{\ast }A)}_{ii}$$=$${\displaystyle \sum _{k=1}^i}{|a_{ki}|}^2,$

${(A{A}^{\ast })}_{ii}$$=$${\displaystyle \sum _{k=i}^n}{|a_{ik}|}^2.$

For $i=1$, we have

$${|a_{11}|}^2={|a_{11}|}^2+{|a_{12}|}^2+\mathrm{\cdots }+{|a_{1n}|}^2.$$

It follows that the only non-zero entry on the first row of $A$ is $a_{11}$.Similarly, for $i=2$, we obtain

$${|a_{12}|}^2+{|a_{22}|}^2={|a_{22}|}^2+\mathrm{\cdots }+{|a_{2n}|}^2.$$

Since $a_{12}=0$, it follows that the only non-zero element on thesecond row is $a_{22}$. Repeating this for all rows,we see that $A$ is a diagonal matrix. Thus any normalupper 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 normalupper triangular matrix. Thus, by the above, $A^{\ast }$ is a diagonal matrix,whence also $A$ is a diagonal matrix. $\mathrm{\square }$