theorem for normal triangular matrices

Theorem 1

([1], pp. 82) A square matrix is diagonal if 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 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 $A{A}^{\ast }$, 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 the second row is $a_{22}$. Repeating this 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. $\mathrm{\square }$

Title	theorem for normal triangular matrices
Canonical name	TheoremForNormalTriangularMatrices
Date of creation	2013-03-22 13:43:35
Last modified on	2013-03-22 13:43:35
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	12
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 15A57
Classification	msc 15-00
Related topic	NormalMatrix