# 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 $A^{\ast}A=AA^{\ast}$. 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

 $\displaystyle(A^{\ast}A)_{ii}$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{i}|a_{ki}|^{2},$ $\displaystyle(AA^{\ast})_{ii}$ $\displaystyle=$ $\displaystyle\sum_{k=i}^{n}|a_{ik}|^{2}.$

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 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. $\Box$

## References

Title theorem for normal triangular matrices TheoremForNormalTriangularMatrices 2013-03-22 13:43:35 2013-03-22 13:43:35 Mathprof (13753) Mathprof (13753) 12 Mathprof (13753) Theorem msc 15A57 msc 15-00 NormalMatrix