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Hometheorem for normal triangular matrices

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# 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 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. $\Box$

# References

- 1
V.V. Prasolov,
*Problems and Theorems in Linear Algebra*, American Mathematical Society, 1994.

## Mathematics Subject Classification

15A57*no label found*15-00

*no label found*

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## Corrections

Proof contained by mathwizard ✓

reword by Mathprof ✓

Use theorem environment by alozano ✓