theorem for normal triangular matricesTheoremForNormalTriangularMatrices2003-06-29 04:06:162006-10-02 18:42:47Theoremtriangular matrix% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thm}{Theorem}
\begin{thm}
(\cite{prasolov}, pp. 82)
A square matrix is diagonal
if and only if it is normal and triangular.
\end{thm}
\emph{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 \PMlinkescapetext{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. $\Box$
\begin{thebibliography}{9}
\bibitem{prasolov} V.V. Prasolov,
\emph{Problems and Theorems in Linear Algebra},
American Mathematical Society, 1994.
\end{thebibliography}