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
${({A}^{\ast}A)}_{ii}$ | $=$ | $\sum _{k=1}^{i}}{|{a}_{ki}|}^{2},$ | ||
${(A{A}^{\ast})}_{ii}$ | $=$ | $\sum _{k=i}^{n}}{|{a}_{ik}|}^{2}.$ |
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. $\mathrm{\square}$
References
- 1 V.V. Prasolov, Problems and Theorems in Linear Algebra, American Mathematical Society, 1994.
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 |