📂Matrix Algebra

Definition of a Normal Matrix

Definition

A square matrix $A \in \mathbb{C}^{n \times n}$ is called a normal matrix ▷normal matrix◁ if it satisfies the following condition: $$ A A^{\ast} = A^{\ast} A $$ Here, $X^{\ast}$ is the conjugate transpose of the matrix $X$.

Properties

Assume $A$ is a square matrix. The necessary and sufficient condition for the triangular matrix $A$ to be a normal matrix is that $A$ is a diagonal matrix: $$ A A^{\ast} = A^{\ast} A \iff \left( A \right)_{ij} = 0 , \forall i \ne j $$

Proof ¹

• $O_{n}$ is the zero matrix with $\left( n \times n \right)$ size. If subscripts are omitted, the size matches that of the matrix inside.
• $\bar{z}$ is the complex conjugate of the complex number $z$.

$(\implies)$

We prove this using mathematical induction. Without loss of generality, let’s consider only the case where $A \in \mathbb{C}^{n \times n}$ is an upper triangular matrix.

If $n = 1$, it trivially holds, and if $n = 2$, then for the upper triangular matrix $$ A = \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} $$ we have $$ \begin{align*} & O_{2} \\ =& A A^{\ast} - A^{\ast} A \\ =& \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \begin{bmatrix} \bar{a} & 0 \\ \bar{b} & \bar{c} \end{bmatrix} - \begin{bmatrix} \bar{a} & 0 \\ \bar{b} & \bar{c} \end{bmatrix} \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \\ =& \begin{bmatrix} |a|^{2} + |b|^{2} & bc \\ c \bar{b} & |c|^{2} \end{bmatrix} - \begin{bmatrix} |a|^{2} & \bar{a} b \\ a \bar{b} & |b|^{2} + |c|^{2} \end{bmatrix} \\ =& \begin{bmatrix} |b|^{2} & b \left( c - \bar{a} \right) \\ \bar{b} \left( c - a \right) & - |b|^{2} \end{bmatrix} \end{align*} $$ which implies $b = 0$. In other words, $A$ is a diagonal matrix. Now assume that $$ A A^{\ast} = A^{\ast} A \implies \left( A \right)_{ij} = 0 , \forall i \ne j $$ holds for $A \in \mathbb{C}^{(n-1) \times (n-1)}$. Using two matrices $B \in \mathbb{R}^{1 \times (n-1)}$ and $C \in \mathbb{C}^{(n-1) \times (n-1)}$, expressed in terms of block matrices, $$ A = \begin{bmatrix} a & B \\ O & C \end{bmatrix} $$ we have $$ \begin{align*} & O_{n} \\ =& A A^{\ast} - A^{\ast} A \\ =& \begin{bmatrix} a & B \\ O & C \end{bmatrix} \begin{bmatrix} \bar{a} & O \\ B^{\ast} & C^{\ast} \end{bmatrix} - \begin{bmatrix} \bar{a} & O \\ B^{\ast} & C^{\ast} \end{bmatrix} \begin{bmatrix} a & B \\ O & C \end{bmatrix} \\ =& \begin{bmatrix} |a|^{2} + B B^{\ast} & B C \\ C B^{\ast} & C C^{\ast} \end{bmatrix} - \begin{bmatrix} |a|^{2} & \bar{a} B \\ a B^{\ast} & B^{\ast} B + C^{\ast} C \end{bmatrix} \\ =& \begin{bmatrix} B B^{\ast} & B C - \bar{a} B \\ C B^{\ast} - a B^{\ast} & C C^{\ast} - C^{\ast} C - B^{\ast} B \end{bmatrix} \end{align*} $$ and the condition $B B^{\ast} = \sum_{k=1}^{n-1} \left( B_{1k} \right)^{2} \in \mathbb{C}^{1 \times 1}$ that $0$ implies that all components of $B$ equal $0$. Meanwhile, $C C^{\ast} - C^{\ast} C = O_{n-1}$ must also hold, making $C$ a normal matrix, and hence $C$ must also be a diagonal matrix. Consequently, the matrix $A$ defined as follows is a diagonal matrix. $$ A = \begin{bmatrix} a & B \\ O & C \end{bmatrix} = \begin{bmatrix} a & O^{\ast} \\ O & C \end{bmatrix} $$

$(\impliedby)$

If $A$ is a diagonal matrix, then the following trivially holds for $k = 1 , \cdots , n$: $$ \left( A A^{\ast} \right)_{kk} = \left( A \right)_{kk}^{2} = \left( A^{\ast} A \right)_{kk} $$

■