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Normal matrix

by , PhD

A matrix is normal if and only if either pre-multiplying or post-multiplying it by its conjugate transpose gives the same result. It turns out that a matrix is normal if and only if it is unitarily similar to a diagonal matrix.

In other words, not only normal matrices are diagonalizable, but the change-of-basis matrix used to perform the diagonalization is unitary. Said in layman terms, normal matrices are those that behave nicely as far as diagonalization is concerned.

Table of Contents

Definition

Let us start with a formal definition of normality.

Definition A $K	imes K$ matrix A is said to be normal if and only if[eq1]where $A^{st }$ denotes the conjugate transpose of A.

A simple example follows.

Example Define[eq2]The conjugate transpose of A is[eq3]The product of $A^{st }$ and A is[eq4]The product of A and $A^{st }$ is[eq5]Therefore, [eq6] and A is normal.

Unitary matrices are normal

Several important kinds of matrices are normal.

Remember that a matrix is unitary if its inverse is equal to its conjugate transpose.

Proposition Let A be a $K	imes K$ matrix. If A is unitary, then it is normal.

Proof

By the definition of unitary matrix, we have[eq7]where I is the identity matrix.

Hermitian matrices are normal

Remember that a matrix is Hermitian if and only if it is equal to its conjugate transpose. Since complex conjugation leaves real numbers unaffected, a real matrix is Hermitian when it is symmetric (equal to its transpose).

Proposition Let A be a $K	imes K$ matrix. If A is Hermitian, then it is normal.

Proof

By the definition of Hermitian matrix, we have[eq8]

Skew-Hermitian matrices are normal

A matrix A is said to be skew-Hermitian if and only if[eq9]

Proposition Let A be a $K	imes K$ matrix. If A is skew-Hermitian, then it is normal.

Proof

By the definition of skew-Hermitian matrix, we have[eq10]

Diagonal matrices are normal

Another useful fact follows.

Proposition Let A be a $K	imes K$ matrix. If A is diagonal, then it is normal.

Proof

If A is diagonal, the products $A^{st }A$ and $AA^{st }$ are diagonal. The main diagonals of these products contain entries of the form[eq11]Therefore, [eq6].

Unitary similarity preserves normality

Remember that two matrices A and $B$ are said to be unitarily similar if and only if[eq13]where $P$ is a unitary matrix.

Proposition Let A be a $K	imes K$ matrix. Let $B$ be unitarily similar to A. If A is normal, then $B$ is normal.

Proof

The proof is as follows:[eq14]where: in steps $rame{A}$ and $rame{C}$ we have used the fact that $PP^{st }=I$ since $P$ is unitary; in step $rame{B}$ we have used the fact that A is normal.

A normal triangular matrix is diagonal

The following proposition will be used below to prove the main result about the diagonalization of normal matrices.

Proposition A triangular matrix is normal if and only if it is diagonal.

Proof

We will prove the proposition for upper triangular matrices. The proof is by induction on the dimension of the matrix. A $1	imes 1$ matrix is diagonal by definition and normal because the product of scalars is commutative. Now, suppose that [eq15] upper triangular matrices are normal if and only if they are diagonal. We need to prove that the claim is true for $K	imes K$ matrices. In order to prove the "only if" part, we choose a $K	imes K$ triangular and normal matrix A. We partition A so as to form the following block-matrix:[eq16]where $lpha $ is a scalar, 0 is a [eq17] vector of zeros, $eta $ is a [eq18] vector and $gamma $ is a [eq19] matrix. Then,[eq20]and[eq21]Since A is normal, [eq6], which implies that each block of $A^{st }A$ must be equal to the corresponding block of $AA^{st }$. The equality[eq23]implies[eq24]which, in turn, implies $eta =0$ by the positive definiteness of the norm. Moreover, the equality[eq25]implies[eq26]which means that $gamma $ is normal. Furthermore, $gamma $ is the diagonal block of an upper triangular matrix. As a consequence, it is upper triangular and, by the induction hypothesis, diagonal. Thus, we have[eq27]which is a diagonal matrix because $lpha $ is a scalar and $gamma $ is diagonal. This proves the "only if" part. The "if" one is trivial because we have already proved a proposition that states that diagonal matrices are normal. The proof for lower triangular matrices is analogous.

Unitary diagonalization

We are now ready to prove the most important result.

Remember that a matrix A is said to be diagonalizable if and only if there exists an invertible matrix $P$ such that[eq28]and $D$ is diagonal. In other words, A is similar to a diagonal matrix $D$.

It turns out that the diagonal entries of $D$ are the eigenvalues of A and the columns of $P$ are the eigenvectors of A.

When $P$ is unitary, the diagonalization becomes[eq29]and we say that A is unitarily diagonalizable.

Proposition A $K	imes K$ matrix A is unitarily diagonalizable if and only if it is normal.

Proof

We first prove the "only if" part, starting from the hypothesis that A is unitarily diagonalizable, that is, unitarily similar to a diagonal matrix $D$. We have proved above that 1) unitary similarity preserves normality and 2) diagonal matrices are normal. As a consequence, A must be normal. We can now prove the "if" part, starting from the hypothesis that A is normal. The Schur decomposition theorem says that any square matrix A is unitarily similar to an upper triangular matrix $T$:[eq30]Since A is normal and unitary similarity preserves normality, $T$ must be normal. But an upper triangular matrix can be normal only if it is diagonal. Therefore, $T$ must be diagonal.

Recall that a $K	imes K$ diagonalizable matrix A is not defective, that is, it possesses K linearly independent eigenvectors. In the case of a normal matrix A, the matrix of eigenvectors $P$ is unitary, which means that the columns of $P$ are orthonormal. In other words, a $K	imes K$ normal matrix A possesses a set of K orthonormal eigenvectors. Said differently, there is a basis of orthonormal vectors for the eigenspace of a normal matrix.

Orthogonal diagonalization of symmetric real matrices

When the matrix A being diagonalized is real and symmetric, then both the matrix of eigenvalues $D$ and the change-of-basis matrix $P$ are real.

Proposition Let A be a $K	imes K$ real and symmetric matrix. Then it can be diagonalized as[eq31]where both $D$ and $P$ are real, $D$ is diagonal and $P$ is orthogonal.

Proof

In the lecture on the properties of eigenvalues and eigenvectors, we have shown that all the eigenvalues of a symmetric real matrix A are real. This implies that for any eigenvalue $lambda $, we can find a real eigenvector by searching for a real solution $x
eq 0$ to the equation[eq32]The solution is guaranteed to exist because $A-lambda I$ is rank-deficient by the definition of eigenvalue. The real eigenvectors thus found can be used in the algorithm that leads to the Schur decomposition. In particular, if we look into the proof of the Schur decomposition, we can see that the Householder matrices used there are real if we choose real eigenvectors for A. As a consequence, the Schur decomposition[eq33]can be performed in such a way that the matrices $D$ and $P$ are real. The matrix $P$ is unitary (i.e., $P^{st }P=I$), but since it is also real, we have [eq34]and[eq35]that is, $P$ is orthogonal. Moreover, since A is real and symmetric, it is Hermitian and therefore normal. From the proof of the previous proposition, we know that the matrix $D$ in the Schur decomposition is diagonal when A is normal.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Is the matrix[eq36]unitarily diagonalizable?

Solution

The matrix A is upper triangular. If it was unitarily diagonalizable, it would be normal. But an upper triangular matrix is normal only if it is diagonal. The matrix A is not diagonal, hence it is not normal and it is not unitarily diagonalizable.

Exercise 2

Is the matrix[eq37]unitarily diagonalizable?

Solution

The matrix A is symmetric (it is equal to its transpose). Since it is real, it is also Hermitian. We know that Hermitian matrices are normal. Therefore, A is normal and unitarily diagonalizable.

Exercise 3

Check whether the matrix [eq38]is normal.

Solution

The conjugate transpose of A is[eq39]We need to compute the products[eq40]and[eq41]Thus, [eq42] and A is not normal.

How to cite

Please cite as:

Taboga, Marco (2017). "Normal matrix", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/normal-matrix.

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