Chapter 6

Normal Matrices

Normal matrices are matrices that include Hermitian matrices and enjoy several of the same properties as Hermitian matrices. Indeed, while we proved that Hermitian matrices are unitarily diagonalizable, we did not establish any converse. That is, if a matrix is unitarily diagonalizable, then does it have any special property involving for example its spectrum or its adjoint? As we shall see normal matrices are unitarily diagonalizable.

6.1 Introduction to Normal matrices

Definition 6.1.1. AmatrixA Mn is called normal if A A = AA . ∈ ∗ ∗

Proposition 6.1.1. A Mn is normal if and only if every matrix unitarily equivalent to A is normal.∈

Proof. Suppose A is normal and B = U ∗AU,whereU is unitary. Then B∗B = U ∗A∗AU = U ∗AA∗U = U ∗AUU ∗A∗U = BB∗.IfU ∗AU is normal then it is easy to see that U ∗AA∗U = U ∗A∗AU. Multiply this equation on the right by U ∗ andontheleftbyU to obtain AA∗ = A∗A.

Examples.

(1) Unitary matrices are normal (U ∗U = I = UU∗).

2 (2) Hermitian matrices are normal (AA∗ = A = A∗A).

2 (3) If A∗ = A,wehaveA∗A = AA∗ = A . Hence matrices for which A = A−,calledskew-Hermitian, are− normal. ∗ − 197 198 CHAPTER 6. NORMAL MATRICES

Example 6.1.1. Consider the arbitrary matrix N M2 (R) , written as ab ∈ N = . If we suppose that N is normal then cd · ¸ T ab ab a2 + c2 ab + cd N N = = ∗ cd cd ab + cd b2 + d2 · ¸ · ¸ · ¸ ab abT a2 + b2 ac + bd NN = = ∗ cd cd ac + bd c2 + d2 · ¸· ¸ · ¸ From this we conclude that b2 = c2,orb = c. Consider the cases in turn. (i) If c = b,thenN is Hermitian and thus± normal. (ii) If c = b =0, then (N N) = ab + cd = b (a d). On the other − 6 ∗ 12 − hand (NN∗)12 = ac + bd =(d a) b.Forb (a d)=(d a) b,wemust have a = d. This gives that real− 2 2 normal matrices− are either− symmetric or have the form × ab N = ba · − ¸ Note this form includes both rotations and skew-symmetric matrices.

Recall the definition of a unitarily diagonalizable matrix: A matrix A Mn is called unitarily diagonalizable if there is a unitary matrix U for which∈ U ∗AU is diagonal. A simple consequence of this is that if U ∗AU = D (where D = diagonal and U = unitary), then

AU = UD and hence A has n orthonormal eigenvectors. This is just a part of the spectral theorem for normal matrices.

Theorem 6.1.1 (Spectral theorem for normal matrices). If A Mn ∈ has eigenvalues λ1 ...λn, counted according to multiplicity, the following statements are equivalent.

(a) A is normal.

(b) A is unitarily diagonalizable.

n n 2 n 2 (c) aij = λj . i=1 j=1 | | j=1 | | (d) ThereP P is an orthonormalP set of n eigenvectors of A. 6.1. INTRODUCTION TO NORMAL MATRICES 199

Proof. (a) (b). If A is normal, then AA is Hermitian and therefore ⇒ ∗ unitarily diagonalizable. Thus U ∗A∗AU = D = U ∗AA∗U.Also,A, A∗, A∗A = AA∗ form a commuting family. This implies that eigenvectors of A∗A are also eigenvectors of A.SinceA∗A has a complete orthonormal set we know that U ∗AU is also diagonal. It is easy to see also that (b) (a) We also note that (a) (d). ⇒ ⇒ (b) (c). Suppose U AU = D.ThenU A U = D and U A AU = D D. ⇒ ∗ ∗ ∗ ∗ ∗ ∗ ∗ By Corollary 3.5.2 similarly preserves the trace. We know trace of A∗A n n n n n n 2 is tr (A∗A)= j=1 k=1 ajk∗ akj = j=1 k=1 a¯kjakj = j=1 k=1 akj . 2 | | Since the trace of D D is Σ λj , the result follows. P P∗ | | P P P P (c) (b). We know that A is unitarily equivalent to a upper triangular ⇒ 2 matrix T . We also know that if A B are unitarily equivalent aij = ∼ ij | | 2 bij . Application of this equality to the upper triangular matrixP T yields ij | | P 2 2 2 2 aij = λj + tij = λj . | | | | | | | | i,j j>i X X X X Thus tij =0forj>i.ThusA is uniformly diagonalizable. (d) (b). Trivial. ⇒ Corollary 6.1.1. Let A Mn and A is normal. If U is unitary and if ∈ U ∗AU is upper triangular then U ∗AU is diagonal.

Theorem 6.1.2. Let N Mn(R).ThenN is normal if and only if there ∈ is a real orthogonal matrix Q Mn(R) such that ∈

A1 A T 2 Q NQ =  . ° (1) ..    An  °    where Ai is 1 1 (real) or Ai is 2 2 (real) of the form × ×

αi βj Ai = . βj αi ·− ¸ Proof. First of all, any matrix A of the form given by (1) is normal, and therefore so also is any matrix unitarily similar (real orthogonally similar in this case) to it. 200 CHAPTER 6. NORMAL MATRICES

To prove the converse we assume that N Mn(R)isnormal.Weknow that N is unitarily diagonalizable. That is, there∈ is a unitary matrix U such that U ∗NU = D, the diagonal matrix of its eigenvalues. Because N is real, all complex eigenvalues occur in complex conjugate pairs. Arrange them as successive diagonal entries in D.Ifλ is a real eigenvalue, we can assume without loss of generality that the corresponding eigenvector is real. For complex eigenvalues, the corresponding eigenvectors also occur in conjugate pairs. Thus if α + iβ is an eigenvector of N with corresponding eigenvector written in real and complex parts u = ur + ius.SinceN is real we have that α iβ is also an eigenvector of N with corresponding eigenvectoru ¯ = − ur ius.BythefactthatN is unitarily diagonalible, these vectors are − orthogonal. This means ur,us =0. h i Replace the eigenvectors ur ius by the real an imaginary parts in U . This gives the matrix Q. Now compute± QT NQ. It is easy to see that com- pute Nur = αur βvs and Nus = αus+βvr. When the first of these vectors − T T (αur βvs) is multiplied by Q we obtain the vector [0,..., α, β, 0,...0] . Multiplication− by the second gives the vector [0,...,β, α, 0,...− 0]T .Inthis αβ way the components arise. βα · − ¸ Corollary 6.1.2. (a) A Mn is symmetric if and only if (1) holds with all blocks 1 1 (and real). ∈ (b) AAT×= I if and only if (1) has the form

λ1 . ..  λp  ∗ A1  .   ..   ∗   Ak    cos θj sin θj where λj = 1 and Aj = − θj R. ± sin θj cos θj ∈ h i 6.2 Exercises

1. If A and B commute and if A is normal, then A∗ and B commute.