Población Comunidad 17

Let A be a complex M byN matrix. It is often helpful to know how large the two-normkAxk2 can be, relative tokxk2; that is, we want to find

a constantaso that

kAxk2/kxk2≤a,

for allx6= 0. We can reformulate the problem by asking how largekAuk2 2

can be, subject tokuk2 = 1. Using Lagrange multipliers, we discover that

a unit vectoruthat maximizeskAuk2

2 has the property that

A†Au=λu,

for some constantλ. This leads to the more general problem discussed in this section.

Definition 5.2 Given anN byNcomplex matrixS, we say that a complex numberλis aneigenvalueofSif there is a nonzero vectoruwithSu=λu. The column vector u is then called an eigenvector of S associated with eigenvalueλ.

Clearly, ifuis an eigenvector ofS, then so iscu, for any constantc6= 0; therefore, it is common to choose eigenvectors to have norm equal to one. Ifλis an eigenvalue ofS, then the matrixS−λIfails to have an inverse, since (S−λI)u= 0 butu6= 0, and so its determinant must be zero. If we treatλas a variable and compute thecharacteristic polynomialofS,

P(λ) = det(S−λI),

we obtain a polynomial of degreeN in λ. Its rootsλ1, ..., λN are then the

eigenvalues ofS. If ||u||2

2=u†u= 1 thenu†Su=λu†u=λ. Note that the

eigenvalues need not be real, even ifS is a real matrix.

Ex. 5.4 Prove that the eigenvalues of an upper triangular matrix T are the entries of its main diagonal, so that the trace of T is the sum of its eigenvalues.

Ex. 5.5 Prove that, if S is square,U is unitary, andU†_{SU =T is upper}

triangular, then the eigenvalues of S and T are the same and S and T

have the same trace. Hint: use the facts thatdet(AB) = det(A) det(B)and Equation (3.22).

Ex. 5.6 Use the two previous exercises to prove that, for any square matrix

We know that a square matrix S is invertible if and only if Sx = 0 implies thatx= 0. We can say this another way now:S is invertible if and only ifλ= 0 is not an eigenvalue ofS.

Ex. 5.7 Compute the eigenvalues for the real square matrix

S= 1 2 −2 1 . (5.2)

Note that the eigenvalues are complex, even though the entries of S are real.

The eigenvalues of the Hermitian matrix

H = 1 2 +i 2−i 1 (5.3) are λ = 1 +√5 and λ = 1−√5, with corresponding eigenvectors u = (√5,2−i)T _{and v = (}√_{5, i−2)}T_{, respectively. Then ˜H, defined as in}

Equation (3.31), has the same eigenvalues, but both with multiplicity two. Finally, the associated eigenvectors of ˜B are

u1 u2 , (5.4) and −u2 u1 , (5.5) forλ= 1 +√5, and v1 v2 , (5.6) and −v2 v1 , (5.7) forλ= 1−√5.

Definition 5.3 Thespectral radiusofS, denotedρ(S), is the largest value of|λ|, where λdenotes an eigenvalue ofS.

Ex. 5.8 Use the facts that λ is an eigenvalue ofS if and only if det(S−

λI) = 0, and det(AB) =det(A)det(B) to show thatλ2 _{is an eigenvalue of}

S2 _{if and only if eitherλor−λis an eigenvalue ofS. Then use this result}

5.4.1

The Hermitian Case

LetHbe anNbyNHermitian matrix. As we just saw, there is a unitary matrixU such thatU†HU =D is real and diagonal. ThenHU =U D, so that the columns ofU are eigenvectors of H with two-norms equal to one, and the diagonal entries ofDare the eigenvalues ofH. SinceU is invertible, its columns form a set ofN mutually orthogonal norm-one eigenvectors of the Hermitian matrix H; call them {u1_{, ..., u}N_{}. We denote by λ}

n, n =

1,2, ..., N, theN eigenvalues, so thatHun_=λ

nun. This is the well known eigenvalue-eigenvector decomposition of the matrix H. Not every square matrix has such a decomposition, which is why we focus on Hermitian

H. The singular-value decomposition, which we discuss shortly, provides a similar decomposition for an arbitrary, possibly non-square, matrix.

The matrix H can also be written as

H=

N

X

n=1

λnun(un)†,

a linear superposition of thedyadmatricesun(un)†. The Hermitian matrix

H is invertible if and only if none of theλare zero and its inverse is

H−1= N X n=1 λ−_n1un(un)†. We also haveH−1_{=U L}−1_U†_.

Ex. 5.9 Show that if z = (z1, ..., zN)T is a column vector with complex entries andH =H† is anN byN Hermitian matrix with complex entries then the quadratic form z†Hz is a real number. Show that the quadratic formz†Hzcan be calculated using only real numbers. Let z=x+iy, with

xandyreal vectors and letH =A+iB, whereAandB are real matrices. Then show thatAT _=A,BT _=−B,xT_{Bx= 0 and finally,}

z†Hz= xT yT A −B B A x y .

Use the fact thatz†Hzis real for every vectorzto conclude that the eigen- values ofH are real.

Ex. 5.10 Show that the eigenvalues of a Hermitian matrix H are real by computing the conjugate transpose of the1 by1 matrixz†Hz.

Definition 5.4 A Hermitian matrixQ is said to be nonnegative-definite

if all the eigenvalues of Qare nonnegative, and positive-definite if all the eigenvalues are positive.

Proposition 5.2 A Hermitian matrix Q is a nonnegative-definite matrix if and only if there is another matrixC, not necessarily square, such that

Q=C†C.

Proof: Assume that Qis nonnegative-definite and letQ=U LU† _{be the} eigenvalue/eigenvector decomposition ofQ. Since the eigenvalues ofQare nonnegative, each diagonal entry of the matrixLhas a nonnegative square root; the matrix with these square roots as entries is called√L. Using the fact thatU†U =I, we have

Q=U LU†=U√LU†U√LU†;

we then take C = U√LU†, so C† = C. This choice of C is called the

Hermitian square rootofQ.

Conversely, assume now thatQ=C†C, for some arbitrary, possibly not square, matrix C. Let Qu=λu, for some non-zero eigenvectoru, so that

λis an eigenvalue ofQ. Then λkuk22=λu†u=u†Qu=u†C†Cu=kCuk 2 2, so that λ=kCuk22/kuk 2 2≥0.

If N is a square complex matrix with N = U DU†, where, as above,

U†U = I and D is diagonal, but not necessarily real, then we do have

N†N = N N†; then N is normal, which means that NT_{N =N N}T_{. The}

matrixN will be Hermitian if and only if Dis real. It follows then that a real normal matrixN will be symmetric if and only if its eigenvalues are real, since it is then Hermitian and real.

The normal matrices are precisely those for which such an eigenvector- eigenvalue decomposition holds, as we saw above. In the appendix on Her- mitian and Normal Linear Operators we prove this result again, as a state- ment about operators on a finite-dimensional vector space.

The following exercise gives an example of a matrix N that is real, normal, not symmetric, and has non-real eigenvalues. The matrix NT_N

has repeated eigenvalues. As we shall see in Theorem 5.4, if a real, normal matrix is such that NT_{N does not have repeated eigenvalues, thenN is}

symmetric and so the eigenvalues ofN are real.

Ex. 5.11 Show that the 2 by2 matrix N =

0 1

−1 0

is real, normal, and has eigenvalues±i. Show that the eigenvalues ofNT_{N are both1.}