Eigenvalues and Eigenvectors of Symmetric Matrices

On the basis of (I.60) it is easy to show that, in the case of a symmetric matrix A, в = 0 and b = 0:

Theorem I.34: The eigenvalues of a symmetric n x n matrix A are all real valued, and the corresponding eigenvectors are contained in Кп.

Proof: First, note that (I.60) implies that, for arbitrary Ц є К,

/ bT / A — a In в In (a

0 £a) —в In A — a iJ bJ

= Ц aT Ab + bT Aa — abTa — ЦaaTb + в bTb — ^a^a.

Next observe that bTa = aTb and by symmetry, bTAa = (bTAa)T = aTATb = aT Ab, where the first equality follows because bT Aa is a scalar (or1 x 1 matrix). Then we have for arbitrary Ц є К,

(Ц + 1)aT Ab — a(t; + 1)aTb + в (bTb — Ц aTa) = 0. (I.61)

If we choose Ц = —1 in (I.61), then в(bTb + aTa) = в ■ x ||2 = 0; conse­quently, в = 0 and thus X = a є К. It is now easy to see that b no longer matters, and we may therefore choose b = 0. Q. E.D.

There is more to say about the eigenvectors of symmetric matrices, namely,

14 Recall (see Appendix III) that the length (or norm) of a complex number x = a + i ■ b, a, b є К, is defined as |x| = ^(a + i ■ b) ■ (a — i ■ b) = Va2 + b2. Similarly, in the vector case x = a + i ■ b, a, b є Кп, the length of x is defined as ||x|| = у/(a + i ■ b)T(a — i ■ b) = Va1 a + bTb.

Theorem I.35: The eigenvectors of a symmetric n x n matrix A can be chosen orthonormal.

Proof: First assume that all the eigenvalues k1,k2,…,k„ of A are different. Let x1, x2,…,xn be the corresponding eigenvectors. Then for i = j, x] Axj =

кjxjxj andxjAxi = kixjxj; hence, (k – kj)xjxj = 0 because, by symmetry,

xT Axj = (xT Axj )T = xjATxi = xjAxi.

Because ki = к j, it follows now that xjxj = 0. Upon normalizing the eigen­vectors as qj = \xj ||-1 xj, we obtain the result.

The case in which two or more eigenvalues are equal requires a com­pletely different proof. First, normalize the eigenvectors as qj = \xj ||-1 xj. Using the approach in Section I.10, we can always construct vectors y2,…, yn є К” such that q1, y2,…, yn is an orthonormal basis of Rn. Then Q1 = (q1, y2,…, yn) is an orthogonal matrix. The first column of QTAQ1 is QTAq1 = к QTq1. But by the orthogonality of Q1, qTQ1 = qT(q1, y2,…, yn) = (q1Tq1, qTy2,…, qTyn) = (1, 0, 0,…, 0); hence, the first col­umn of QTAQ1 is (k1, 0, 0,…, 0)T and, by symmetry of QTAQ1, the first row is (k1,0, 0,…, 0). Thus, QTAQ1 takes the form

Подпись: QT AQ1k1 0M

0 An-1 /

Next, observe that

det (QTAQ1 – kIn) = det (QTAQ1 – kQTQ1)

= det (Qt)A – kIn)Q1]

= det (QT) det(A – k In) det( Q1) = det(A – kIn),

Подпись: k2 0T . 0 An-2
Подпись: Q2T An-1 Q 2

and thus the eigenvalues of QTAQ1 are the same as the eigenvalues of A; consequently, the eigenvalues of An-1 are k2,…,kn. Applying the preceding argument to An-1, we obtain an orthogonal (n – 1) x (n – 1) matrix Q| such that

Hence, letting

a=(0 Ql

which is an orthogonal n x n matrix, we can write

Подпись: Ql QT AQ1Q2 ="Л2 O

O An-2,
where Л2 is a diagonal matrix with diagonal elements к and к2. Repeating this procedure n — 3 more times yields

QT… QTQTAQ1Q2… Qn = Л,

where Л2 is the diagonal matrix with diagonal elements kt, k2,…,kn.

Note that Q = Q t Q2 … Qn, is an orthogonal matrix itself, and it is now easy to verify that the columns of Q are the eigenvectors of A. Q. E.D.

In view of this proof, we can now restate Theorem I.35 as follows:

Theorem I.36: A symmetric matrix A can be written as A = QЛ QT, where Л is a diagonal matrix with the eigenvalues of A on the diagonal and Q is the orthogonal matrix with the corresponding eigenvectors as columns.

This theorem yields several useful corollaries. The first one is trivial:

Theorem I.37: The determinant of a symmetric matrix is the product of its eigenvalues.

The next corollary concerns idempotent matrices (see Definition I.12):

Theorem I.38: The eigenvalues of a symmetric idempotent matrix are either 0 or 1. Consequently, the only nonsingular symmetric idempotent matrix is the unit matrix I.

Proof: Let the matrix A in Theorem I.36 be idempotent: A ■ A = A. Then, A = QЛ QT = A ■ A = QЛ QTQЛ QT = QЛ2 QT; hence, Л = Л2. Because Л is diagonal, each diagonal element к j satisfies к j = к2; hence, к j (1 — к j) = 0. Moreover, if A is nonsingular and idempotent, then none of the eigenvalues can be zero; hence, they are all equal to 1: Л = I. Then A = QIQT = A = QQT = I. Q. E.D.

Leave a reply

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>