Orthogonal Diagonalization Math 422
Problem 1 Given a real n n matrix A; under what conditions does there exist an orthogonal matrix P that diagonalizes A?
Proposition 2 An orthogonally diagonalizable real matrix A has the following properties:
1. A is symmetric. 2. A has an orthonormal set of eigenvectors.
Proof. Since A is orthogonally diagonalizable, there exists an orthogonal matrix P and a diagonal matrix D such that P T AP = D: (1)
T T (1) Solving for A we have A = PDP T : Thus AT = PDP T = P T DT P T = PDP T = A; and A is symmetric. (2) Multiplying both sides of (1) on the left by P gives AP = PD: Let P = [p1 pn] = [pij] : Then j j
AP = A [p1 pn] = [Ap1 Apn]; j j j j on the other hand,
1 0 0 p p p p 11 1n . 11 1 1n n . . 2 0 2 . 3 . . PD = . . = . . = [1p1 npn] : 2 . . 3 . . 2 . . 3 j j 6 . .. 0 7 pn1 pnn 6 7 pn11 pnnn 6 7 6 0 0 n 7 6 7 4 5 6 7 4 5 4 5 Therefore Api = ipi for all i; and each pi is an eigenvector for A: Furthermore, the pi’s are orthonormal by assumption. These properties are also su¢ cient.
Proposition 3 A matrix with an orthonormal set of eigenvectors is orthogonally diagonalizable.
Proof. Suppose that A has an orthonormal set of eigenvectors p1;:::; pn . Then Api = ipi for each i; f g and the matrix P = [p1 pn] diagonalizes A: Since p1;:::; pn is an orthonormal set, we have j j f g 1 0 0 pT p p p p 1 1 1 1 n . T . . . 2 0 1 . 3 P P = . [p1 pn] = . . = = I: 2 . 3 j j 2 . . 3 . . T 6 . .. 0 7 pn pn p1 pn pn 6 7 6 7 6 7 6 0 0 1 7 4 5 4 5 6 7 4 5 Therefore P is orthogonal matrix. The proof that a real symmetric matrix is orthogonally diagonalizable requires two preliminary results.
Proposition 4 A real symmetric matrix has real eigenvalues.
Proof. Let A be an n n real symmetric matrix and suppose that Ax = x with x = 0 and possibly complex. To isolate ; we …rst note that 6
xT (Ax) = xT (x) = xT x:
On the other hand, xT (Ax) = (Ax)T x
1 (since uT v = vT u for complex vectors u and v). Therefore xT x = (Ax)T x = xT AT x = xT Ax; where the last equality follows from the symmetry of A: Now A = A since A is real, hence xT x = xT Ax = xT Ax = xT x = xT x = xT x: Since xT x = xT x = 0 we have 6 0 = xT x xT x = xT x: Therefore = so that R: 2 Theorem 5 If a real matrix A has real eigenvalues, there is an orthogonal matrix Q and an upper triangular matrix T such that QT AQ = T: Proof. We …rst prove this result for 2 2 real matrices A. Choose an eigenvalue and an associated unit eigenvector u; then Au = u: Since solutions of the homogeneous system (A I) x = 0 are vectors in R2; the particular solution u R2: Choose a unit vector v R2 orthogonal to u and let Q = [u v] : Then 2 2 j uT uT uT Au uT Av uT u uT Av QT AQ = A [u v] = [Au Av] = = vT j vT j vT Au vT Av vT u vT Av uT Av = 0 vT Av is the desired upper triangular matrix T . Now continue by induction on matrix dimension n: Assume the conclusion holds for all n n matrices with real eigenvalues. Let A be an (n + 1) (n + 1) real matrix with real eigenvalues. Choose an eigenvalue and an associated unit eigenvector u1; then Au1 = u1: Choose a n+1 basis for R containing u1 and apply Gram-Schmidt to obtain an orthonormal basis u1; u2;:::; un+1 : f g Let Q = u1 u ; then j j n+1 T T T T u u u Au1 u Aun+1 1 1 1 1 T . . . Q AQ = 2 . 3 A u1 un+1 = 2 . 3 [Au1 Aun+1] = 2 . 3 T j j T j j T T u u u Au1 u Aun+1 6 n+1 7 6 n+1 7 6 n+1 n+1 7 4 5 4 5 T 4 T 5 u Au2 u Aun+1 1 1 T T T u1 u1 u1 Au2 u1 Aun+1 2 3 . . = . . = T T 2 . . 3 6 0 u2 Au2 u2 Aun+1 7 T T T 6 . . . 7 un+1u1 un+1Au2 un+1Aun+1 6 . . . 7 6 7 6 . . . 7 4 5 6 T T 7 6 0 u Au2 u Aun+1 7 6 n+1 n+1 7 4 5 = : 0 B Now suppose for the moment that the n n matrix B has real eigenvalues. Then by assumption, there is an orthogonal matrix S and an upper triangular matrix T such that ST BS = T: Consider the (n + 1) (n + 1) block matrix 1 01 n R = 0n 1 Sn n and note that R is orthogonal since 1 0 1 0 1 0 1 0 RT R = = = = I: 0 ST 0 S 0 ST S 0 I Furthermore, since the product U = QR is orthogonal we have 1 0 1 0 1 0 U T AU = RT QT AQR = = 0 ST 0 B 0 S 0 ST 0 BS = = ; 0 STBS 0 T
2 which is upper triangular. So to complete the proof, we must show that the eigenvalues of B are real. Let 0 be an eigenvalue of B: Then
T 0 det Q AQ 0I = det = ( 0) det (B 0I) = 0: 0 B 0I T Hence 0 is also an eigenvalue of Q AQ; i.e.,
T Q AQx =0x (2) for some non-zero vector x: Multiplying both sides of (2) on the left by Q gives
A (Qx) =0 (Qx) :
But Qx = 0 since Q is non-singular, and it follows that 0 is also an eigenvalue of A: Therefore 0 R. 6 2 Theorem 6 A real symmetric matrix is orthogonally diagonalizable.
Proof. Let A be a real symmetric matrix. Then Proposition 4 implies that the eigenvalues of A are real and Theorem 5 implies that A is upper triangularizable, i.e., there is an orthogonal matrix Q and an upper triangular matrix M such that QT AQ = M: Applying the transpose and the fact that A is symmetric we obtain T T M T = QT AQ = QT AT QT = QT AQ = M: Thus M is a symmetric upper triangular matrix and is consequently a diagonal matrix. We collect and sumarize these results in the folowing theorem:
Theorem 7 Let A be a real n n matrix. The following are equivalent: 1. A is symmetric. 2. A has an orthonormal set of eigenvectors. 3. A is orthogonally diagonalizable.
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