8.2 Orthogonal Diagonalization

1. 8.2 Orthogonal Diagonalization

2. Theorem 1 The following conditions are equivalent for an (n x n) matrix P: 1. P is invertible and P-1 = PT 2. The rows of P are orthonormal. 3. The columns of P are orthonormal. Proof: (1) (2) Given P-1 = PT, PPT = I Let X1,…,Xn be rows of P. Then XjT is jth col of PT So (i,j) entry of PPT is Xi•Xj Since PPT=0, we know that all (i,j) entries where i≠j will be 0, so Xi•Xj=0 (rows are orthog) if i≠j and also Xi•Xj=1 if i=j (rows are orthonormal) (1/3 equiv is same idea)

3. Definition An (n x n) mtx P is called an orthogonal matrix if it satisfies one (and so all) of the conditions in Theorem 1.

4. Example Is the following matrix orthogonal for any angle? Show rows are orthonormal.

5. Example If P and Q are orthogonal matrices, show that PQ is also orthogonal, and that P-1=PT is orthogonal. Solution: P and Q are invertible, so PQ is also (PQ)-1 = Q-1P-1=QTPT = (PQ)T so PQ is orthogonal. Also (P-1)-1 = P = (PT)T = (P-1)T so P-1 is orthogonal.

6. Definition An (n x n) matrix A is orthogonally diagonalizable when an orthogonal matrix P can be found such that P-1AP=PTAP is diagonal. (We will see that this only happens with symmetric matrices.)

7. Theorem 2-Principal Axes Theoremor Real Spectral Theorem The following conditions are equivalent for A (n x n): 1. A has an orthonormal set of n eigenvectors. (w/ eigenvalues that we call the spectrum of the matrix. 2. A is orthogonally diagonalizable. 3. A is symmetric. Proof: (1)(2) P = [X1 … Xn] (n x n) P is orthog iff {X1,…,Xn} is orthonormal set in n (by thm1) P-1AP is diag iff {X1,…,Xn} consists of the eigenvectors of A (from 6.2)

8. Proof-cont. (2)(3) If PTAP = D is diagonal where PT=P-1(given from 2), then A = PDP-1=PDPT So, AT = (PDPT)T=PTTDTPT=PDTPT=PDPT = A (since D diag) So A is symmetric. (3)(2) very tough to prove -- omitted

9. Definition Principal Axes: A set of orthonormal eigenvectors of a symmetric matrix A.

10. Example Find an orthogonal matrix P such that P-1AP is diagonal. We know that A is orthogonally diagonalizable since it is symmetric (Thm 2) 1. Find eigenvalues and eigenvectors 2. Make sure you get an orthogonal matrix for P (may need to use G-S alg) 3. Normalize

11. Theorem 3 If A is an (n x n) symmetric matrix, then (AX)• Y = X•(AY) for all columns X and Y in n. Proof: X • Y = XTY for columns X and Y. So: (AX) •Y = (AX)TY (by def of dot product) =XTATY = XTAY (since A is symmetric) = X • (AY) 

12. Theorem 4 If A is a symmetric matrix, then eigenvectors of A corresponding to distinct eigenvalues are orthogonal. Proof: Let AX = X and AY = Y where  ≠ (so 2 distinct eigenvalues). Then (using thm 3): (X•Y) = (X)•Y=(AX)•Y = X•(AY) = X•(Y) = (X•Y) So: (- ) (X•Y) = 0, so (X•Y) = 0 since ≠  which tells us that the eigenvectors are orthogonal. 

13. Example Diagonalize: 1. Clearly orthogonally diagonalizable since symmetric 2. Find eigenvalues 3. Find eigenvectors 4. Eigenvectors of one eigenvalue are orthogonal by thm 4, but if an eigenvalue has multiplicity > 1, then it also has >1 eigenvector. Its diff’t eigenvalues may not be orthogonal, so you may need to use G-S alg. 5. Normalize to get orthonormal vectors.

14. Theorem 5-Triangulation Thm If A is an (n x n) matrix with real eigenvalues, an orthogonal matrix P exists such that PTAP is upper triangular. Proof: much like the proof we omitted earlier.

15. Corollary Since a matrix and its diagonalized form are similar, they will have the same determinant and trace, we can say: If A is an (n x n) matrix with real eigenvalues 1, 2,…, n (may not all be distinct), then detA= 1 2... n and tr A= 1+2+…+ n. (since these are the values we get in the for det and tr of the diagonalized matrix)