Linear Algebra 20 Symmetric Matrices and Quadratic Forms

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1 Linear Algebra 20 Symmetric Matrices and Quadratic Forms Wei-Shi Zheng, What Do You Learn from This Note Do you still remember the following equation or something like that I have mentioned in the class f = x T A x. (1) It is exactly a quadratic form function which is going to detail in the following. This connects the eigen-analysis and orthogonal matrix very well. Basic Concept:symmetric matrix( 对称矩阵 ), quadratic form, quadratic function, positive definitely matrix( 正定矩阵 ), Spectral Decomposition ( 谱分解 ), singular value decomposition ( 奇异值分解 ) 2 What is symmetric matrix Definition 1 (symmetric matrix). A square matrix A is said to be symmetric iff A T = A (or equivalently, [A] ij = [A] ji ). Examples: 0 n n, I n, diagonal matrices, etc.. Theorem 2. All square matrices in R n n form a vector space. Denote this vector space as M n. 1

2 Theorem 3. The set of all symmetric matrices of size n, denoted by Sym n, forms a subspace of M n. Proof. Let A and B be symmetric matrices of size n and c a scalar. Then 1. 0 n n is symmetric obviously; 2. A + B is symmetric since (A + B) T = A T + B T = A + B; 3. ca is symmetric since (ca) T = ca T = ca. So Sym n is a subspace of M n. 3 Eigenvectors for Symmetric Matrix Theorem 4. Let A Sym n (R). Also let λ 1, λ 2 R be distinct eigenvalues of A with eigenvectors v 1, v 2 respectively. Then v 1 v 2. Proof. We have λ 1 ( v 1 v 2 ) = λ 1 v 1 v 2 = Av 1 v 2 = v T 1 A T v 2 = v T 1 Av 2 = v 1 Av 2 = v 1 λ 2 v 2 = λ 2 ( v 1 v 2 ). So (λ 1 λ 2 )( v 1 v 2 ) = 0. But λ 1 λ 2 0 since they are distinct, which forces that v 1 v 2 = 0. Theorem 5. Let A Sym n (R) and λ a complex eigenvalue of A. Then λ R, that is λ is a real eigenvalue of A. Proof. Let v C n be a λ eigenvector. Then A v = λ v. Also, A T = A. So λ(v T v) = v T (λ v) = v T A v = A v T v = (λ v T ) v = λ( v T v). So (λ λ) v T v = 0. Since v 0, it follows that v T v = v 1 v v n v n = v v n 2 > 0 where ( v 1 v n ) T = v, which forces that λ λ = 0. So λ is real. 2

3 4 Orthogonality & Symmetry Theorem 6. Let A be any symmetric function and U any orthogonal matrix. Then U 1 AU is still a symmetric matrix. Proof. Notice that U T = U 1. So (U 1 AU) T = U T A T (U 1 ) T = U 1 AU. We can now prove the major result on symmetric matrices. Theorem 7 (orthogonally diagonalizable). Let A be a square matrix. Then A is a symmetric matrix if and only if there exists an orthogonal matrix U such that U 1 AU is diagonal. Proof. Firstly, it is easy to verify that if there exists an orthogonal matrix U such that U 1 AU is diagonal, then A is symmetric. It is because let the diagonal matrix be D, then A = UDU 1 Sym n (R). Secondly, or say reversely, if A is symmetric, then we need to prove there exists an orthogonal matrix U such that U 1 AU is diagonal. It can be done by induction on n as follows. 1. Step 1: Base Case: n = 1. Trivial. 2. Step 2: Inductive Hypotheses: Assume the result holds for n Step 3: Inductive Step: Let λ 1 be an eigenvalue of A, which is real by Theorem 5, and u 1 a unit λ 1 eigenvector. Then we can extend { u 1 } to a basis of R n, say { u 1, v 2,..., v n }. Furthermore, by applying the Gram Schmidt process, we obtain an orthonormal basis U 1 = { u 1, u 2,..., u n } of R n. Define U 1 = ( u 1 u n ) O n (R), which is the matrix from basis U 1 to basis E. Let F = U 1 1 AU 1 It is not hard to see that the first column of F is (λ 1 0 0) since U1 1 = U T. So F has the form ( ) λ1 0 F = T n 1 0 n 1 A, 3

4 where A Sym n 1 (R). By Inductive Hypotheses, there exists ( ) 1 0 U O n 1 (R) such that U 1 A U is diagonal. Define U 2 = T n 1 0 n 1 U, which is an orthogonal matrix. Then ( ) U2 1 U1 1 AU 1 U 2 = U2 1 λ1 0 F U 2 = T n 1 0 n 1 U 1 A U, which is diagonal. So take U = U 1 U 2, then U is an orthogonal matrix and U 1 AU is diagonal. Remarks (Spectral Decomposition, 谱分解 ): If A is orthogonally diagonalizable, and A = P DP T = ( λ ) 1 u T 1 u 1 u n...., then A = λ 1 u 1 u T λ n u n u T n is the spectral decomposition of matrix A. 5 Singular Value Decomposition Question: If matrix A is not symmetric, it may not be orthogonally diagonalizable. However, A T A is symmetric and can be orthogonally diagonalized as P DP 1. Then is it possible to find some orthogonal matrix U and Σ such that P DP 1 = P Σ T U T UΣP 1, A = UΣP 1 = UΣV T? λ n u T n Answer: Yes. It can be. The Singular Value Decomposition can realize this!!! Question: What is singular values? Definition 8. Let A be m n matrix. Then A T A can be orthogonally diagonalized. Let { v 1,, v n } be the orthogonal eigenvector basis of matrix A T A, where their corresponding eigenvalues are λ 1 λ 2 λ n 0. Then we say σ i = λ i are the singular values of matrix A. 4

5 Theorem 9. Let A be m n matrix, where r = rank(a). Then there exists a m m orthogonal ( matrix U and a n ) n orthogonal matrix V and a m n D O matrix Σ = r (n r), where D = diag(σ 1, σ 2,, σ r ) O (m r) r O (m r) (n r) σ 1 σ 2 σ r > 0 are singular values of A, such that A = UΣV. Proof. STEP 1: Let { v 1,, v n } be the orthogonal eigenvector basis of matrix A T A, where their corresponding eigenvalues are λ 1 λ 2 λ r > λ r+1 = λ r+2 = = λ n = 0. Then we can prove that {A v 1,, A v r } are orthogonal basis of ColA. STEP 2: Normalize all vectors in the set {A v 1,, A v r } to obtain orthonormal basis { u 1,, u r }, where u i = 1 A v i A v i = 1 σ i A v i. That is A v i = σ i u i, 1 i r. STEP 3: Extend { u 1,, u r } to be the orthonormal basis of R m { u 1,, u r, u r+1,, u n }. STEP 4: Let U = [ u 1,, u n ] and V = [ v 1,, v n ]. Then they are orthogonal matrices and AV = [A v 1,, A v r, 0 0] = [σ 1 u 1,, σ r u r, 0 0]. Then we can easily have A = UΣV T. Example: ( Compute ) the singular value decomposition for matrix A = Applications: Image Processing ( 见演示 ) 5

6 6 Application: Quadratic Form( 二次型 ) Definition 10 (quadratic form). A real quadratic form q(x 1,..., x n ) in n variables x 1,..., x n is a polynomial over R having the form q(x 1,..., x n ) = c ij x i x j. 1 i j n Examples: 0, 8x 2 1, x 2 x 7, x x x 2 3, x 1 x x 2 5, 2x 3 x 4 4x 2 1 are all quadratic forms. Now we are given a quadratic form q(x 1,..., x n ) = 1 i j n c ijx i x j. We can define a symmetric matrix A such that c ij if i = j; 1 [A] ij = c 2 ij if i < j; 1 c 2 ji if i > j. Then it is easy to verify that q(x 1,..., x n ) = (x 1 x n )A(x 1 x n ) T. Conversely, for any A Sym n (R), q(x 1,..., x n ) = (x 1 x n )A(x 1 x n ) T = a ii x 2 i + 2a ij x i x j 1 i n 1 i<j n is a quadratic form, which is called the quadratic form corresponding to A and is denoted by q A (x 1,..., x n ). Thus, any quadratic form in n variables can be represented by a symmetric matrix of size n uniquely. Definition 11 (quadratic form function). Let q(x 1,..., x n ) be a real quadratic form. Then the function f q : R n R, f q ( x) = q(x 1,..., x n ), where x = (x 1 x n ) T, is called the quadratic form function induced by q. If q corresponds to A Sym n, then we write f A for f q and f A ( x) = x T A x. 6

7 Question: Can we have a more simplified quadratic form for the same quadratic function? Recall that for any x R n, we have x = [x] E, where E is the standard basis. Suppose that B is another basis of R n. Then x = [ x] E = P [ x] B, where P = [B] E. So for any quadratic form function f A we have f A ( x) = f A (P y) = y T P T AP y = f P T AP ( y), where y = [ x] B. So f A and f P T AP can be regarded as the same quadratic form function with respect to different bases. Therefore for the sake of convenience, it is quite natural to choose an invertible matrix P such that f P T AP has a simple form. Also, under the postulates of Euclidean Geometry, only orthogonal coordinate transformation is admitted, that is P is required to be orthogonal. Now by Theorem 6, we can choose an orthogonal matrix U such that U T AU is diagonal. So we have Theorem 12. Let f A ( x) = x T A x be a quadratic form function. Then there exists U O n such that f A ( x) = f U T AU( y) = y T U T AU y = λ 1 y λ n y 2 n, where y = (y 1 y n ) T and λ 1,..., λ n are all eigenvalues of A counting multiplicity. 7

8 6.1 Positive definite quadratic form Definition 13 (positive definite quadratic form). Let q be a real quadratic form in n variables. Then q is said to be 1. positive definite ( 正定的 ) if f q ( x) > 0 for all x 0; 2. negative definite ( 负定的 ) if f q ( x) < 0 for all x 0; 3. indefinite ( 不定的 ) if the values of f q ( x) can be both positive and negative. Theorem 14. Let q A be the quadratic form corresponding to A Sym n (R) and λ 1,..., λ n all the eigenvalues of A. Then 1. q A is positive definite iff λ i > 0 for all i = 1,..., n; 2. q A is negative definite iff λ i < 0 for all i = 1,..., n; 3. q A is indefinite iff there exist i, j such that λ i > 0 and λ j < 0. Proof. Let U be an orthogonal matrix such that f A ( x) = f U T AU( y) = λ 1 y λ n y 2 n. Then the range of f A ( x) is the same as that of f U T AU. So the result follows easily by arguing the range of λ 1 y λ n y 2 n. Theorem 15. Let A Sym 2 (R) and λ 1 λ 2 be eigenvalues of A. Then equation f A ( x) = 1 represents a conic ( 圆锥 ) section, which can be classified as: 1. an ellipse ( 椭圆 ) if λ 1 λ 2 > 0; 2. a hyperbola ( 双曲线 ) if λ 1 > 0 > λ 2 ; 3. empty set if 0 λ 1 λ 2 ; 4. a pair of parallel lines if λ 1 > λ 2 = 0. Proof. Let U O 2 be such that f A ( x) = f U T AU( y) = λ 1 y λ 2 y 2 2. Then the original equation turns out to be λ 1 y λ 2 y 2 2 = 1 and the result follows easily. 8

9 7 Geometric Understanding of Symmetric Matrix Theorem 16. Let A be a symmetric matrix, then we have M = max{ x T A x x = 1} = the maximum eigenvalue of A, n = min{ x T A x x = 1} = the minimum eigenvalue of A. The Cafe Terrace on the Place du Forum, by Van Gogh 9