Lecture 28: The Spectral Theorem

Lecture 28: The Spectral Theorem 28 28.1 The Spectral Theorem Review: Hermitian Spaces Last time, we discussed Hermitian spaces which were just complex vector spaces with a positive defnite Hermitian form. Given a linear operator T , we defned the adjoint T ∗ , which had the property that ⟨v, T ∗ w⟩ = ⟨T v, w⟩. We called a linear operator T normal if T T ∗ = T ∗ T . We then were able to state the Spectral Theorem. 28.2 The Spectral Theorem The Spectral Theorem demonstrates the special properties of normal and real symmetric matrices. Theorem 28.1 Given a Hermitian space V , for any normal linear operator T , there exists an orthonormal eigenbasis of V : {u1 , · · · , un }. In matrix form, for any normal matrix M ∈ GLn (C), there exists a unitary matrix P such that P −1 M P is diagonal. The real version states that for a Euclidean vector space V and a symmetric linear operator T , there exists an orthonormal eigenbasis; equivalently, for any symmetric matrix M ∈ GLn (R), there exists an orthogonal matrix P such that P −1 M P is diagonal. All eigenvalues of real symmetric matrices are real. Example 28.2     3 −1 1 1 √ Say we take the symmetric matrix M = . One eigenvector is 2 with eigenvalue λ1 = 2. The 1 −1 3   1 other eigenvector should be orthogonal, so it is √12 with eigenvalue λ2 = 4. −1 In 2 dimensions, the fact that our change of basis is coming from an orthogonal basis tells us that the change in coordinates is just a rotation. This rotation makes the operator diagonalizable. This is called the Spectral Theorem because the eigenvalues are often referred to as the spectrum of a matrix. Any theorem that talks about diagonalizing operators is often called a spectral theorem. Now we will state some lemmas in order to prove the Spectral Theorem. Lemma 28.3 (Lemma 1) First, for a linear operator T : V −→ V where V is Hermitian, and a subspace W ⊂ V such that T (W ) ⊂ W, then T ∗ (W ⊥ ) ⊂ W ⊥ . Proof. Take some u ∈ W ⊥ . For any w ∈ W, then ⟨w, T ∗ u⟩ = ⟨T w, u⟩ = 0, by the defnition of the adjoint linear operator, so T ∗ u ∈ W ⊥ . Lemma 28.4 (Lemma 2) If T v = λv, then T ∗ v = λv. This just means that T and T ∗ have the same eigenvectors, and eigenvalues related by complex conjugation. Proof. We can frst consider a special case. • Special Case: λ = 0. We want to show that T v = 0 implies that T ∗ v = 0. ⟨T ∗ v, T ∗ v⟩ = ⟨T v, T v⟩ = 0 by normality (this was a property shown last time). Since T is positive defnite, this implies that T ∗ v = 0. This also implies that if a vector v ∈ ker T , then v ∈ ker T ∗ as well. Lecture 28: The Spectral Theorem • General Case: λ ∈ C. We want to show that T v = λv implies that T ∗ v = λv. In this case, create a new linear operator S = T − λI. Then, Sv = 0. Also, the adjoint is S ∗ = T ∗ − λI and SS ∗ = S ∗ S, since T T ∗ = T ∗ T. By the frst case, since v ∈ ker S, we also have that v ∈ ker S ∗ so T ∗ = λv as needed. Proof of the Spectral Theorem. Now we can prove the Spectral Theorem by induction on the dimension of V. The general process will be breaking V up into a direct sum and fnding orthonormal eigenbases over each part. Since the feld is F = C, it is always possible to fnd an eigenvector w ∈ V such that T w = λw. By normalizing w, assume that ⟨w, w⟩ = 1. Then, let W = Span(w); then V = W ⊕ W ⊥ since all subspaces of Hermitian spaces are non-degnerate. In addition, since w is an eigenvector, T W ⊂ W. If we can show that T W ⊥ preserves W ⊥ , then we can induct. By Lemma 2, w is also an eigenvector of T ∗ , so T ∗ (W ) ⊂ W, and by Lemma 1, (T ∗ )∗ (W ⊥ ) = T (W ⊥ ) ⊂ W ⊥ . Then V = W ⊕ W ⊥ , and T maps each part of the splitting to itself, so by induction there exists u2 , · · · , un an orthonormal eigenvasis for W ⊥ , and {w, u2 , · · · , un } is an orthonormal eigenbasis for V. Student Question. Why is the equivalent of the theorem not true over the real numbers? Answer. Most of this argument works, except in the very frst step, where we found an eigenvector and eigenvalue. We cannot guarantee this will happen with normal linear operators over the real numbers. However, as we found last week, for symmetric (and Hermitian) matrices, the eigenvalues are all real, and in particular it is always possible to fnd one eigenvector w ∈ V with real eigenvalue to allow the induction to occur. 2 2 As an application, suppose  we have  a quadratic function f (x, y) = ax + bxy + cy . Let this function be a b/2 represented by a matrix ; then b/2 c   x f (x, y) = (x, y)M = ⟨v, T v⟩ y for a linearoperatorT given by M. By the Spectral Theorem, there an change of coordinates  exists   orthogonal  λ1 0 x x′ T P MP = , where P is an orthogonal matrix. It takes =P . Then 0 λ2 y y′      ′ x λ1 x ′ ′ f (x, y) = (x, y)M = (x , y ) = λ1 (x′ )2 + λ2 (y ′ )2 . y λ2 y′ Example 28.5 If f (x, y) = 3x − 2xy + 3y , the matrix would be M = 2 2  3 −1  −1 . Then 3 3x2 − 2xy + 3y 2 = 2(x′ )2 + 4(y ′ )2 , for some change of coordinates taking x and y to x′ and y ′ , since we saw earlier that the eigenvalues were 2 and 4. In n variables, let f (x1 , · · · , xn ) = a11 x21 + · · · + ann x2n +   a11 · · · aij  ..  ..  .  .  . (aij ) =   . ..  aji  ann P i 0 • One line. x2 = 0 • One point. x2 + y 2 = 0 • The empty set ∅. x2 + y 2 = −1 Theorem 28.6 After an isometry, all curves of the form ax2 + bxy + cy 2 + dx + ey + f = 0 look like one of these options. #» #»T #» Proof. The  curve be  rewritten as v A v + B v + f = 0. After an orthogonal change of basis, we can assume λ1 0 that A = Then λ1 x2 + λ2 y 2 + b1 x + b2 y + f = 0. 0 λ2 . b1 b2 • Nonzero eigenvalues. If λ1 , λ2 = ̸ 0, then we can complete the square, taking x ⇝ x+ 2λ and y ⇝ y+ 2λ . 1 2 Then λ1 x2 + λ2 y 2 = c. If c = 0, it is a single point or two intersecting lines, and if c ̸= 0, it is an ellipse or a hyperbola. • Zero eigenvalues. The other cases. The details of this aren’t that important, and we glossed over much of it at the end. The main idea to take away is just the power of the Spectral Theorem to take a complicated quadratic form and simplify into something more readily analyzed. One idea where the Spectral Theorem is at play is in multivariable calculus with the second derivative test. Hidden behind the test is that we have a symmetric matrix and we are trying to fgure out what the signs of the eigenvalues are, as that will tell us whether our critical point is a minimum, maximum, or saddle point.