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BY DEREK JUNG, ADAPTED FROM NOTES BY UCLA PROF. PETER PETERSEN

Proposition 7.3.5. (Characterization of the unitary group) Let B ∈ Mn (C). Then the following are equivalent: • B ∈ Un . • B ∗ B = Id. • |By| = |y| for all y ∈ Cn . • The columns of B form an orthonormal basis of Cn . • The rows of B form an orthonormal basis of Cn . 7.4. The spectral theorem for self-adjoint operators. In math, we are often interested in studying maps that have nice forms. For example, we showed in the last chapter that every square complex matrix has an almost unique Jordan canonical form. In this section, we will prove the spectral theorem, which states that every self-adjoint operator has a diagonal matrix representation. We will assume that V is a finite-dimensional complex inner product space (so F = C). The case F = R will follow with similar (and usually simpler) proofs.   0 1 Example 7.4.1. A := is not diagonalizable. Equivalently, if we define T : R2 → 0 0 R2 by T (x) = Ax, there does not a basis {v1 , v2 } of R2 for which [T ]{v1 ,v2 } is diagonal. Lemma 7.4.2. Suppose T : V → V is self-adjoint. If λ is an eigenvalue of T , then λ ∈ R. Proof. Let x 6= 0 be an eigenvector corresponding to λ. Then ¯ xi. λhx, xi = hT x, xi = hx, T xi = λhx, As x 6= 0, it follows that λ is real.



The proof of the following proposition is a bit more involved, so I will omit it. A proof can be found in Petersen’s notes ([6], Theorem 32, pages 205-206). Proposition 7.4.3. (Existence of eigenvalues for self-adjoint operators) Every self-adjoint operator on V has a real eigenvalue. This gives us by induction the spectral theorem for self-adjoint operators. Theorem 7.4.4. (Spectral theorem) Let T : V → V be a self-adjoint operator. Then there exists an orthonormal basis e1 , . . . , en of eigenvectors. Moreover, all of the eigenvalues of T are real. Proof. The second part of the theorem follows from Lemma 7.4.2. By Proposition 7.4.3, there exists an eigenvector e1 of T with corresponding eigenvalue λ1 . By dividing by its norm if necessary, we may assume e1 has norm 1. Consider the subspace {e1 }⊥ = {x ∈ V : hx, e1 i = 0}. We show {e1 }⊥ is T -invariant. For each x ∈ {e1 }⊥ , hT (x), e1 i = hx, T (e1 )i = λ¯1 hx, e1 i = 0. Note by Proposition 6.4.4, with M = span{e1 }, dim{e1 }⊥ = dim(V ) − 1. Thus, we may restrict T to T |{e1 }⊥ and apply induction on the dimension of V to conclude there is an orthonormal basis of eigenvectors for T . For a more constructive argument, we can choose an eigenvector e2 of T |{e1 }⊥ : {e1 }⊥ → {e1 }⊥ , and continue choosing eigenvectors dim(V ) − 2 more times. 

SWILA NOTES

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Recall that a basis of eigenvectors leads us to a diagonal matrix representation (see Section 5.4). Thus, the spectral theorem immediately gives us the following two corollaries. Corollary 7.4.5. Let T : V → V be a self-adjoint operator. Then there exists an orthonormal basis B = {e1 , . . . , en } and a real n × n diagonal matrix D such that [T ]B = D. Here, the k th diagonal entry of D is the eigenvalue corresponding to ek . Definition 7.4.6. We say that a matrix A ∈ Mn (C) is unitarily similar to B ∈ Mn (C) if there exists a unitary matrix U ∈ Un such that A = U BU −1 . Note Corollary 7.4.5 gives that every self-adjoint matrix is unitarily similar to a diagonal matrix: Corollary 7.4.7. Let A ∈ Mn (C) be a self-adjoint matrix. Then there exists an orthonormal basis B = {e1 , . . . , en } and a real n × n diagonal matrix D such that    ∗ | | | | A =  e1 · · · en  D  e1 · · · en  . | | | | Proof. The spectral theorem gives us an orthonormal basis of eigenvectors such that    −1 | | | | A =  e1 · · · en  D  e1 · · · en  . | | | | As {e1 , . . . , en } is an orthonormal basis of Cn , the corollary follows from Proposition 7.3.5 (the matrix with column vectors e1 , . . . , en is unitary).  Remark 7.4.8. We can now finish the proof from Example 6.1.5 that the trace is an inner product on matrices. Suppose 0 6= A ∈ Mn (C). As A∗ A is self-adjoint, there is a unitary matrix U and a diagonal matrix D such that A∗ A = U DU ∗ = U DU −1 . Here, the diagonal entries of D are the eigenvalues λ1 , . . . , λn of A∗ A (possibly with repetition). By Lemma 2.5.2, n X ∗ hA, Ai = tr(A A) = tr(D) = λi . i=1 ∗

We must have that λi 6= 0 for some i. Otherwise, A A(x) = 0 for all x ∈ Cn , which would imply A = 0. Thus, it suffices to show λi ≥ 0 for all i. For 1 ≤ i ≤ n, suppose xi is a (non-zero) eigenvector corresponding to λi . Then λi ||xi ||2 = hA∗ Axi , xi i = ||Axi ||2 . It follows that λi ≥ 0, whence the trace is an inner product on Mn (C). Random Thought 7.4.9. A physicist, a mathematician, and a statistician are sitting on a train. The physicist is playing with a rubber band, the mathematician is eating dinner, and the statistician is reading a newspaper. Who is the conductor? Not the rubber band... because rubber isn’t a conductor!

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BY DEREK JUNG, ADAPTED FROM NOTES BY UCLA PROF. PETER PETERSEN

Lemma 7.4.10. Let T : V → V be a self-adjoint operator. Suppose x, y are eigenvectors with distinct eigenvalues λ, µ, respectively. Then x is orthogonal to y. Proof. By Lemma 7.4.2, λ and µ are both real. Observe λhx, yi = hT x, yi = hx, T yi = hx, µyi = µhx, yi, which implies that hx, yi = 0.



Theorem 7.4.11. Let T : V → V be a self-adjoint operator and λ1 , . . . , λk the distinct eigenvalues of T . Then 1V = projker(T −λ1 1V ) + · · · + projker(T −λk 1V ) and Here, projker(T −λi 1V )

T = λ1 projker(T −λ1 1V ) + · · · + λk projker(T −kλk 1V ) . is the orthogonal projection onto ker(T − λi 1V ) (see Definition 6.4.1).

Proof. By Lemma 7.4.10, the subspaces ker(T − λj 1V ) are mutually orthogonal. This implies that if xi ∈ ker(T − λi 1V ), then  xi , if i = j, projker(T −λj 1V ) (xi ) = 0, if i 6= j. As we may write x = x1 + · · · + xk ∈ V , xi ∈ ker(T − λi 1V ), this implies 1V = projker(T −λ1 1V ) + · · · + projker(T −λk 1V ) . Moreover, T (x) = λ1 x1 + · · · + λk xk = λ1 projker(T −λ1 1V ) (x) + · · · + λk projker(T −λk 1V ) (x). This proves the theorem.



The final corollary is the only result in this section that solely makes sense for F = C. Corollary 7.4.12. (Spectral theorem for skew-adjoint operators) Let T : V → V be a skew-adjoint operator (T ∗ = −T ). Then there exists an orthonormal basis of eigenvectors e1 , . . . , ek with corresponding eigenvalues ia1 , . . . , iak , aj ∈ R. Proof. This follows from the spectral theorem for self-adjoint operators after noting that iT is self-adjoint.  7.5. Normal operators and the spectral theorem for normal operators. In the previous section, we learned that self-adjoint operators are diagonalizable. In other words, every self-adjoint operator has an orthonormal basis of eigenvectors. A natural question to ask is whether we can characterize those linear operators which have an orthonormal basis of eigenvectors. And is it even normal for a linear operator to be diagonalizable? Well, I’m not sure about that question, but the linear operators which are diagonalizable are exactly the normal ones. We will typically assume V is a finite-dimensional complex inner product space in this section. Definition 7.5.1. We call a linear operator T : V → V normal if T ∗ T = T T ∗ . Remark 7.5.2. Note that self-adjoint and skew-adjoint operators are normal.