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7. Linear operators on IPS: The love story of a couple linear transformations who respected their own personal (inner product) space Other title contenders for after the colon were “Saving (inner product) space for Hilbert” and “Can keep my hands to myself, by Selena Gomez”. I remember this one time I was doing my homework, finding conjugate transposes of matrices when I decided to be cool and buy some cigarettes. Long story short, I chickened out, got arrested for trying to litter the cigarettes, and only missed out on jail time by doing a hundred miserable hours of community service. And that’s the last time I ever avoided adjoint. In this chapter, we will continue to assume V is a finite-dimensional inner product space and F = R or C unless otherwise noted. 7.1. Self-adjoint maps. In Section 6.5, given a linear operator T : V → W , we constructed another linear operator T ∗ : W → V satisfying hT x, yiV = hx, T ∗ yiV

for all x, y ∈ V.

∗

We called T the adjoint of T . Definition 7.1.1. We say a linear operator T : V → V is self-adjoint (skew-adjoint) if T ∗ = T (T ∗ = −T ). We say a matrix A ∈ Mn (F) is self-adjoint (skew-adjoint) if A∗ = A (A∗ = −A).   0 −b Example 7.1.2. The matrix is skew-adjoint for any b ∈ R. b 0   a −ib Example 7.1.3. The matrix is self-adjoint for any a, b ∈ R. ib a Example 7.1.4. Given any linear map T : V → W , T T ∗ : W → W and T ∗ T : V → V are self-adjoint. The next lemma and proposition are from Friedberg, Insel, and Spence’s Linear algebra ([1], Corollary to Theorem 6.5 and Theorem 6.10). Lemma 7.1.5. Let T be a linear operator on V and B = {v1 , . . . , vn } an orthonormal basis for V . If A = [T ]B , then Aij = hT vi , vj i. Proof. By Corollary 6.3.8, for each i = 1, . . . , n, T vi = hT vi , v1 iv1 + · · · + hT vi , vn ivn . The lemma follows.



Proposition 7.1.6. Let B = {v1 , . . . , vn } be an orthonormal basis for V and T : V → V a linear operator. Then [T ∗ ]B = [T ]∗B . Proof. Let A = [T ]B and B = [T ∗ ]B . By Lemma 7.1.5, Bij = hT ∗ vi , vj i = hvi , T vj i = hT vj , vi i = Aji . The proposition follows. The following corollary is then immediate by writing things out.



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Corollary 7.1.7. Let B be an orthonormal basis for V and T : V → V a linear operator. Then T is self-adjoint if and only if the matrix [T ]B is self-adjoint. Remark 7.1.8. This result does not hold if we remove the restriction that B is orthonormal. For example, let V = R2 , F = R, B = {(1, 0), (1, 1)}, and   2 1 T (x) = x. 1 2 We end this section with an important result for self-adjoint and skew-adjoint operators. We introduce the following property of some linear operators. Definition 7.1.9. We say a linear operator T : V → V is reducible if every invariant subspace W ⊆ V has a complementary invariant subspace. In other words, for every subspace W ⊆ V satisfying T (W ) ⊆ W , there exists another subspace W 0 ⊆ V such that T (W 0 ) ⊆ W 0 and V = W 0 ⊕ W . Proposition 7.1.10. All self-adjoint operators and skew-adjiont operators are reducible. More specifically, given a linear operator T : V → V that is self-adjoint or skew-adjoint, and an invariant subspace M ⊂ V , the orthogonal complement M ⊥ is also invariant. Proof. Assume T : V → V is self-adjoint or skew-adjoint. Assume T (M ) ⊂ M . Let x ∈ M and z ∈ M ⊥ . Since L(x) ∈ M , 0 = hz, T (x)i = hT ∗ (z), xi = ±hT z, xi. It follows that T (z) ∈ M ⊥ .



7.2. Isometries. The idea of polarization identities is to rewrite inner products as a linear combinations of norms squared. If we have information about how linear operators behave with norms, we could then deduce how they behave with linear operators. Recall the polarization identities stated in Lemma 6.3.11, which we rewrite here for convenience: Fix an inner product space V and x, y ∈ V . If F = R,
1 hx, yi = ||x + y||2 − ||x − y||2 . 4 If F = C, 4


1X j 1 ||x + y||2 − ||x − y||2 + i||x + iy||2 − i||x − iy||2 . hx, yi = i ||x + ij y||2 = 4 j=1 4 Note that a linear operator T = 0 if and only if hT (x), yi = 0 for all x, y ∈ V . We can improve this for self-adjoint operators. Proposition 7.2.1. Let T : V → V be self-adjoint. Then T = 0 if and only if hT x, xi = 0 for all x ∈ V . Proof. (⇒) is clear. Now assume hT x, xi = 0 for all x ∈ V . For any x, y ∈ V , 0 = hT (x + y), x + yi = hT x, xi + hT x, yi + hT y, xi + hT y, yi = 0 + hx, T ∗ yi + hT y, xi + 0 (T is self-adjoint)

= hx, T yi + hT y, xi = 2RehT y, xi.

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Letting x = T y, we see T y = 0 by non-degeneracy.



Remark 7.2.2. By a similar proof, you can show that any ring satisfying x2 = x for all x is commutative. One can extend the above proposition to all linear operators when F = C. Proposition 7.2.3. Let T : V → V be a linear operator on a complex inner product space. Then T = 0 if and only if hT x, xi = 0 for all x ∈ V . Proof. Fix x, y ∈ V . One can show 0 = hT (x + y), x + yi = hT x, yi + hT y, xi and 0 = hT (x + iy), x + iyi = −ihT x, yi + ihT y, xi. This implies  

1 1 As −i i y = T x.

1 1 −i i



hT x, yi hT y, xi



 =

0 0

 .

 is invertible, we may conclude hT x, yi = 0. The exercise follows letting 

Random Thought 7.2.4. There once was a French political party named Lagrange’s Multipliers, which was a group of ardent supporters of the operation of multiplication. It disbanded quickly however, due to some ... division in the party. Theorem 7.2.5. Let T : V → W be a linear map between inner product spaces. Then the following are equivalent: (1) ||T x|| = ||x|| for all x ∈ V . (2) hT x, T yi = hx, yi for all x, y ∈ V . (3) T ∗ T = 1V . (4) T takes orthonormal sets of vectors to orthonormal sets of vectors. Proof. (1) ⇔ (2): This follows from the polarization identities. (1) ⇔ (3): Assume (1) holds. Then hx, xi = hT x, T xi = hT ∗ T x, xi

for all x ∈ V.

By Proposition 7.2.1, T ∗ T = 1V . This proves (1) ⇒ (3). A similar calculation shows (3) ⇒ (1). (2) ⇒ (4): Fix an orthonormal basis B = {v1 , . . . , vn } of V . Then hT vi , T vj i = hvi , vj i = δij . (4) ⇒ (1): Assume (4) holds. Fix x ∈ V with ||x|| = 1. Then we may complete {x} to an orthonormal basis of V by the Gram-Schmidt Process. By (4), we have in particular ||T x|| = 1. By linearity, we may conclude ||T y|| = ||y|| for all y ∈ V .  We leave this as an exercise to the reader. Recall that a linear operator T is an isometry if it preserves norms. Corollary 7.2.6. (Characterization of isometries) Let T : V → W be an isomorphism. Then T is an isometry if and only if T ∗ = T −1 .

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Remark 7.2.7. Endow R2 with the normal Euclidean norm and note the linear transformation T : R → R2 given by T x = (x, 0) satisfies ||T x|| = |x|. By Theorem 7.2.5, T ∗ T = 1R . However, T is clearly not an isomorphism. Thus a converse of the previous corollary does not hold. We end this section by defining orthogonal transformations. Definition 7.2.8. Fix an inner product space V . A linear operator T : V → V is called an orthogonal transformation if hT x, T yi = hx, yi for all x, y ∈ V. Note that Theorem 7.2.5 and Corollary 7.2.6 combine to give several equivalent definitions of orthogonal transformations. 7.3. The orthogonal and unitary groups. Recall the definition of a group (see Definition 2.7.9). We now define two important groups of matrices. Definition 7.3.1. We define the orthogonal group On to be the collection/group of matrices A ∈ Mn (R) satisfying At A = Idn . We define the unitary group Un to be the collection/group of matrices B ∈ Mn (C) satisfying B ∗ B = Id. Remark 7.3.2. Note that if a square matrix A ∈ Mn (F) satisfies A∗ A = Idn , then A is invertible with A∗ = A−1 (see Proposition 2.7.8). Thus, we indeed have that On ⊂ GLn (R) and Un ⊂ GLn (C). However, these are strict containments as orthogonal and unitary matrices have determinant with absolute value 1. Remark 7.3.3. Recall that matrices in Mn (F) may be naturally identified with linear operators of Fn . Thus, we have equivalent defintions of On and Un . The orthogonal group On is the collection of linear maps T : Rn → Rn satisfying T ∗ T = 1Rn . The unitary group Un is the collection of linear maps T : Cn → Cn satisfying T ∗ T = 1Cn . Recall that the columns of a matrix A ∈ Mn (F) are Ae1 , Ae2 , . . . , Aen . The following two propositions follow from Theorem 7.2.5. Proposition 7.3.4. (Characterization of the orthogonal group) Let A ∈ Mn (R). Then the following are equivalent: • • • • •

A ∈ On . At A = Id. |Ax| = |x| for all x ∈ Rn . The columns of A form an orthonormal basis of Rn . The rows of A form an orthonormal basis of Rn .

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 .