Lecture 3u Properties of Complex Inner Product Spaces (pages 426-429) When doing the problems assigned for the previous lecture, you hopefully no- ticed that the standard inner product for complex numbers is not symmetrical (that is, that h~z; ~wi 6= h~w; ~zi). So, right away we know that our definition of an inner product will have to be different than the one we used for the reals. But, hopefully you also noticed that h~z; ~wi = h~w; ~zi, so this is yet another case where we need to introduce conjugation to extend a result from the reals to the complex numbers. Bilinearity also needs some adjustments in the complex numbers. Let's take a look at the definition of a (generic) inner product on Cn. Definition: Let V be a vector space over C.A complex inner product on V is a function h ; i : V × V ! C such that (1) For all z 2 V, we have that hz; zi is a non-negative real number, and hz; zi = 0 if and only if z = 0. (2) For all w; z 2 V, hz; wi = hw; zi (3) For all u; v; w; z 2 V and all α 2 C we have (i) hv + z; wi = hv; wi + hz; wi (ii) hz; w + ui = hz; wi + hz; ui (iii) hαz; wi = αhz; wi (iv) hz; αwi = αhz; wi Property (1) is still the same as in Rn, and is still referred to as being \positive definite”. Property (2) is known as the Hermitian property of the inner prod- uct (instead of the symmetric property). Because the complex inner product is not symmetric, we cannot find a simple counterpart to bilinearity, but we can combine the statements of property (3) into one statement as follows: Let u; v; w; z 2 V and α; β; γ; δ 2 C, and let's expand out hαu + βv; γw + δzi. One use of part (i) gets us to hαu; γw + δzi + hβv; γw + δzi Then we can use part (ii) twice to get hαu; γwi + hαu; δzi + hβv; γwi + hβv; δzi Now, we can use part (iii) four times to get αhu; γwi + αhu; δzi + βhv; γwi + βhv; δzi 1 And lastly, we use part (iv) four times to get αγhu; wi + αδhu; zi + βγhv; wi + βδhv; zi So we see that the inner product is almost bilinear{we simply need to remember to take the conjugate of any scalar we pull out of the right side of the inner product. We won't spend much time on non-standard inner product spaces, but we should at least verify that the standard inner product we defined is in fact an inner product! Example: Show that the standard inner product defined on Cn is a complex inner product. n Pn Pn 2 Property (1) Let ~z 2 C . Then h~z; ~zi = j=1 zjzj = j=1 jzjj . Since this is the sum of non-negative real numbers, it must be a non-negative real number. Pn 2 2 Moreover, the only way to have that h~z; ~zi = j=1 jzjj = 0, is to have jzjj = 0 for all 1 ≤ j ≤ n, and since the only complex number with a modulus of 0 is 0, we see that h~z; ~zi = 0 if and only if ~z = ~0. Property (2) Let ~z; ~w 2 Cn. Then we can use properties of the conjugate to see that Pn h~w; ~zi = j=1 wjzj Pn = j=1 wj zj Pn = j=1 zjwj = h~z; ~wi Property (3i) Let ~v; ~w; ~z 2 Cn. Then Pn h~v + ~z; ~wi = j=1(vj + zj)wj Pn = j=1 vjwj + zjwj Pn Pn = j=1 vjwj + j=1 zjwj = h~v; ~wi + h~z; ~wi Property (3ii) Let ~u;~w; ~z 2 Cn. Then 2 Pn h~z; ~w + ~ui = j=1 zj(wj + uj) Pn = j=1 zj(wj + uj) Pn = j=1 zjwj + zjuj Pn Pn = j=1 zjwj + j=1 zjuj = h~z; ~wi + h~z; ~ui Property (3iii) Let ~w; ~z 2 Cn and α 2 C. Then Pn hα~z; ~wi = j=1 αzjwj Pn = α j=1 zjwj = αh~z; ~wi Property (3iv) Let ~w; ~z 2 Cn and α 2 C. Then Pn h~z; α ~wi = j=1 zjαwj Pn = j=1 zj(α wj) Pn = j=1 zj(α)(wj) Pn = α j=1 zjwj = αh~z; ~wi There are two additional properties that hold of the complex inner product: the Cauchy-Schwarz Inequality and the Triangle Inequality. Theorem 9.5.1 Let V be a complex inner product space with inner product h ; i. Then, for all w; z 2 V, we have Cauchy-Schwarz Inequality: jhz; wij ≤ jjzjj jjwjj Triangle Inequality: jjz + wjj ≤ jjzjj + jjwjj For once, we cannot simply copy the proof from the proof used in the reals. The textbook provides a proof of the Cauchy-Shwarz Inequality, so I'll only prove the Triangle Inequality. 3 Proof of the Triangle Inequality: Note that the Triangle Inequality is equivalent to the statement jjz + wjj2 ≤ (jjzjj + jjwjj)2 or that jjz + wjj2 − (jjzjj + jjwjj)2 ≤ 0 Let's expand the left side: jjz + wjj2 − (jjzjj + jjwjj)2 = jjz + wjj2 − (jjzjj2 + 2jjzjj jjwjj + jjwjj2) = hz + w; z + wi − hz; zi − hw; wi − 2jjzjj jjwjj = hz; zi + hz; wi + hw; zi + hw; wi − hz; zi − hw; wi − 2jjzjj jjwjj = hz; wi + hw; zi − 2jjzjj jjwjj = hz; wi + hz; wi − 2jjzjj jjwjj = 2Re(hz; wi) − 2jjzjj jjwjj So we need to show that 2Re(hz; wi) − 2jjzjj jjwjj ≤ 0 which is the same as showing that Re(hz; wi) ≤ jjzjj jjwjj We will make use of the Cauchy-Schwarz Inequality, by first showing that Re(hz; wi) ≤ jhz; wij. To see this, we first note that jhz; wij2 = (Re(hz; wi))2 + (Im(hz; wi))2 Since (Im(hz; wi))2 ≥ 0, we see that (Re(hz; wi))2 ≤ jhz; wij2 And thus we have that jRe(hz; wi)j ≤ jhz; wij But since Re(hz; wi) ≤ jRe(hz; wi)j, we have shown that 4 Re(hz; wi) ≤ jhz; wij And the Cauchy-Shwarz Inequality tells us that jhz; wij ≤ jjzjj jjwjj, so we see that Re(hz; wi) ≤ jhz; wij ≤ jjzjj jjwjj which completes our proof of the Triangle Inequality. 5.