# The Unapologetic Mathematician

## Complex Inner Products

Now consider a complex vector space. We can define bilinear forms, and even ask that they be symmetric and nondegenerate. But there’s no way for such a form to be positive-definite. Indeed, we saw that there isn’t even a notion of “order” on the field of complex numbers. They do contain the real numbers as a subfield, but we can’t manage to stay in the positive real numbers. Indeed, if we have $\langle v,v\rangle=a+0i$ for some real $a\geq0$, then we also have $\langle iv,iv\rangle=i^2(a+0i)=-a+0i$. So it seems we aren’t going to get the same geometric interpretations this way.

But let’s slow down and look at a one-dimensional complex vector space — the field of complex numbers itself. We do have a notion of length here. We define the length of a complex number $z=a+bi$ as the square root of $\bar{z}z=(a-bi)(a+bi)=a^2+b^2$. This quantity is always a positive real number, and thus always has a square root. And it looks sort of like how we compute the squared length of a vector with a bilinear form. Indeed, if we think of $\mathbb{C}$ as a real vector space with basis $\{1,i\}$, it’s exactly the norm we get when we define this basis to be orthonormal. The only thing weird is that conjugation.

Well, let’s run with this a while. Given a complex vector space $V$, we want a form $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$ which is

• linear in the second slot — $\langle u,av+bw\rangle=a\langle u,v\rangle+b\langle u,w\rangle$
• conjugate symmetric — $\langle v,w\rangle=\overline{\langle w,v\rangle}$

Conjugate symmetry implies that the form is conjugate linear in the first slot — $\langle av+bw,u\rangle=\bar{a}\langle v,u\rangle+\bar{b}\langle w,u\rangle$ — and also that $\langle v,v\rangle=\overline{\langle v,v\rangle}$ is always real. This makes it reasonable to also ask that the form be

• positive definite — $\langle v,v\rangle>0$ for all $v\neq0$

This mixture of being linear in one variable and “half-linear” in the other makes the whole form “one and a half” times linear, or “sesquilinear”.

Anyhow, now we do get a notion of length, defined by setting $\lVert v\rVert^2=\langle v,v\rangle$ as before. What about angle? That will depend directly on the Cauchy-Schwarz inequality, assuming it holds. We’ll check that now.

Our previous proof doesn’t really work, since our scalars are now complex, and we can’t argue that certain polynomials have no zeroes. But we can modify it. We start similarly, calculating

$\displaystyle0\leq\langle v-tw,v-tw\rangle=\langle v,v\rangle-t\langle v,w\rangle-\bar{t}\langle w,v\rangle+\bar{t}t\langle w,w\rangle$

Now the Cauchy-Schwarz inequality is trivial if $w=0$, so we may assume $\langle w,w\rangle\neq0$, and set $t=\frac{\langle w,v\rangle}{\langle w,w\rangle}$. Then we see

\displaystyle\begin{aligned}0&\leq\langle v,v\rangle-\frac{\langle w,v\rangle\langle v,w\rangle}{\langle w,w\rangle}-\frac{\langle v,w\rangle\langle w,v\rangle}{\langle w,w\rangle}+\frac{\langle w,v\rangle\langle v,w\rangle}{\langle w,w\rangle}\\&=\langle v,v\rangle-\frac{\lvert\langle v,w\rangle\rvert^2}{\langle w,w\rangle}\end{aligned}

Multiplying through by $\langle w,w\rangle$ and rearranging, we find

$\displaystyle\lvert\langle v,w\rangle\rvert^2\leq\langle v,v\rangle\langle w,w\rangle$

which is the complex version of the Cauchy-Schwarz inequality. And then just as in the real case we can write it as

$\displaystyle\frac{\lvert\langle v,w\rangle\rvert^2}{\lVert v\rVert^2\lVert w\rVert^2}\leq1$

which implies that

$\displaystyle-1\leq\frac{\langle v,w\rangle}{\lVert v\rVert\lVert w\rVert}\leq1$

which we can again interpret as the cosine of an angle.

So all the same notions of length and angle can be recovered from this sort of complex inner product.

April 22, 2009 Posted by | Algebra, Linear Algebra | 13 Comments