The Unapologetic Mathematician

Mathematics for the interested outsider

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


  1. […] Polarization Identities If we have an inner product on a real or complex vector space, we get a notion of length called a “norm”. It turns out that the norm […]

    Pingback by The Polarization Identities « The Unapologetic Mathematician | April 23, 2009 | Reply

  2. […] Gram-Schmidt Process Now that we have a real or complex inner product, we have notions of length and angle. This lets us define what it means for a […]

    Pingback by The Gram-Schmidt Process « The Unapologetic Mathematician | April 28, 2009 | Reply

  3. Sweet looking blog. Great job with the formulas, notation, and linking to background topics.

    I have a question in regard to this topic. At the end of the post, you point out that

    -1\leq\frac{\lvert\langle v,w\rangle\rvert}{\lVert v\rVert\lVert w\rVert}\leq1

    Now, all of \lvert\langle v,w\rangle\rvert, \lVert v\rVert, and \lVert w\rVert are positive definite, so can it not be written as

    0\leq\frac{\lvert\langle v,w\rangle\rvert}{\lVert v\rVert\lVert w\rVert}\leq1

    If so, what are the effects on interpretation as cosine of an angle?

    Comment by John Freeman | April 30, 2009 | Reply

  4. Oh, I should write that differently… Thanks for catching it.

    Comment by John Armstrong | April 30, 2009 | Reply

  5. […] start with either a bilinear or a sesquilinear form on the vector space . Let’s also pick an arbitrary basis of . I want to emphasize that […]

    Pingback by Matrices and Forms I « The Unapologetic Mathematician | June 24, 2009 | Reply

  6. […] which we can also think of as a linear map from to its dual space . We can also consider a sesquilinear form on a complex vector space, which is equivalent to an antilinear (conjugates scalars) map from […]

    Pingback by Matrices and Forms II « The Unapologetic Mathematician | June 25, 2009 | Reply

  7. “So it seems we aren’t going to get the same geometric interpretations this way.”

    If you use Geometric (Clifford) Algebra you get the correct geometric interpretations without all the fiddling around with complex numbers.

    This extends right through Dirac, Pauli and a guage theory of gravity.

    Comment by Maya Incaand | July 9, 2009 | Reply

  8. Sure, Maya. I’m not ready to throw out Clifford Algebras quite yet, but that’s another way to go.

    Comment by John Armstrong | July 9, 2009 | Reply

  9. […] for our new bilinear form to be an inner product it must be symmetric (or conjugate-symmetric). This is satisfied by picking our transformation to be symmetric (or hermitian). But we also need […]

    Pingback by Positive-Definite Transformations « The Unapologetic Mathematician | July 13, 2009 | Reply

  10. […] carrying a representation of a group is an “invariant form”. To start with, this is a complex inner product , which we recall means that it […]

    Pingback by Invariant Forms « The Unapologetic Mathematician | September 27, 2010 | Reply

  11. […] space of class functions also has a nice inner product. Of course, we could just declare the basis to be orthonormal, but that’s not quite what […]

    Pingback by Class Functions « The Unapologetic Mathematician | October 15, 2010 | Reply

  12. […] how does this apply to Maschke’s theorem? Well, given a -module , the collection of sesquilinear forms on the underlying space forms a vector space itself. Indeed, such forms correspond to correspond […]

    Pingback by Projecting Onto Invariants « The Unapologetic Mathematician | November 13, 2010 | Reply

  13. This is wrong. There is no (good enough) ordering of the complex numbers,
    so the last inequality is meaningless.

    Comment by Websailor32 | July 30, 2013 | Reply

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