The Unapologetic Mathematician

Mathematics for the interested outsider

Positive-Definite Transformations

We’ll stick with the background vector space V with inner product \langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle. If we want another inner product to actually work with, we need to pick out a bilinear form (or sesquilinear, over \mathbb{C}. So this means we need a transformation to stick between bras and kets.

Now, 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 our form to be “positive-definite”. That is, we need

\displaystyle\langle v\rvert B\lvert v\rangle\geq0

for all vectors v\in V, and for equality to obtain only when v=0.

So let’s look at this condition on its own, first over \mathbb{R}. If A is antisymmetric, then we see \langle v\rvert A\lvert v\rangle=-\langle v\rvert A\lvert v\rangle by taking the adjoint, and thus A must be zero. But an arbitrary transformation B can be split into a symmetric part S=\frac{1}{2}\left(B+B^*\right) and an antisymmetric part A=\frac{1}{2}\left(B-B^*\right). It’s easy to check that S+A=B. So the antisymmetric part of B must be trivial, and the concept of being “positive-definite” only makes real sense for symmetric transformations.

What happens over \mathbb{C}? Now we want to interpret the positivity condition as saying that \langle v\rvert B\lvert v\rangle is first and foremost a real number. Then, taking adjoints we see that \langle v\rvert B^*\lvert v\rangle=\langle v\rvert B\lvert v\rangle. Thus the transformation B-B^* must always give zero when we feed it two copies of the same vector.

But now we have the polarization identities to work with! The real and imaginary parts of \langle v\rvert B-B^*\lvert w\rangle are completely determined in terms of expressions like \langle u\rvert B-B^*\lvert u\rangle. But since these are always zero, so is the rest of the form. And thus we conclude that B=B^*. That is, positive-definiteness only really makes sense for Hermitian transformations.

Actually, this all sort of makes sense. Self-adjoint transformations (symmetric or Hermitian) are analogous to the real numbers sitting inside the complex numbers. Within these, positive-definite matrices are sort of like the positive real numbers. It doesn’t make sense to talk about “positive” complex numbers, and it doesn’t make sense to talk about “positive-definite” transformations in general.

Now, there are three variations that I should also mention. The most obvious one is for a transformation to be “negative-definite”. In this case, we have \langle v\rvert H\lvert v\rangle\leq0, with equality only for v=0. We can also have transformations which are “positive-semidefinite” and “negative-semidefinite”. These are just the same as the definite versions, except we don’t require that equality only obtain for v=0.

July 13, 2009 - Posted by | Algebra, Linear Algebra


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  2. […] this last inequality holds because the bra-ket pairing is an inner product, and is thus positive-definite. Indeed, a positive-definite (or negative-definite) form must be nondegenerate. Thus it is […]

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  3. […] off, note that no matter what we use, the transformation in the middle is self-adjoint and positive-definite, and so the new form is symmetric and positive-definite, and thus defines another inner product. […]

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  4. […] if is not only self-adjoint, but positive-definite? We would like the determinant to actually be a positive real […]

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  5. […] going to want to save and to denote transformations). We also have its adjoint . Then is positive-semidefinite (and thus self-adjoint and normal), and so the spectral theorem applies. There must be a unitary […]

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  6. […] that unitary transformations are like unit complex numbers, while positive-semidefinite transformations are like nonnegative real numbers. And so this “polar decomposition” is like the polar form of a complex number, where we […]

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  7. […] Here’s a neat thing we can do with the spectral theorems: we can take square roots of positive-semidefinite transformations. And this makes sense, since positive-semidefinite transformations are analogous to nonnegative […]

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  8. […] if all of these eigenvalues are positive at a critical point , then the Hessian is positive-definite. That is, given any direction we have . On the other hand, if all of the eigenvalues are negative, […]

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  9. […] already seen that the composition of a linear transformation and its adjoint is self-adjoint and positive-definite. In terms of complex matrices, this tells us that the product of a matrix and its conjugate […]

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