# The Unapologetic Mathematician

## 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 $v=0$.

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Posted by John Armstrong | Algebra, Linear Algebra

## 9 Comments	»

1. [...] Forms I The notion of a positive semidefinite form opens up the possibility that, in a sense, a vector may be “orthogonal to itself”. That [...]

Pingback by | July 15, 2009 | Reply

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 [...]

Pingback by | July 17, 2009 | Reply

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. [...]

Pingback by | July 27, 2009 | Reply

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

Pingback by | August 3, 2009 | Reply

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 [...]

Pingback by | August 17, 2009 | Reply

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 [...]

Pingback by | August 19, 2009 | Reply

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 [...]

Pingback by | August 20, 2009 | Reply

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, [...]

Pingback by | November 24, 2009 | Reply

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 [...]

Pingback by | October 13, 2010 | Reply

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