The Determinant of a Positive-Definite Transformation

Let’s keep pushing the analogy we’ve got going.

First, we know that the determinant of the adjoint of a transformation is the complex conjugate of the determinant of the original transformation (or just the same, for a real transformation). So what about self-adjoint transformations? We’ve said that these are analogous to real numbers, and indeed their determinants are real numbers. If we have a transformation $H$ satisfying $H^*=H$ , then we can take determinants to find

$\displaystyle\det(H)=\det(H^*)=\overline{\det(H)}$

and so the determinant is real.

What if $H$ is not only self-adjoint, but positive-definite? We would like the determinant to actually be a positive real number.

Well, first let’s consider the eigenvalues of $H$ . If $\lvert v\rangle$ is an eigenvector we have $H\lvert v\rangle=\lambda\lvert v\rangle$ for some scalar $\lambda$ . Then we can calculate

$\displaystyle\langle v\rvert H\lvert v\rangle=\langle v\rvert\lambda\lvert v\rangle=\lambda\langle v\vert v\rangle=\lambda\lVert v\rVert^2$

If $H$ is to be positive-definite, this must be positive, and so $\lambda$ itself must be positive. Thus the eigenvalues of a positive-definite transformation are all positive.

Now if we’re working with a complex transformation we’re done. We can pick a basis so that the matrix for $H$ is upper-triangular, and then its determinant is the product of its eigenvalues. Since the eigenvalues are all positive, so is the determinant.

But what happens over the real numbers? Now we might not be able to put the transformation into an upper-triangular form. But we can put it into an almost upper-triangular form. The determinant is then the product of the determinants of the blocks along the diagonal. The $1\times1$ blocks are just eigenvalues, which still must be positive.

The $2\times2$ blocks, on the other hand, correspond to eigenpairs. They have trace $\tau$ and determinant $\delta$ , and these must satisfy $\tau^2<4\delta$ , or else we could decompose the block further. But $\tau^2$ is definitely positive, and so the determinant $\delta$ has to be positive as well in any of these blocks. And thus the product of the determinants of the blocks down the diagonal is again positive.

So either way, the determinant of a positive-definite transformation is positive.

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August 3, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

7 Comments »

Just a question: doesn’t the real case follow from the complex case (just treat the real transformation as a complex one in which all numbers happen to be real, the upper-triangularizing basis might not be real but who cares: the determinant is basis-independent)?

Comment by Piotr | August 3, 2009 | Reply
The line between “useful technique” and “dirty trick” is fine indeed.

Comment by John Armstrong | August 3, 2009 | Reply
@Piotr: since we seem to be talking about self-adjoint and positive definite transformations, in both the real and complex cases it turns out that such a transformation is diagonal with respect to some orthonormal basis. So certainly in the real case we can put the transformation into upper triangular and even diagonal form.

Comment by Todd Trimble | August 4, 2009 | Reply
Todd, that’s right, but I haven’t shown it yet, so I can’t use it. The easy way to show it, though, would use even more examination of eigenpairs than I did in the post.

Comment by John Armstrong | August 4, 2009 | Reply
Yeah, I figured you hadn’t (I tried the search terms “spectral” and “diagonalization” just to make sure). But I didn’t want Piotr to come away with a wrong impression that “Now we might not be able to put the transformation into an upper-triangular form” means literally what it says — there is an implicit preface “for all we know from our development thus far” in there.

Comment by Todd Trimble | August 4, 2009 | Reply
There’s an interesting philosophical point there: the difference between mathematics-as-completed-entity and mathematics-situated-in-context.

I think that the transition is a subtle one, but an important one in terms of the growth of a mathematical student. Through most of school, mathematics can be (and is) taken as a completed thing given down from on-high. It takes a definite shift in perspective to see it as incomplete and in-development.

In a sense, one could say that the mathematics I’m talking about here is different than, say, the mathematics in Axler’s text, because I started at a different place and made different decisions along the way. There’s a sort of “path-dependency” to mathematics. The question, then, is how all these paths converge on the same results? How can the Secret Blogging Seminar doesn’t have to start at the beginning, but can assume a general, common background?

Comment by John Armstrong | August 4, 2009 | Reply
[…] is a unit complex number, and the determinant of a positive-semidefinite transformation is a nonnegative real number. If is nonsingular, so is actually positive-definite, then will be strictly positive, so the […]

Pingback by Polar Decomposition « The Unapologetic Mathematician | August 19, 2009 | Reply

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