# The Unapologetic Mathematician

## 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.

August 3, 2009 Posted by | Algebra, Linear Algebra | 7 Comments