# The Unapologetic Mathematician

## Eigenpairs

Well, Wednesday I was up at the University of Pennsylvania again, and yesterday I was making arrangements for a visit to San Diego in a couple weeks. And next week is an exam week, so I’ll have to inch forward today.

We’ve seen a lot about Jordan normal forms, which can pretty much capture the behavior of any single linear transformation over an algebraically closed field. But not all fields are algebraically closed, and one of them is very important to us. We want to investigate the situation over the field $\mathbb{R}$ of real numbers a little more deeply.

The key point about algebraically closed fields is that we can find some upper-triangular matrix. And the crux of that is the fact that any linear transformation has at least one eigenvalue. And that happens because the characteristic polynomial always has a root over an algebraically closed field. So if your field isn’t algebraically closed a characteristic polynomial might not have roots, and your transformation might have no eigenvalues.

And indeed, some real polynomials have no roots. But all is not lost! We do know something about factoring real polynomials. We can break any one down into the product of linear terms like $(X-\lambda)$ and quadratic terms like $(X^2-\tau X+\delta)$. If we’re factoring the characteristic polynomial of a linear endomorphism $T$, then a linear term $(X-\lambda)$ gives us an eigenvalue $\lambda$, so the new and interesting stuff is in the quadratic terms. I’m going to use the nonstandard term “eigenpair” to describe a pair of real numbers $(\tau,\delta)$ that shows up in this way.

If we were working over the complex numbers, we could factor a quadratic term into a pair of linear terms:

$\displaystyle X^2-\tau X+\delta=\left(X-\frac{\tau+\sqrt{\tau^2-4\delta}}{2}\right)\left(X-\frac{\tau-\sqrt{\tau^2-4\delta}}{2}\right)$

which gives us two complex eigenvalues

$\displaystyle\frac{\tau\pm\sqrt{\tau^2-4\delta}}{2}$

This gives us no problem over the real numbers if $\tau^2\geq4\delta$, so an eigenpair must have $\tau^2<4\delta$. In this case the two complex roots are a conjugate pair. Their sum is $\tau$, and their product is $\delta$.

So how can this arise in practice? Well, since it’s a quadratic term it’s the characteristic polynomial of an endomorphism on$\mathbb{R}^2$. So let’s write down a $2\times2$ matrix and take a look:

$\displaystyle\begin{pmatrix}a&b\\c&d\end{pmatrix}$

The characteristic polynomial is the determinant of ${X}$ times the identity matrix minus this matrix. We calculate

$\displaystyle(X-a)(X-d)-(-b)(-c)=X^2-(a+d)X+(ad-bc)$

So we can define $\tau$ to be the trace of this matrix, and $\delta$ to be its determinant. If $\tau^2<4\delta$, we’ve got an eigenpair.

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

## 1 Comment »

1. […] a factor of the characteristic polynomial of this formula is exactly what we defined to be an eigenpair. That is, just as eigenvectors — roots of the characteristic polynomial — correspond to […]

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