# The Unapologetic Mathematician

## Determining Eigenvalues

Yesterday, we defined eigenvalues, eigenvectors, and eigenspaces. But we didn’t talk about how to actually find them (though one commenter decided to jump the gun a bit). It turns out that determining the eigenspace for any given eigenvalue is the same sort of problem as determining the kernel.

Let’s say we’ve got a linear endomorphism $T:V\rightarrow V$ and a scalar value $\lambda$. We want to determine the subspace of $V$ consisting of those eigenvectors $v$ satisfying the equation $T(v)=\lambda v$

First, let’s adjust the right hand side. Instead of thinking of the scalar product of $\lambda$ and $v$, we can write it as the action of the transformation $\lambda1_V$, where $1_V$ is the identity transformation on $V$. That is, we have the equation $T(v)=\left[\lambda1_V\right](v)$

Now we can do some juggling to combine these two linear transformations being evaluated at the same vector $v$: $\left[T-\lambda1_V\right](v)=0$

And we find that the $\lambda$-eigenspace of $T$ is the kernel $\mathrm{Ker}(T-\lambda1_V)$.

Now, as I stated yesterday most of these eigenspaces will be trivial, just as the kernel may be trivial. The interesting stuff happens when $\mathrm{Ker}(T-\lambda1_V)$ is nontrivial. In this case, we’ll call $\lambda$ an eigenvalue of the transformation $T$ (thus the eigenvalues of a transformation are those which correspond to nonzero eigenvectors). So how can we tell whether or not a kernel is trivial? Well, we know that the kernel of an endomorphism is trivial if and only if the endomorphism is invertible. And the determinant provides a test for invertibility!

So we can take the determinant $\det(T-\lambda1_V)$ and consider it as a function of $\lambda$. If we get the value ${0}$, then the $\lambda$-eigenspace of $T$ is nontrivial, and $\lambda$ is an eigenvalue of $T$. Then we can use other tools to actually determine the eigenspace if we need to.

January 27, 2009 Posted by | Algebra, Linear Algebra | 4 Comments