# The Unapologetic Mathematician

## Eigenvectors of an Eigenpair

An eigenvalue $\lambda$ of a linear transformation $T$ is the same thing as a root of the characteristic polynomial of $T$. That is, the characteristic polynomial has a factor $(X-\lambda)$. We can evaluate this polynomial at $T$ to get the linear transformation $(T-\lambda I_V$. Vectors in the kernel of this space are the eigenvalues corresponding to the eigenvector $\lambda$.

Now we want to do the same thing with an eigenpair $(\tau,\delta)$. This corresponds to an irreducible quadratic factor $(X^2-\tau X+\delta)$ in the characteristic polynomial of $T$. Evaluating this polynomial at $T$ we get the transformation $T^2-\tau T+\delta I_V$, which I assert has a nontrivial kernel. Specifically, I want to focus in on some two-dimensional invariant subspace on which $T$ has no eigenvalues. This corresponds to a $2\times2$ block in an almost upper-triangular representation of $T$. So we’ll just assume for the moment that $V$ has dimension $2$.

What I assert is this: if $p$ is a monic polynomial (with leading coefficient ${1}$) of degree two, then either $p$ is the characteristic polynomial of $T$ or it’s not. If it is, then $p(T)=0$, and its kernel is the whole of $V$. If not, then the kernel is trivial, and $p(T)$ is invertible.

In the first case we can just pick a basis, find a matrix, and crank out the calculation. If the matrix of $T$ is

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

then the characteristic polynomial is $X^2-(a+d)X+(ad-bc)$. We substitute the matrix into this polynomial to find

\displaystyle\begin{aligned}&\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}-(a+d)\begin{pmatrix}a&b\\c&d\end{pmatrix}+(ad-bc)\begin{pmatrix}{1}&{0}\\{0}&{1}\end{pmatrix}=\\&\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}-\begin{pmatrix}a^2+ad&ab+bd\\ac+cd&ad+d^2\end{pmatrix}+\begin{pmatrix}ad-bc&{0}\\{0}&ad-bc\end{pmatrix}=\\&\begin{pmatrix}{0}&{0}\\{0}&{0}\end{pmatrix}\end{aligned}

On the other hand, if $q$ is the characteristic polynomial and $p$ is any other monic polynomial of degree two, then $q(T)=0$, as we just showed. Then we can calculate

$\displaystyle p(T)=p(T)-q(T)=aT+bI_V$

for some constants $a$ and $b$, at least one of which must be nonzero. If $a=0$, then $p(T)$ is a nonzero multiple of the identity, which is invertible as claimed. On the other hand, if $a\neq0$, then

$\displaystyle p(T)=a(T-\frac{b}{a}I_V)$

which must be invertible since we assumed that $T$ has no eigenvalues.

So for any $2\times2$ block, the action of an irreducible quadratic polynomial in $T$ either kills off the whole block or has a trivial kernel. This makes it reasonable to define an eigenvector of the eigenpair $(\tau,\delta)$ to be a vector in the kernel $\mathrm{Ker}\left(T^2-\tau T+\delta I_V\right)$, in analogy with the definition of an eigenvector of a given eigenvalue.

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April 3, 2009 - Posted by | Algebra, Linear Algebra

## 2 Comments »

1. [...] no eigenvalues, then the generalized eigenspace of any eigenvalue is trivial. On the other hand we’ve seen that the kernel of is either the whole of or nothing, and the former case happens exactly when [...]

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2. [...] then is either one- or two-dimensional. Then the statement reduces to what we worked out by cases earlier. So from here we’ll assume that , and that the statement holds for all matrices with [...]

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