4
$\begingroup$

Is it possible for a matrix with characteristic polynomial $(λ−a)^3$ to have an eigenline (one-dimensional eigenspace)?

I know that geometric multiplicity can generally be smaller than algebraic multiplicity. But I was wondering if algebraic multiplicity $n$ might be a special case. This question is motivated by my earlier one The greatest possible geometric multiplicity of an eigenvalue, where I learnt that $A$ has an $n$-dimensional eigenspace iff. $A=\lambda I$.

  • 0
    @Martin Thanks, am now made aware. Don't think I will be near hitting the limit though. And I will try to be judicious. :)2012-08-07

3 Answers 3

4

Sure. Consider the matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ which has a single eigenvalue $1$ of multiplicity $2$, yet the only solutions to the equation $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} x +y \\ y\end{pmatrix}=\begin{pmatrix} x \\ y\end{pmatrix}$ come when $y=0$, and so the eigenspace is $1$-dimensional.

  • 0
    Weren't you equating $Ax=kx$? Oh nevermind, this is inconsequential.2012-08-07
1

Yes. Take for example $A = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ whose characteristic polynomial is $(\lambda-1)^3$, but whose eigenspace is one-dimensional (spanned by $(1,0,0)$.)

If the matrix is symmetric, this can't happen though. (This is a nice exercise to show.)

1

Algebraic multiplicity $n$ is not a special case. Take the following operator for example:

$ T(w, z) = (z, 0) $

It has the following matrix in the standard base:

$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $

The characteristic polynomial is $z^2$. Yet, $\operatorname{null}T$ is one-dimensional (spanned by $(1, 0)$).

Algebraic multiplicity of an eigenvalue $\lambda$ is equal to $\operatorname{null}(T - \lambda I)^{\operatorname{dim}V}$. In this example, $\operatorname{null}T^2$ does indeed have 2 dimensions.