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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. $A=\lambda I$.

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I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware about the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. – Martin Sleziak Aug 7 '12 at 8:58
@Martin Thanks, am now made aware. Don't think I will be near hitting the limit though. And I will try to be judicious. :) – Ryan Aug 7 '12 at 9:13
up vote 4 down vote accepted

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.

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Note that the same type of example (an upper-triangular matrix of all ones) works for any $n$. – Alex Becker Aug 7 '12 at 8:59
Thanks! Small typo in your answer, I think (missing the scalar multiple on the RHS of the second eqn) – Ryan Aug 7 '12 at 9:08
@Ryan Since the scalar is $1$, it can be omitted. :) – Alex Becker Aug 7 '12 at 9:12
Weren't you equating $Ax=kx$? Oh nevermind, this is inconsequential. – Ryan Aug 7 '12 at 13:27

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.)

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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.

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