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I'm reading my textbook and I'm really confused about geometric multiplicity. I've read the definition and they have given an example but I'm still lost. I've tried looking it up on other websites. This helped me make sense of algebraic multiplicity but I don't understand geometric multiplicity at all. I was wondering if someone could explain this as simply as possible.

ayv2
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  • Could you please add the definition that you know about this multiplicity? Did you define it in terms of the Jordan normal form or in terms of dimension of eigenspaces and generalized eigenspaces or just with the characteristic and minimal polynomial? – Lutz Lehmann Feb 25 '14 at 07:14
  • The book says that geometric multiplicity of an eigenvalue lambda to be dimE(lambda), the dimension of its corresponding eigenspace. I have no idea what this definition is saying. – ayv2 Feb 25 '14 at 07:17
  • Geometric multiplicity, as you say, is the number of linearly independent eigenvectors related to a given eigenvalue. Whereas the algebraic multiplicity is the dimension of the invariant subspace. In general $1\le$ geometric $\le$ algebraic. Think for example of the basic 2 times 2 nilpotent matrix. It has only one eigenvector but the algebraic multiplicity is 2. – lcv Feb 25 '14 at 07:31
  • then also this might help to get an understanding for it: http://en.wikipedia.org/wiki/Jordan_normal_form – Max Feb 25 '14 at 07:31
  • See http://www.axler.net/DwD.html. Funny fact: the algebraic multiplicity is so "geometric" as the geometric multiplicity. – Martín-Blas Pérez Pinilla Feb 25 '14 at 07:47

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The geometric multiplicity of an eigenvalue is the dimension of its corresponding eigenspace. For example, let $$ A= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ Then $\operatorname{char}_A(\lambda)=(\lambda-1)^2$ so the only eigenvalue of $A$ is $\lambda_1=1$. Since the degree of the $(\lambda-1)$ term of the characteristic polynomial is $2$, we say that $\lambda_1$ has algebraic multiplicity two.

Now, we wish to find all of the eigenvectors of $A$ corresponding to $\lambda_1$. That is, we want all vectors $v$ such that $Av=v$ (see if you can do this yourself). By inspecting the equation $Av=v$, we see that all of the eigenvectors of $A$ corresponding to $\lambda_1$ are of the form $$v=a\cdot\begin{pmatrix}1\\ 0\end{pmatrix}\qquad a\in\mathbb R$$ That is, the eigenspace of $A$ corresponding to $\lambda_1$ is $$ E_{\lambda_1}=\operatorname{Span}\left\{\begin{pmatrix}1\\ 0\end{pmatrix}\right\} $$ Hence $\dim E_{\lambda_1}=1$ and the geometric multiplicity of $\lambda_1$ is one (note that this is different from the algebraic multiplicity).

Brian Fitzpatrick
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  • Okay, so they have this example in the book where the span for one root is [0 1 0] and [1 0 1]. The second root had a span of [-1 3 1]. They said that the geometric multiplicity is 2. So, why am I not counting span[-1 3 1]? – ayv2 Feb 25 '14 at 07:30
  • Because you treat each root differently. The geometric multiplicity of the second root would be 1. – Brian Fitzpatrick Feb 25 '14 at 07:30
  • oh, okay. Thank you for the help. – ayv2 Feb 25 '14 at 07:32