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Geometric multiplicity of an eigenvalue of a matrix is the dimension of the corresponding eigenspace. The algebraic multiplicity is its multiplicity as a root of the characteristic polynomial.

It is known that the geometric multiplicity of an eigenvalue cannot be greater than the algebraic multiplicity. This fact can be shown easily using the Jordan normal form of a matrix.

I was wondering if there is a more elementary way to prove this fact, possibly longer but without using the Jordan normal form? (This is an exercise in Kreyszig's book on functional analysis, and given the author's style, I suspect that he did not intend the solution to use Jordan form, because otherwise I guess he would have given a hint about that. But I might be wrong.)

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up vote 17 down vote accepted

You don't need the Jordan form: suppose the geometric multiplicity of $\lambda$ is $k$, and let $\gamma=\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ be a basis for the corresponding eigenspace. Extend the basis $\gamma$ to a basis $\beta$ for $F^n$, and let $Q$ be the change-of-basis matrix. Then the characteristic polynomials of $A$ and $Q^{-1}AQ$ are the same. The upper left $k\times k$ block of $Q^{-1}AQ$ is simply $\lambda I_k$, and, the $(n-k)\times k$ block under it is all zeroes. So the characteristic polynomial of $Q^{-1}AQ$ is a multiple of $(\lambda - t)^k$ hence the algebraic multiplicity of $\lambda$ is at least $k$.

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Seems like this is exactly what I was looking for. Thanks! – AgCl Oct 8 '10 at 4:37

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