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I was wandering if there is a chance that two real matrices with the same characteristic polynomial have a different rank? I tried to prove it, but i failed. any suggestions?

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Take the zero two by two matrix and a nonzero zero two by two nilpotent matrix. –  Pierre-Yves Gaillard Sep 18 '11 at 15:51
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Here's a related question: Can two real matrices with the same minimal polynomial have different rank? –  alex.jordan Sep 18 '11 at 18:50

2 Answers 2

$\pmatrix{0 & 1\\ 0&0}$ and $\pmatrix{0&0\\0&0}$ both have characteristic polynomial $\lambda^2$.

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The dimension of the nullspace of a matrix is equal to the geometric multiplicity of eigenvalue 0 (this is rather easy to prove). So, if the geometric multiplicity was always equal to the algebraic multiplicity, you would have your result. But as you probably know, this is not always the case.

Counterexamples can be constructed as Jordan matrices. For the area that corresponds to eigenvalue 0, form different number of Jordan blocks. Since similar matrices have the same characteristic polynomial and all matrices are similar to a Jordan matrix, this accounts for all cases (up to similarity transform).

For example $$\begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad \begin{pmatrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ both have characteristic polynomial $2\lambda^2 - \lambda^3$ but the first have a two-dimensional nullspace and the second a one-dimensional nullspace.

However, if your matrices have eigenvalue 0 with algebraic multiplicity at most one, they have the same rank.

Here's an example of two matrices having the same minimal polynomial but different rank:

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

Both matrices have minimal polynomial $\lambda^2$. The first matrix has a 3-dimensional nullspace and the second a 2-dimensional nullspace.

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