Does the characteristic polynomial encode the rank?

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SUMMARY

The discussion centers on whether the rank of a matrix is encoded in its characteristic polynomial. It is established that for a matrix A, the characteristic polynomial can be expressed as x^rP(x), where P is a polynomial of degree n-r, indicating that A has rank n-r. However, this relationship holds true primarily for diagonalizable matrices. The algebraic multiplicity of eigenvalues must be considered, as demonstrated by the example of the matrix with a characteristic polynomial of x², which has a single eigenvector, indicating a rank of 1 despite the polynomial suggesting otherwise.

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Similar matrices share certain properties, such as the determinant, trace, eigenvalues, and characteristic polynomial. In fact, all of these properties can be determined from the character polynomial alone.

However, similar matrices also share the same rank. I was wondering if the rank is also encoded in the characteristic polynomial of the matrix.

In other words, if two matrices have the same characteristic polynomial, need their rank be the same?

I'd like to know the answer, so that I can decide whether to prove or to cook up a counterexample.

Thanks!

BiP
 
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Well, there is the obvious: the characteristic polynomial of matrix A is x^rP(x) where P is a n- r degree polynomial, if and only if A has rank n- r.
 
HallsofIvy said:
Well, there is the obvious: the characteristic polynomial of matrix A is x^rP(x) where P is a n- r degree polynomial, if and only if A has rank n- r.

This seems to be obviously true only for diagonalizable matrices. What if the matrix ##A## is not diagonalizable?

BiP
 
The algebraic multiplicity of eigenvalues are at least as large as the geometric multiplicities, so if 0 is an eigenvalue with dimension k eigenspace, then the characteristic polynomial has at least a factor of xk. It's not exactly xk though, for example the matrix

0 1
0 0

has characteristic polynomial x2 but 0 only has a single eigenvector
 

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