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Let $M$ be a complex square matrix such that it has the following:

Minimal polynomial: $\mu(\lambda)= (\lambda-3)^2(\lambda+2)^3(\lambda-5)$

Characteristic polynomial: $\chi(\lambda)=(\lambda-3)^4(\lambda+2)^5(\lambda-5)^2$

$\dim(E_3)=2$ and $\dim(E_{-2})=3$, where $E_\lambda$ is the eigenspace for $\lambda \in \mathbb{R}\,$.

Now, my question is this:
If $N$ is another complex square matrix with the same characteristic polynomial, minimal polynomial, and geometric multiplicities as stated above, must there be an invertible matrix $Q$ such that $Q^{-1}MQ=N$ ?

If this is true, I would like to understand why. And if not, a counter-example would be much appreciated.

Hanno
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FireFenix777
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2 Answers2

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All of the information that you have given can be interpreted as a description of the Jordan normal form of $M$. In particular:

Characteristic polynomial:

  • $M$ has size $\deg(\chi) = 11$
  • The sum of the sizes of the Jordan blocks associated with $\lambda = 3$ (i.e. the number of times $3$ appears on the diagonal) is $4$. This sum is $5$ for $\lambda = -2$ and $2$ for $\lambda = 5$.

Minimal polynomial:

  • The largest block associated with $\lambda = 3$ has size $2$. Respectively, $3$ for $\lambda = -2$ and $1$ for $\lambda = 5$.

Algebraic multiplicity:

  • There are $2$ Jordan blocks associated with $\lambda = 3$ and $3$ for $\lambda = -2$.

Let $J_k(\lambda)$ denote the Jordan block of size $k$ associated with $\lambda$, and let $\oplus$ denote a direct sum. Verify that with the information given above, the Jordan form of $M$ must be (up to a permutation of the blocks)

$$ J = [J_2(3) \oplus J_2(3)] \oplus [J_3(-2) \oplus J_1(-2) \oplus J_1(-2)] \oplus [J_1(5) \oplus J_1(5)]. $$

Two matrices are similar if and only if they have the same Jordan form. However, any matrix $N$ that has all of the listed characteristics must also have this Jordan form. It follows that for any such $N$ is similar to $M$, i.e. there is a matrix $Q$ for which $ Q^{-1}MQ = N$.

Ben Grossmann
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  • I'm confused, doesn't the post you linked Hanno say that this need not be true for $n >6$? – FireFenix777 Jan 07 '21 at 22:46
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    @FireFenix777 It says that the characteristic polynomial and minimal polynomial being the same isn't always enough, which is why we had to interpret the information given to make sure that it really is enough in our case. – Ben Grossmann Jan 07 '21 at 23:36
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    @FireFenix777 In other words, we could find pairs of non-similar matrices that have the same characteristic and minimal polynomials and the same geometric mulitplicities, but we can't find two non-similar matrices that have the particular polynomials and multiplicities listed in your question. – Ben Grossmann Jan 07 '21 at 23:38
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Warning: This answer is incorrect, as currently given.
Correction is underway.

If you are given both the characteristic and the minimal polynomial of a matrix, then you know its Jordan canonical form (which is usually not unique because permutations of Jordan blocks are still allowed).
A matrix $M$ and its Jordan normal form $J$ are similar, which means $Q^{-1}MQ=J$ for some invertible matrix $Q$.

Your question is answered in the positive, because $M$ and $N$, sharing characteristic and minimal polynom, then also share the Jordan normal form, and both $M,N$ are similar to it. Now recall that similarity is an equivalence relation.

If you strive for uniqueness, then look after the Frobenius normal form of a square matrix.

Hanno
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