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Let say I have a characteristic polynomial e.g. $(\lambda-1)^4(\lambda-2)^3$. How can I find the number of non-similar matrices with this characteristic equation?

hardmath
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Shubham Raghuvanshi
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  • If two matrices share the same set of distinct eigenvalues, they are similar, see this. – Bobson Dugnutt May 17 '17 at 12:33
  • @Lovsovs I don't think that that fact helps here, since the given eigenvalues are not distinct. – florence May 17 '17 at 12:36
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    @florence I interpreted OP's question as being about any characteristic polynomial (hence the "let's say"), and my link covers the case where the eigenvalues are distinct. – Bobson Dugnutt May 17 '17 at 12:38
  • When I saw the phrase "number of matrices" in the title, I expected to see a problem involving finite fields. But you make no statement about which field is used. – hardmath May 17 '17 at 12:41
  • @hardmath I think we can allow answer like $+\infty$ here. – Alex Vong May 17 '17 at 12:59
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    Number of matrices will be finite since we are not counting similar matrices. And it has nothing to do with finite fields it is just simple counting problem. – Shubham Raghuvanshi May 17 '17 at 13:07
  • @ShubhamRaghuvanshi: While it may seem irrelevant to you, I suspect this is because you assume the entries of the matrix are real numbers and thus as well the coefficients of the characteristic polynomial. If this is your assumption, please include that in the Question, esp. if you are asking about a general treatment. – hardmath May 17 '17 at 14:47

1 Answers1

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Since your characteristic polynomial "splits" into linear factors,

$$ (\lambda - 1)^4 (\lambda - 2)^3 $$

we can find a Jordan normal form (over the complex field) or a "real" Jordan normal form (over the field of real numbers) to represent each possible class of similar matrices having this characteristic polynomial.

For each eigenvalue $\lambda =1,2$ the algebraic multiplicity can be partitioned in several ways. The multiplicity $4$ of the first eigenvalue can be expressed:

$$ 4 = 1+1+1+1 $$ $$ 4 = 1+1+2 $$ $$ 4 = 1+3 $$ $$ 4 = 2+2 $$ $$ 4 = 4 $$

In terms of the Jordan blocks, there could be four blocks of size $1$ (geometric multiplicity $4$), two blocks of size $1$ and one of size $2$, one block of size $1$ and one of size $3$, two blocks each of size $2$, or a single block of size $4$.

Similarly the second eigenvalue with algebraic multiplicity $3$ can be expressed in partitions:

$$ 3 = 1+1+1 $$ $$ 3 = 1+2 $$ $$ 3 = 3 $$

So we have five possibilities for the Jordan blocks of $\lambda = 1$ and three possibilities for the Jordan blocks of $\lambda = 2$. Altogether there are fifteen ($5\times 3$) possible similarity classes if the matrix is considered over the real or the complex numbers.

hardmath
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