Let $F=\mathbb{R}$ or $\mathbb{C}$.

I am wondering whether a set of invariants (characteristic polynomial, a set of eigen-values, the set of multiplicity of each eigen-values) exactly characterizes the similar relation of matrix.

More precisely, let $A,B\in M_{n \times n}(F)$ be the two matrices whose characteristic polynomials and total set of eigen-values are same. Furthermore, if $\lambda\in F$ is any common eigen-value of them, then we suppose that the algebraic multiplicity of $\lambda$ of $A$ and $B$ are same and ditto for their geometric multiplicities.

In this situation, can we say that $A$ and $B$ are similar? That is, is there an invertible matrix $U\in M_{n \times n}(F)$ such that $U^{-1}AU=B$?

I think it may not true. But I have some trouble in finding counter-example.

Any comments on this will be highly appreciated.

  • $\begingroup$ I also know the minimal polynomial is also invariants. But I am not sure it is independent with the above set of invariants. If it is independent invariants, then the set of invariants including the above and was added 'minimal polynomial' completely determines similarity relation? $\endgroup$ – user29422 May 10 '17 at 9:36
  • $\begingroup$ The answer lies in the world of "Jordan normal form". (Maybe there are other normal forms, also, I am not sure). $\endgroup$ – Giuseppe Negro May 10 '17 at 9:40
  • $\begingroup$ You show good taste by homing in on the characteristic and the minimal polynomials: these are the two ends of a set of polynomials called the invariant factors of the matrix which are what you want. This works ovre any field. Rational Canonical Form is the code. $\endgroup$ – ancient mathematician May 10 '17 at 9:58
  • $\begingroup$ Thanks all. I didn't think Jordan canonical form. I found that all invariants I wrote above including minimal polynomial does not completely characterize similarity, $\endgroup$ – user29422 May 10 '17 at 11:37

No, you can not. If you take all the invariants you gave, you have to make your matrices a little bigger, but there are still counterexamples (I think $8 \times 8$ should work, but I'm not sure). An invariant is the Jordan normal form. There is also a generalized Jordan normal form which will work over the reals. But taking characteristic polynomial, minimal polynomial, eigenvalues, multiplicities, etc. will not work if the matrices get to big. In fact, you can generate counterexamples by using Jordan normal forms, you just have to know how to read off all these invariants from the normal form.

  • $\begingroup$ Thanks. After studying Jordan normal form, I constructed the counter example like $A= \left[ {\begin{array}{cc} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array} } \right]$ and $$B= \left[ {\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{array} } \right] $$. Then they are not similar but their invariants I said above are all same. So I found that the set of invariants I wrote above does not completely determines similarity. $\endgroup$ – user29422 May 10 '17 at 11:31
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    $\begingroup$ These matrices are similar, because they have the same Jordan normal from. So it is not the counterexample. $\endgroup$ – Hedgehog May 10 '17 at 15:02
  • $\begingroup$ @Hedgehog, Really? I thought they are already in Jordan normal form. If not, what are their Jordan normal forms? $\endgroup$ – user29422 May 11 '17 at 6:32
  • $\begingroup$ Your matrices have the same two blocks, just swapped. As permuting the blocks gives equivalent matrices (just take the permutation matrix to conjugate them), the Jordan form has, somewhere in its definition, a unique ordering of blocks (e.g. from small to big). Thus these two matrices either "are" the same Jordan form, or one of them is not in normal form, you should consult your lecture notes about that. $\endgroup$ – Dirk May 11 '17 at 9:39
  • $\begingroup$ @Bemte, Thanks. I got to know that Jordan normal form of a matrix is determined upto permutation of its blocks. Then what you meant in the above is that to construct a counter-example of matrices which has same invariants all I wrote but not similar, should I go up to at least ($8 \times 8$) size matrix? That's formidable. If we exclude minimal polynomial invariants, I could find counter-example in $(3\times 3)$ matrix. $\endgroup$ – user29422 May 11 '17 at 9:49

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