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Let $A$ and $B$ be matrices with coefficients in some field $k$. If they both have characteristic polynomial $x^5-x^3$ and minimal polynomial $x^4-x^2$, then they must be similar.

I've thought about this for awhile now, and I still have not made any progress.

GillyB
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  • Hint: you can prove $\det(xI-A)$ equals $\det(xI-P^{-1}AP)$ for an invertible matrix $P$ – mate89 Sep 27 '17 at 01:48
  • Yes. Similar matrices have the same characteristic polynomial, but I'm not sure what you're getting at. – GillyB Sep 27 '17 at 01:51
  • Yeah, that's not helpfull here. Hmmm the case is always true for $n \le 3$. See http://math.stackexchange.com/a/56725 – mate89 Sep 27 '17 at 02:22

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Do you know about Jordan Canonical Form?

Two matrices are similar if and only they have the same Jordan Canonical Forms (same meaning up to rearrangements of the Jordan Blocks).

So we first factor $x^5 - x^3 = x^3 \cdot (x-1)(x+1)$ which means they have eigenvalues (including multiplicity, obviously) $0, 0, 0, 1, -1$.

If your field is not of characteristic $2$: Since $1$ and $-1$ appear in the characteristic polynomial once, their Jordan blocks must have size $1$.$0$, a priori, could have a block of size $3$, a block of size $2$ and a block of size $1$, or three blocks of size $1$.

If your field is of characteristic $2$: Your eigenvalues are $0, 0, 0, 1, 1$. Block sizes for $0$ are the same, block sizes for $1$ are one block of size $2$ or two blocks of size $1$.

We factor their minimal polynomial as $x^4 - x^2 = x^2(x+ 1)(x-1)$. The minimal polynomial tells us the largest Jordan Block corresponding to the eigenvalue $0$ is of size exactly $2$. If we have one Jordan block of size exactly $2$, the other Jordan block is determined to be $1$.

If your field is of characteristic $2$, then you also get that the block with ones on the diagonal must be of size $2$ since the factor appears twice in the minimal polynomial.

So there is only one option for the Jordan Canonical form of a matrix with this characteristic polynomial and minimal polynomial,

In characteristic not $2$, a block of size $1$ with $1$ on the diagonal, a block of size $1$ with $-1$ on the diagonal, a block of size $2$ with $0$ on the diagonal, and a block of size $1$ with $0$ on the diagonal.

In characteristic $2$, a block of size $2$ with $1$ on the diagonal, a block of size $2$ with $0$ on the diagonal, and a block of size $1$ with $0$ on the diagonal.

As there is only one option for the Jordan Forms, they must have the same Jordan form, hence are similar.

Christian
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  • I'm aware of Jordan Canonical Forms for algebraically closed fields. Does this solution generalize to arbitrary fields? – GillyB Sep 27 '17 at 03:08
  • @GillyB Well Jordan Canonical form is only guaranteed to exist for algebraically closed fields, but if all of the roots of the minimal/characteristic polynomial lie in the field you're working in, the matrix can still be put into Jordan form, and results of uniqueness still hold. The field being algebraically closed only ensures that your field has all of the eigenvalues (This is the only place the algebraically closed condition is used). I just realized that what I gave doesn't work for Characteristic $2$ since $1 = -1$, but it is easily modifiable. – Christian Sep 27 '17 at 03:13
  • Interesting. For some reason this never occurred to me. – GillyB Sep 27 '17 at 03:20
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    You can also just extend your base field to an algebraically closed field, since extending the base field doesn't change whether two matrices are similar. – Eric Wofsey Sep 27 '17 at 03:21