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I gotta show if or if not those $ 2 $ matrices are similar:

$$ \left(\begin{matrix} 3 & 2 & -2 \\ 1 & 4 & 0 \\ -2 & 1 & -1 \\ \end{matrix}\right) $$

$$ \left(\begin{matrix} 1 & 3 & -1 \\ 3 & 3 & 1 \\ -2 & 1 & 2 \\ \end{matrix}\right) $$

I already calculated their determinant but it's equal so nothing there. Is it possible to show it based on their polynominal?

Bach
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xim
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  • What are their eigenvalues? – John Douma Jan 21 '19 at 00:47
  • For first matrice, the characteristic polynomial is -λ^3+6λ^2+λ-28 and for second matrice, the characteristic polynomial, is the same, unless i'm wrong. I'm having difficulty solving eigenvalues out of that (degree 3). – xim Jan 21 '19 at 01:06
  • Did you try to see if it has simple roots ? – Dlem Jan 21 '19 at 01:15
  • Similar as matrices over which field? – Adam Higgins Jan 21 '19 at 01:15
  • Then you are done. – John Douma Jan 21 '19 at 01:22
  • @AdamHiggins not sure to understand your question? both of matrices are in a field f and are 3 by 3 i guess? – xim Jan 21 '19 at 01:22
  • @JohnDouma if they have same characteristic polynomial, they're always similar? – xim Jan 21 '19 at 01:22
  • Not in general. Can you diagonalize each matrix? – John Douma Jan 21 '19 at 01:24
  • I'm not sure how to, i never did the eigenvalues with a third degree polynomial. To diagonalize i would need the values. – xim Jan 21 '19 at 01:27
  • @xim Two square $(n \times n)$-matrices $M,N$ are similar over a ring $R$ if and only if there exists an invertible $(n \times n)$-matrix $P$ $\textbf{with entires in}$ $R$ such that $M = P^{-1}NP$. Thus determining whether two matrices are similar requires one to specify which ring you mean. – Adam Higgins Jan 21 '19 at 01:31
  • @AdamHiggins I would be shocked if it wasn't the real numbers. This looks like a first course in Linear Algebra. – John Douma Jan 21 '19 at 01:34
  • @AdamHiggins with real numbers i would of done it pretty easily, i just have no clue on this one i been trying since 2 hours. – xim Jan 21 '19 at 01:37
  • A couple hints. The minimum polynomial of both matrices is $f(x) = -x^{3} + 6x^{2} + x - 28$. Can you see why the fact that $f(4) = 8 > 0$ gives that $f(x)$ has three distinct real roots? And can you then conclude from this that both matrices has a complete set of eigenvectors, with both having the same eigenvalues? And then can you conclude on whether or not they are similar from this? – Adam Higgins Jan 21 '19 at 01:45
  • why did you pick f(4)? i'm not sure what this has to do with the problem i'm a bit lost :D usually i'd find the characteristic polynomial, find the eigenvalue, find the corresponding eigenvectors linked to each eigenvalue. In this case, my polynomial doesn't give me full values of eigenvalues – xim Jan 21 '19 at 02:25

2 Answers2

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Hint: The two matrices have the same characteristic polynomial: $ p(x)=x^3-6x^2-x+28 $ and it has three different roots in $ \mathbb R $(Consider $ p(0) $ and $ p(3) $ and use the intermediate value theorem). So the two matrices are similar.

Bach
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For matrices to be similar, it is necessary but not sufficient for the characteristic polynomials to be the same. It is actually sufficient in this case though, since both matrices appear to be diagonalizable with distinct diagonal entries.

Here's a little background: recall that the obstruction to being diagonalizable over an algebraically closed field is the difference between "algebraic multiplicity" and "geometric multiplicity" of the eigenvalues. Geometric multiplicity is at least one for each eigenvalue, and less than or equal to algebraic multiplicity (degree of the root in the characteristic polynomial). But here, the roots of the characteristic polynomial are all distinct. This makes things much easier for us, since the geometric multiplicity of each eigenvalue is at least one, and there are three distinct eigenvalues. That means the sum of the geometric multiplicities is $3$, so our matrix is diagonalizable. Diagonalizable matrices with the same characteristic polynomial will be both similar to the same diagonal matrix, so this solves the problem.

In general, over $\mathbb{C}$, the Jordan decomposition for each matrix is a complete invariant for similarity of matrices. Over $\mathbb{R}$, the rational canonical form is a complete invariant. That is, the matrices are similar if and only if they have the same decomposition (up to the symmetries like permuting blocks).

Ronald J. Zallman
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