6
$\begingroup$

This question already has an answer here:

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?

$\endgroup$

marked as duplicate by Key Flex, user91500, Lord Shark the Unknown, mrtaurho, metamorphy Jan 21 at 11:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What are their eigenvalues? $\endgroup$ – John Douma Jan 21 at 0:47
  • $\begingroup$ 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). $\endgroup$ – xim Jan 21 at 1:06
  • $\begingroup$ Did you try to see if it has simple roots ? $\endgroup$ – DLeMeur Jan 21 at 1:15
  • $\begingroup$ Similar as matrices over which field? $\endgroup$ – Adam Higgins Jan 21 at 1:15
  • $\begingroup$ Then you are done. $\endgroup$ – John Douma Jan 21 at 1:22
1
$\begingroup$

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.

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.