0
$\begingroup$

For what $x$ are the following two matrices similar?

$$A = \begin{pmatrix} 3&0&-1 \\ -2&1&1 \\ 2&0&0 \end{pmatrix}, B = \begin{pmatrix} 1&x&0 \\ 0&1&0 \\ -1&x&2 \end{pmatrix}$$

I know some necessary conditions for two matrices to be similar. For example, same eigenvalues, same characteristic polynomial. In this case, the characteristic polynomials of both matrices are $(1-\lambda)^2(2-\lambda)$. But this does not tell me anything about $x$?

$\endgroup$
2
$\begingroup$

Since $A$ is similar $B$ so $A,B$ have the same characteristic and also same minimal polynomial.

Check that minimal polynomial of $A$ is $(x-2)(x-1)$.

So $B$ satisfies $(B-2I)(B-I)=0\implies x=0$

$\endgroup$
  • $\begingroup$ How did you find the minimal polynomial of $A$ so quickly (if that's the case)? $\endgroup$ – 3x89g2 Nov 5 '16 at 16:46
  • $\begingroup$ quickly,what? I had to compute it;took 10 minutes $\endgroup$ – Learnmore Nov 5 '16 at 16:50
  • $\begingroup$ fair enough. thank you :) $\endgroup$ – 3x89g2 Nov 5 '16 at 16:50
  • $\begingroup$ @learnmore, x can be anynumber for A and B are similars. $\endgroup$ – retro_var Nov 5 '16 at 19:03
  • $\begingroup$ @testpilot Why? $\endgroup$ – 3x89g2 Nov 5 '16 at 22:24
1
$\begingroup$

There is a general method to determine whether two matrices are similar, which involves computing and comparing the Rational Canonical Form of the two matrices; however this is rather a lot of work (and in addition does not play well with the presence of undetermined entries like $x$) while one can do a lot better in small size examples like this one.

You calculated the characteristic polynomials and found both to be $(X-1)^2(X-2)$, independently of$~x$. The eigenvalue $\lambda=2$ will necessarily given an eigenspace of dimension$~1$ and causes no concern. For the eigenvalue $\lambda=1$ with algebraic multiplicity$~2$ you will have a generalised eigenspace $\ker((M-I)^2)$ that has dimension$~2$ for both matrices$~M$, but two possibilities exist for the (ordinary) eigenspace $\ker(M-I)$: either it is a subspace of dimension$~1$ of the generalised eigenspace, or (much less likely) it has dimension$~2$ and fills the generalised eigenspace. You can check that the latter happens for $M=A$ (the vectors $(0,1,0)$ and $(1,0,2)$ generate the eigenspace, but not for$~B$ (the vector $(1,0,1)$ spans the kernel) except when $x=0$, in which case (for instance) $(0,1,0)$ which always belongs to the generalised eigenspace actually belong to$~\ker(B-I)$.

In conclusion, for $x\neq0$ one has $A$ diagonalisable but $B$ not, so they are not similar. But for $x=0$ both matrices are diagonalisable, and also having the same characteristic polynomial, they are clearly similar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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