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Let $A,B$ be two $n \times n$ complex matrices which have the same minimal polynomial $M(t)$ and the same characteristic polynomial $P(t) = (t-\lambda_1)^{a_1}\cdots(t - \lambda_k)^{a_k},$ where $\lambda_i \neq \lambda_j$ for $i \neq j.$

Prove if $P(t)/M(t) = (t-\lambda_1)\cdots(t-\lambda_k),$ then $A$ and $B$ are similar.

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  • $\begingroup$ I was previously under the impression that if two matrices had the same minimal and characteristic polynomials, they had to be similar (apparently wrong?). I'm not sure why this specific condition implies similarity. I'm still trying to build up intuition on how to start these type of problems. The condition says that the geometric multiplicity of the eigenvalues is one less than algebraic, but I don't see how this helps deduce anything about $A \sim B.$ I've tried writing down everything I know about M and P, properties of similar matrices, and so on, and seeing what implies what. $\endgroup$ – Merkh Apr 27 '16 at 18:23
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We use the following key facts.

i) $P(t)$ is the product of all its invariant factors.

ii) if $f_1, \cdots f_k$ are you invariant factors, we have $f_1 | f_2 | \cdots | f_k$.

ii) The minimal polynomial is the largest invariant factor.

iii) $M(t)$ has the same irreducible factors as $P(t)$.

iv) If two matrices have the same invariant factors, they will have the same Jordan form and thus be similar.

From what you stated, $P(t)/ M(t)$ is a product of distinct linear factors.

What are the possible invariant factors?

$M(t)$ is one of them and from what you have above, $$M(t) = ( t - \lambda_1)^{a_1-1} \cdots (t-\lambda_n)^{a_n-1} $$ The next one must divide $M(t)$.

Using the properties above, what is your only other invariant factor?

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  • $\begingroup$ Im a little shaky on what exactly are invariant factors, but from the properties above, specifically the first one, the only other invariant factor is $(t-\lambda_1)\cdots(t-\lambda_k).$ So from this, we use the fact that A and B have the same invariant factors (M and P/M) to deduce that they have the same Jordan form and we are done. Does that seem like the correct reasoning? I have some reading up to do on invariant factors and such. Do you have an good references where I could read about this stuff? $\endgroup$ – Merkh Apr 27 '16 at 18:36
  • $\begingroup$ Yeah that's what I had in mind. There is this good linear algebra book here google.com/url?sa=t&source=web&rct=j&url=http://… $\endgroup$ – Danny Lara Apr 27 '16 at 18:43

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