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Let $A \in M_n(\mathbb R)$ and suppose its minimal polynomial is: $M_A(t)=\prod_{i=1}^{k}(t-\lambda_i)^{\textstyle s_i}.$

When $\lambda _1,\lambda_2,\lambda _3,......,\lambda _k$ are distinct eigenvalues.

We define a new matrix: $B\in M_{2n}(\mathbb R)$ by: $\left(\begin{matrix} A &I_n \\ 0 & A \end{matrix}\right)$

I need to prove that the minimal polynomial of $B$ is $M_B(t)=\prod_{i=1}^k(t-\lambda_i)^{\textstyle s_{i}+1}.$


Edit: What follows below refered to the original version of this question, in which the definition of $B$ was: $B = \left(\begin{array}{cc}A&I_n\\I_n&A\end{array}\right).$


I tried to do that in induction. I don't understand this basic case: for $1$x$1$ matrix: $\begin{pmatrix} 5 \end{pmatrix}$ we get that $ \begin{pmatrix} 5 & 1\\ 1& 5 \end{pmatrix}$ 's minimal polynomial is $(x-4)(x-6)$ and it doesn't answer the condition. so I have also a problem with understanding the question I guess.

Thanks!!

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    @Joze$f$: Robert Israel has written what I would have written, more or less.2011-08-22

2 Answers 2

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For the matrix $B = \pmatrix{A & I \cr 0 & A\cr}$ and any polynomial $p$, note that p(B) = \pmatrix{ p(A) & p'(A)\cr 0 & p(A)\cr}. If the polynomials $p(t)$ and p'(t) are both divisible by $(t - \lambda)^j$, then $p(t)$ must be divisible by $(t - \lambda)^{j+1}$.

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    @Jozef: It *must* be divisible by $(t-\lambda)^j$; in fact, it's divisible by an even larger power of $(t-\lambda)$. In general: if $g(x)$ is an irreducible factor of $p(x)$, write $p(x) = g(x)^mh(x)$, where $h(x)$ is not divisible by $g(x)$. Use the Product Rule to show that $g^{m-1}(x)$ divides both $g(x)$ and $g'(x)$, and that $g^m(x)$ does not divide $g'(x)$.2011-08-22
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Prove by induction that for any polynomial $P$, $ P(B) = P\left( \begin{bmatrix} A & I \\ I & A \end{bmatrix} \right) = \frac{1}{2} \begin{bmatrix} P(A+I)+P(A-I) & P(A+I)-P(A-I) \\ P(A+I)-P(A-I) & P(A+I)+P(A-I) \end{bmatrix}. $

Hence $P(B)=0 \Leftrightarrow P(A+I) = P(A-I) = 0.$ You take it from here.

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    This answers an earlier version of the question.2011-08-22