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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.


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First: I suspect that the power in what you are trying to prove is supposed to be $s_i + 1$, not $s_{i+1}$ (that is, the old power plus 1, rather than "the next power", which would not make sense for $i=k$). Second, I don't think this is true; are you positive about the definition of $B$? –  Arturo Magidin Aug 22 '11 at 18:36
dear @Arturo: you're right in your first Diagnosis.I'll edit it. Regarding the second, that's the way it is written, but come to think of it, in another version of this ex. I saw what instead of the left $I_n$ there was $0$.. –  Jozef Aug 22 '11 at 18:42
That would make a bit more sense, since then the result of raising $B$ to the $k$th power is a block upper triangular matrix with diagonal blocks equal to $A^k$; but the upper right block would be, if I'm not mistaken, $kA^{k-1}$. Still, you may want to try that. –  Arturo Magidin Aug 22 '11 at 18:48
Looking at my example, we can see that it's not correct, no? –  Jozef Aug 22 '11 at 18:51
I don't think induction is the right way to go; instead, to prove that $B^k = \left(\begin{array}{cc}A^k&kA^{k-1}\\0&A^k\end{array}\right)$, and then show that the given polynomial works by showing that evaluating $B$ at that polynomial is the zero matrix, that the minimal polynomial must be a multiple of $M_A(t)$, and that nothing short of the given $M_B(t)$ will work. –  Arturo Magidin Aug 22 '11 at 19:03

2 Answers 2

up vote 6 down vote accepted

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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What does $p'$ has to do with the question? –  Jozef Aug 22 '11 at 19:22
I don't get it, can you please extend your answer? what is $j$? for one eigenvalue? I'm sorry but it's too little words for me. –  Jozef Aug 22 '11 at 19:25
@Jozef: You could try to compute recursively B^k. For example B^2=[A^2,2A|0,A^2] and the derivative of x^2 is 2x, right? That is what the derivative of the polynomial has to do with it... –  Did Aug 22 '11 at 19:30
Why does $p(t)$ can't be divisible by $(t-\lambda)^{j}$? –  Jozef Aug 22 '11 at 19:31
@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 $g'(x)$. –  Arturo Magidin Aug 22 '11 at 20:35

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. –  Yuval Filmus Aug 22 '11 at 19:12

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