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I am working on a problem that is stated as follows: Let $A$ be an $n\times n$ matrix. We define the set $\mathcal{I}$ to be the space of polynomials $p$ such that $p(A)=0$. We say the minimal polynomial $m_A$ is the polynomial of lowest degree with leading coefficient one. Let $\lambda_1, \lambda_2,...,\lambda_k$ be the distinct eigenvalues of $A$ with index values $d_1,d_2,...,d_j$ respectively. Show that $$m_A(x)=\prod_{i=1}^{k} (x-\lambda_i)^{d_i}$$

I am having trouble understanding the meaning of "index values $d_1, d_2,...,d_j$" of the eigenvalues, what does that mean?

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    I think that the word "space" isn't accurate to describe $\mathcal{I}$, and also that those $d's$ are the algebraic multiplicities of the corresponding eigenvalues – 2017-02-11
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    @mayer_vietoris but if the d's are the algebraic multiplicities of the eigenvalues wouldn't $m_A$ just be the characteristic polynomial? – 2017-02-11
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    let me write a more clear answer – 2017-02-11

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In general given a square matrix $A$ you can define the so-called Jordan normal form. This new matrix is equivalent (when it exists) with $A$ and is a matrix in a block diagonal form, that is of the form $$J_{A}= \begin{pmatrix} J_1 & & & \\ & J_2 & & \\ & & & J_n \\ \end{pmatrix} , $$ where the $J_{i}'s$ are the so-called jordan blocks. Now, how many and big they are, depends on the eigenvalues of $A$ (that's why doesn't always exist). The goemetric multiplicity of an eigenvalue for instance gives the number of Jordan Blocks, that contains the eigenvalue. Whilst, the index of an eigenvalue in defined as the biggest Jordan Block that contains the eigenvalue.

I hope that helps.

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    Ans also you're right about the algebraic multiplicities. Indeed then would imply that this is the characteristic polynomial. And when I read it in the first place, I was thinking that this $\mathcal{I}$ exists because of the characteristic polynomial and replied fast. – 2017-02-11
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    im sorry I still don't understand, the index of an eigenvalue is the biggest Jordan block? – 2017-02-12
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    When I say the biggest Jordan Block that contains this eigenvalue, I mean this natural number $n_{i}$, where the Jordan Block is a $n_i \times n_i$ matrix. – 2017-02-12
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    Let $A$ be the matrix $\begin{pmatrix} 1& 1& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0& 1& 0 \\ 0& 0& 0& 1 ]\end{pmatrix}$ then the characteristic polynomial is $x^4-4x^3+6x^2-4x+1$ and the minimal polynomial is $x^2-2x+1$. – 2017-02-12
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    @MarcBogaerts sorry Marc, but I don't understand what do you mean by this example? – 2017-02-12
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    @mayer_vietoris: It's a remark on a comment made by kevin about the difference between the minimal polynomial and the characteristic polynomial, and I thought it was an illustration of your answer – 2017-02-12
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    @mayer_vietoris: That the characteristic polynomial of a matrix doesn't necessarily equals its minimal polynomial. Kevin cited: *"...wouldn't $m_A$ just be the characteristic polynomial?"* – 2017-02-13
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    @MarcBogaerts: IF these $d_i \thinspace 's$ were indeed the algebraic multiplicities, then over an algebraically closed field this $m_A$ would be the characteristic polynomial. My mistake was that thought something different in the first place and confused Kevin. – 2017-02-13