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How to prove this exercise?

Let $A$ be a $n \times n$ diagonal matrix with characterist polynominal

$$(x-c_1)^{d_1}...(x-c_k)^{d_k}$$

where $c_i$ are distinct. Let $V$ be the space of $n \times n$ matrixes $B$ such that $AB=BA$. Prove that dimension of $V$ is $d_{1}^{2}+...+d_{k}^{2}$.

Thomas Andrews
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Eduardo Silva
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2 Answers2

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Hint: Suppose $A$ acts on the $n$-dimensional space $E$, $E=\sum_iKer(A-c_i)^{d_i}$ An element $B$ of $V$ preserves $V_i=\{x:(A-c_iI)^{d_i}(x)=0\}$, since $A$ is diagonal, the restriction of $A$ to $V_i$ is diagonal, so $M^{d_i\times d_i}$ commute with the restriction of $A$ to $V_i$ so $V=\sum_iM^{d_i\times d_i}$ this gives its dimension

Tsemo Aristide
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  • i dont get it, $x \in V_{i}$ are a $n \times n$ matrix or a $n \times 1$ matrix? can you explain further how to obtain $V_{i}$ and $M^{d_{1}\times d_{1}}$? – Eduardo Silva Jan 22 '16 at 01:29
  • $x$ is an element of the vector space $E$, a $n\times n$ matrix defines a linear application on a $n$-dimensional vector space. – Tsemo Aristide Jan 22 '16 at 01:44
  • but $Ker(A-c_{i}I)^{d_{1}}$ its a aplication from $n$-dimensional to $n$-dimensional space, how does a $n \times n$ matrix can be a vector on this space? $Ker(A-c_{i}I)^{d_{1}}(B) = 0$ implies that $B$ its a eingevector of $c_{i}$ by the aplication $T(B) = AB$ – Eduardo Silva Jan 22 '16 at 20:45
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Let $a_i=A_{ii}$ be the ordered $c_i$.

Then $(AB)_{ij} = a_iB_{ij}$ and $(BA)_{ij} = B_{ij}a_j$.

So these are equal exactly when $B_{ij}=0$ or $a_i=a_j$.

So all the coefficients of $B$ are zero except when the $a_i=a_j$. How many cases of that are there? (Hint: there are $d_j$ values of $a_i$ that are equal to $c_j$...

Thomas Andrews
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  • so are $d_{1} + ... + d_{k}$ values non zero in $B$ then a basis of V contain $n$ elements, but how did a get to $d_{1}^{2} + ... + d^{2}_{k}$ – Eduardo Silva Jan 22 '16 at 01:24