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Let A be a diagonal matrix whose characteristic polynomial is

$P(x) = (x − 15)(x − 14)^{2}(x-13)^{3}......(x-2)^{14}(x-1)^{15}$ Let V be the set of all $120 × 120$ matrices commuting with A. what is dimension of V ?

My answer : dimension of $V = 1^2 +2^2 +3^2+.......+15^2 = \frac{15(15+1) (2.15+1)}{6}$

is Its True /false ???

  • http://math.stackexchange.com/questions/1476010/computing-the-dimension-of-a-vector-space-of-matrices-that-commute-with-a-given – Will Jagy Apr 24 '18 at 18:02
  • @WillJagy...is my answer is correct or not ??? pliz tell me –  Apr 24 '18 at 18:07
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    Looks good, doesn't it? I assume you had goofdreason to simply square and add the dimensions of the eigenspaces. – Hagen von Eitzen Apr 24 '18 at 18:09
  • ya ..@HagenvonEitzen sir...i directly compute .as i take reference from Hoffman kunz linear algebra –  Apr 24 '18 at 18:11
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    How about if you do an example 3 by 3, let it be diagonal also, and let the diagonal entries be (2,1,1). Write a general 3 by 3 matrix $M,$ call the entries a,b,c,d,e,f,g,h,i, explicitly solve the (linear) conditions that cause $M$ to commute with the fixed diagonal matrix. – Will Jagy Apr 24 '18 at 18:13
  • Yes, your answer is correct. Though I hope you understand you why that formula is correct – Naweed G. Seldon Apr 24 '18 at 19:03

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I thought it would be appropriate to see how this formula comes about. There are 2 key elements to this:

$1.$ For a diagonal matrix of the from $A=aI$, for any $a$ in your base field. Every matrix obviously commutes with $A$. Therefore the dimension of the space of matrices commuting with $A$ in this case is nothing but $n^2$, if you're working over the space of $n \times n$ matrix.

$2$. Now if you have a diagonal matrix of the form $$A = \begin{pmatrix} a\\ &a\\ &&a\\ &&&b\\ &&&&b\\ &&&&&c \end{pmatrix} = \begin{pmatrix} aI_3\\ &bI_2\\ &&cI_1 \end{pmatrix}$$

(The blank spaces are $0$s and the $0-$matrix) Then any matrix commutes with $A$ if and only if it commutes with the each submatrix ($I_n$ is the identity matrix of order $n$), hence the dimension in this case is nothing but $3^2 + 2^2 + 1^2$

In our case, we have $15$ submatrices, which are $I_{15}, 2I_{14}, 3I_{13},\cdots,14I_2,15I_1$, giving us the dimension as $15^2 + 14^2 + \cdots + 3^2 + 2^2 + 1^2$

Note, if the diagonal elements of $A$ were distinct, then the only matrices commuting with $A$ would be all diagonal matrices, giving the dimension simply $n$

Naweed G. Seldon
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