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 ???
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 $n$