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Let $A$ be a $6\times 6$ matrix with charpoly $x(x+1)^2(x-1)^3$. We need to find the dimension of $$\gamma=\{B\in M_n(\mathbb{R}):AB=BA\}.$$

What is the relation of charpoly of $A$ with dimension of the space? Please give me some hints to proceed.

Myshkin
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  • I surmise one of the $(x-1)^p$ factors should have been $(x+1)^p$. The characteristic polynomial gives you almost the Jordan normal form, and for matrices in Jordan normal form, the set of commuting matrices is easier to determine. – Daniel Fischer Aug 27 '13 at 13:59
  • Is there any other way other than Jordan Form? – Myshkin Aug 27 '13 at 14:26
  • But by jordan normal form we will get possibilities of dimension.....how to find exact dimension? – neelkanth May 21 '15 at 02:04
  • You can look at the second answer of this: http://mathoverflow.net/questions/105040/centralizer-of-a-matrix-over-a-finite-field – Sungjin Kim Oct 02 '16 at 14:22

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As I commented, Centralizer of a Matrix is very helpful. Let $\mathbb{F}$ be a field and $A\in M_n(\mathbb{F})$. Denote by $C_A=\{B\in M_n(\mathbb{F}) | AB=BA\}$ the centralizer of $A$. the dimension of such space $C_A$ over a field $\mathbb{F}$ is given by $$ \sum_p (\mathrm{deg}(p)) \sum_{i,j}\min (\lambda_{p,i}, \lambda_{p,j}), $$ where $p$ is an irreducible polynomial that divides the characteristic polynomial of $A$, and $\lambda_p= \sum \lambda_{p,i}$ with $\lambda_{p,i}$ being the powers of $p$ in the primary decomposition of the $\mathbb{F}[x]$-module $\mathbb{F}^n$ acting by $A$-multiplication.

In your case, $\mathrm{deg}(p)$ is always $1$, $n=6$, and $\mathbb{F}=\mathbb{R}$. Then the primary decomposition of $\mathbb{R}[x]$-module acting by $A$-multiplication is one of: $$ \frac{\mathbb{R}}{(x)} \oplus \begin{cases} \frac{\mathbb{R}}{(x+1)}\oplus \frac{\mathbb{R}}{(x+1)} \oplus \begin{cases} \frac{\mathbb{R}}{(x-1)}\oplus \frac{\mathbb{R}}{(x-1)}\oplus \frac{\mathbb{R}}{(x-1)} : \mathrm{case \ I}\\ \frac{\mathbb{R}}{(x-1)}\oplus \frac{\mathbb{R}}{(x-1)^2}: \mathrm{case \ II} \\ \frac{\mathbb{R}}{(x-1)^3} : \mathrm{case \ III}\end{cases}\\ \frac{\mathbb{R}}{(x+1)^2} \oplus \begin{cases} \frac{\mathbb{R}}{(x-1)}\oplus \frac{\mathbb{R}}{(x-1)}\oplus \frac{\mathbb{R}}{(x-1)} : \mathrm{case \ IV}\\ \frac{\mathbb{R}}{(x-1)}\oplus \frac{\mathbb{R}}{(x-1)^2} : \mathrm{case \ V}\\ \frac{\mathbb{R}}{(x-1)^3} : \mathrm{case \ VI}\end{cases}\end{cases} $$

Denote by $J(\lambda, r)$ the Jordan block of size $r\times r$ with eigenvalue $\lambda$. Also, $I_r$ is the identity matrix of size $r\times r$.

Now, the following characterizes all possibilities:

Case I: The matrix $A$ is diagonalizable and similar to $$ 0I_1 \oplus (-1)I_2 \oplus 1I_3. $$ The minimal polynomial of $A$ is $x(x+1)(x-1)$. The dimension of $\gamma$ is $1+2^2+3^2 = 15$.

Case II: The matrix $A$ is similar to $$ 0I_1 \oplus (-1)I_2 \oplus I_1\oplus J(1,2). $$ The minimal polynomial of $A$ is $x(x+1)(x-1)^2$. The dimension of $\gamma$ is $1+2^2+5 = 10$.

Case III: The matrix $A$ is similar to $$ 0I_1 \oplus (-1)I_2 \oplus J(1,3). $$ The minimal polynomial of $A$ is $x(x+1)(x-1)^3$. The dimension of $\gamma$ is $1+2^2+3=8$.

Case IV: The matrix $A$ is similar to $$ 0I_1\oplus J(-1,2) \oplus I_3. $$ The minimal polynomial of $A$ is $x(x+1)^2(x-1)$. The dimension of $\gamma$ is $1+2+3^2=12$.

Case V: The matrix $A$ is similar to $$ 0I_1 \oplus J(-1,2) \oplus I_1 \oplus J(1,2). $$ The minimal polynomial of $A$ is $x(x+1)^2(x-1)^2$. The dimension of $\gamma$ is $1+2+5=8$.

Case VI: The matrix $A$ is similar to $$ 0I_1 \oplus J(-1, 2) \oplus J(1,3). $$ The minimal polynomial of $A$ is $x(x+1)^2(x-1)^3$. The dimension of $\gamma$ is $1+2+3=6$.

This case VI is when the characteristic polynomial and the minimal polynomial coincide. In general, with $A\in M_n(\mathbb{F})$, the centralizer of $A$ has minimal dimension $n$ in case the characteristic polynomial and the minimal polynomial of $A$ are identical.

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