0

Given an $n\times n$ complex matrix $A$, let $\mathbb{C}[A]$ be the image of the evaluation map $\mathbb{C}[x]\to \text{Mat}_{n\times n}(\mathbb{C})$ given by $x\to A$. Let $Z(A)$ be the subring of $\text{Mat}_{n\times n}(\mathbb{C})$ consisting of all matrices that commute with $A$: $Z(A)=\{B\in\text{Mat}_{n\times n}(\mathbb{C}):AB=BA\}$.

I am trying to prove that $\text{dim}_{\mathbb{C}}(Z(A))=n$ if and only if $\text{dim}_{\mathbb{C}}(\mathbb{C}[A])=n$. The $\leftarrow$ direction is proven in this post.

Also, I realize that I may assume WLOG that $A$ is in its Jordan Normal form because if $A=PJP^{-1}$ with $J$ being $A$'s Jordan Normal form, then $AB=BA$ if and only if $(P^{-1}BP)J=J(P^{-1}BP)$.

To prove $\rightarrow$ direction, I am assuming $\text{dim}_{\mathbb{C}}(Z(A))=n$ and $\text{dim}_{\mathbb{C}}(\mathbb{C}[A])<n$. Then this implies that the minimal polynomial of $A$ isn't equal to the characteristic polynomial of $A$. To reach a contradiction, I think I need to come up with $n+1$ many $\mathbb{C}-$linearly independent matrices of $Z(A)$ using that min poly of $A$ isn't equal to char poly of $A$, (or equivalently, using that there is at least two Jordan blocks in $J$ that has same eigen values), but I am having hard time doing this. Any advice is appreciated!

mathlearner98
  • 772

1 Answers1

1

On a piece of paper I'd explain this in about a minute. Drawing graphics of matrices and vectors in MathJax is difficult. So I'll explain using symbols and let you draw the matrices and vectors yourself.

Let in $N(k) \in M_k(\Bbb{C})$ be the nilpotent Jordan block of dimension $k$. This is matrix with $N(k)_{i,i+1} = 1$ for $1\le i < k$ and all other entries zero. Verify for yourself the following statements

  1. The non-unital subalgebra generated by $N(k)$ are the strictly upper triangular matrices in $M_k(\Bbb{C})$ that are constant on rightward-downward pointing diagonals. It has dimension $k-1$.
  2. The unital subalgebra generated by $N(k)$ are the upper triangular matrices in $M_k(\Bbb{C})$ that are constant on rightward-downward pointing diagonals. It has dimension $k$.
  3. $N(k)$ has a cyclic vector ($(0,0,\dots,0,1)^t$), so by the post you linked to, the centralizer of $N(k)$ in $M_k(\Bbb{C})$ is the same as the unital subalgebra generated by it and has dimension $k$. (Centralizer or commutant is the thing you denote by $Z()$).

Now let's move to dimension $n$ and let $J\in M_n(\Bbb{C})$ be a matrix in Jordan canonical form. For some Jordan block $J_*$ of $J$ I'll denote by $\dim(J_*)$ its dimension.

Now let's first analyze the matrices in $Z(J)$ that have the same block-diagonal structure as $J$. If $J_*$ is any Jordan block in $J$ and $\dim(J_*) = k$, then working just in that block we get a dimension of $k$ in $Z(J)$ from matrices that commute with $J_*$ inside that block and have zero in all the other blocks. Adding together the dimensions of Jordan blocks we see that $\dim(Z(J))\ge n$. Morever, the part of $Z(J)$ constructed by block diagonal matrices this way is a commutative sub-algebra of $Z(J)$.

Now we need to show we can get matrices outside the block-diagonal subalgebra so the dimension of $Z(J)\ge n+1$. For this end, let $J_1, J_2$ be two Jordan blocks in $J$ associated to the same eigenvalue, and such that $\dim(J_2)\ge \dim(J_1)$. One way to choose these is to let $J_1$ be the smallest block associated to an eigenvalue that has at least two blocks.

Let $l_1=\dim(J_1), l_2=\dim(J_2)$. We assumed $l_1\le l_2$. Let $u_1, \dots, u_{l_1}$ be the Jordan basis vectors associated to block $J_1$. Let $v_1, \dots, v_{l_2}$ be the Jordan basis vectors associated to block $J_2$. Now define a matrix $B$ that works as follows

  1. $B(u_i) := v_i$ for $1\le i \le l_1$.
  2. $B(x) := 0$ for all other vectors of a Jordan basis (including the $v_*$-s.)

The matrix $B$ is not any of the $n$-dimensional space of block diagonal matrices in $Z(J)$ constructed earlier - because it mixes Jordan blocks. You can verify by direct computation that it commutes with $J$ so $B\in Z(J)$.

Chad K
  • 4,873