Matrix Algebras: Generator

Question

Problem

Given the algebra $\mathcal{M}_\mathbb{C}(2)$.

Denote the normals: $$\mathcal{N}:=\{N\in\mathcal{M}_\mathbb{C}(2):N^*N=NN^*\}$$

And their calculus: $$\mathcal{N}(N):=\{\eta(N):\eta\in\mathcal{B}(\mathbb{C})\}$$

Regard commuting ones: $$N,N'\in\mathcal{N}:\quad N'N=NN'$$

Do they admit one: $$N_0\in\mathcal{N}:\quad\mathcal{N}(N)\cup\mathcal{N}(N')\subseteq\mathcal{N}(N_0)$$

Are there counterexamples?

Attempt

Choose one nondiagonal: $$N=U^*DU\quad N'=D'$$

There exists one with: $$\eta_0(N)=\eta_0(U^*DU)\neq\eta_0(D)$$

But how to proceed then?

Reference

I need this for: Superalgebra

Maybe I misread your definitions but $N_+$ and $N_-$ both lie in the subalgebra generated by $\mathrm{diag}(-1,1)$. — Sebastian Schoennenbeck, Jul 20 '15 at 13:24
Actually I am fairly certain that it is impossible to construct two such matrices, since each $2 \times 2$-matrix is either scalar or generates its own centralizer. — Sebastian Schoennenbeck, Jul 20 '15 at 13:27
@SebastianSchoennenbeck: I just noticed that $N'=1-N$ and $N=1-N'$. So it is $\mathcal{N}(N)=\mathcal{N}(N')$. — C-star-W-star, Jul 20 '15 at 13:51
Yep, and this will always happen as stated in my second comment — Sebastian Schoennenbeck, Jul 20 '15 at 13:53
@SebastianSchoennenbeck: Yes right. Let me try another one. (Modifying my thread.) — C-star-W-star, Jul 20 '15 at 13:54
Hm, I would guess that $\mathrm{diag}(1,2,3,4)$ will generate a subalgebra containing your two matrices. The generated subalgebra has dimension $4$ (look at the minimal polynomial) and is hence the algebra of diagonal matrices containing your two chosen ones. — Sebastian Schoennenbeck, Jul 20 '15 at 14:03
Additionally you actually cannot choose them nondiagonal since two commuting normal matrices can be brought into diagonal form simultaneously. — Sebastian Schoennenbeck, Jul 21 '15 at 06:16

score 1 · Accepted Answer · answered Jul 21 '15 at 07:42

1

This should never be possible:

Let $N_1,N_2 \in \mathbb{C}^{n\times n}$ be two commuting normal matrices. Then (by the spectral theorem and the fact that they commute) there is a unitary matrix $P$ such that $PN_1P^*=D_1,PN_2P^*=D_2$ are diagonal.

Take now any diagonal matrix $D$ with pairwise distinct diagonal entries. Then $D$ generates an algebra of dimension $n$ (look at its minimal polynomial) and each element of this algebra is diagonal, so $D$ generates the subalgebra of diagonal matrices. In particular $D_1$ and $D_2$ lie in it and $N_1,N_2$ hence lie in the subalgebra generated by $P^*DP$.

answered Jul 21 '15 at 07:42

Sebastian Schoennenbeck

2,686

How do you know that they are diagonalized by the same matrix? – C-star-W-star Jul 21 '15 at 08:34
Because they commute. Any set of commuting matrices each of which is diagonalizable itself can be brought into diagonal form simultaneously. See (e.g.) here: http://math.stackexchange.com/questions/56307/simultaneous-diagonalization – Sebastian Schoennenbeck Jul 21 '15 at 08:36
Ah ok. Thanks for the link. – C-star-W-star Jul 21 '15 at 08:53

Matrix Algebras: Generator

1 Answers1