Are simultaneously diagonalizable matrices dense in $(M_{n\times n})^k$?

Let $M_{n\times n}$ be the space of $n\times n$ matrices (over algebraically closed field). Then I know that the diagonalizable matrices are dense in $M_{n\times n}$ (because $M_{n\times n}$ is irreducible wrt. Zariski topology and the diagonalizable matrices are the open subset defined by the nonvanishing of the discriminant). If $(M_{n\times n})^k$ consists of $k$-tuples of matrices, do the tuples of simultaneously diagonalizable matrices form a dense subset?

No, quite the opposite in fact, even for $k=2$.

Two diagonalizable matrices are simultaneously diagonalizable iff they commute. This is a relatively rare situation, not a common one: it is defined by the simultaneous vanishing of all the entries in the matrix $AB-BA$. Instead of a Zariski-open condition, you have a Zariski-closed one, defining a variety of a fairly small dimension.

• ah right, thank you! – anon Aug 15 '18 at 21:54

Let $S \subseteq (M_{n \times n})^k$ be the set of simultaneously diagonalizable matrices. For the trivial cases $n = 0,1$ or $k = 0$ we have that $S = (M_{n \times n})^k$, and $S$ is still dense in $(M_{n \times n})^k$ for $k = 1$.

But for $n \geq 2$ and $k \geq 2$ the set $S$ is contained in the proper closed subset $$V = \{ (A_1, \dotsc, A_n) \in (M_{n \times n})^k \mid \text{A_i A_j - A_j A_i = 0 for all i,j = 1, \dotsc, k} \},$$ because simultaneously diagonalizable matrices commute, and is therefore not dense.

Let $T\subset {M_n}^k$ be the considered algebraic set of $k$ simultaneously diagonalizable matrices.

$\textbf{Proposition 1}$. $n^2+kn-n\leq dim(T)\leq n^2+kn$.

$\textbf{Proof}$. The second inequality comes from the @A. Pongrácz 's post.

For the first one. Let $A\in M_n$ s.t. $A$ has distinct eigenvalues. Then $AB=BA$ iff $B$ is a polynomial of degree $n-1$ in $A$.

Consider $U=(A,B_1,\cdots,B_{k-1})\in T$. Then, in a neighborhood of $U$, the local dimension of $T$ is $n^2+(k-1)n$.

$\textbf{Proposition 2}$. One has $dim(T)=n^2+kn-n$.

$\textbf{Proof}$. We use the A. Pongrácz ' proof.

Let $\Delta=\{A\in M_n;A$diagonal $\},\Delta_G=\{A\in GL_n;A$ diagonal$\}$; note that $\Delta_G$ is a non-normal subgroup of $GL_n$. We consider the pseudo-parametrization of $T$ (with $im(f)=T$)

$f:(A_1,\cdots,A_k,S)\in\Delta^k\times GL_n\rightarrow (S^{-1}A_1S,\cdots,S^{-1}A_kS)\in T$.

We consider the quotient $GL_n/\Delta_G$ defined by the equivalence relation $P\sim Q$ iff $PQ^{-1}\in\Delta_G$. If $P\sim Q$, then $P=UQ$ where $U\in\Delta_G$; if $D\in\Delta$, then $P^{-1}DP=Q^{-1}DQ$.

Finally $\bar{f}:(A_1,\cdots,A_k,S)\in\Delta^k\times GL_n/\Delta_G\rightarrow (S^{-1}A_1S,\cdots,S^{-1}A_kS)\in T$

satisfies $im(\bar{f})=T$. Since $GL_n/\Delta_G$ is an algebraic set of dimension $n^2-n$, we deduce that $dim(T)\leq n^2+kn-n$. $\square$

The dimension of $M_{n\times n}^k$ is $kn^{2}$.

The set of tuples of diagonal matrices is a $kn$-dimensional subspace. Then you can also pick a matrix $S$ for the base transition, and conjugate all these tuples of diagonal matrices with all $S\in GL_{n\times n}$. So basically, you get $n^2$ new degree of freedom. Altogether, this means that the dimension of the manifold $T$ of your tuples is at most $kn+n^2< kn^2$ for $k>1, n>1$ (except for the $k=n=2$, which should be easy to cover), so this closure is quite small inside $M_{n\times n}^k$ in general.

(In fact, I think one should be able to tell the dimension of $T$ precisely.)

• $M_n^k$ has dimension $kn^2$ and the sub-vector space constituted of $k$ diagonal matrices has dimension $kn$. Then $dim(T)\leq kn+n^2$. – loup blanc Aug 16 '18 at 11:49
• My goodness, I must have been half asleep. Thanks a lot! (Fixed.) – A. Pongrácz Aug 16 '18 at 11:58