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Is there any lower bound known for the minimal number of generators needed to generate the full matrix algebra of real $n\times n$ matrices — when using only symmetric matrices for the generators?

Analogous question for complex matrices — when using only Hermitian matrices for the generators.

I am aware that $3$ generators suffice when using only idempotent generators. This is a result of Naum Krupnik (Minimal number of idempotent generators of matrix algebras over arbitrary field, Comm. Algebra 20 (1992), no. 11, 3251–3257). (Tandfonline link, restricted access)

I am not familiar with this type of results, so this might be well known or easy. Thanks for any tips.

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  • $\begingroup$ Thanks a lot for the detailed answer, it fully answers my question. This is indeed a nice use of Burnside' theorem. $\endgroup$ – Monique Laurent Sep 2 '12 at 19:23
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It is a well-known result of Burnside that a set of (say, real) $n \times n$ matrices generate the full matrix algebra if and only if they do not have a common non-trivial invariant subspace of $\mathbb C^n$. Now, if a matrix $A$ is diagonalizable with distinct eigenvalues the only subspaces of $\mathbb C^n$ that are invariant under $A$ are the ones spanned by sets of eigenvectors. So you can take two orthonormal bases of $\mathbb R^n$, such that no non-trivial subspace of $\mathbb C^n$ is generated by a subset of the first and a subset of the second; then take two matrices with distinct real eigenvalues, and the elements of the first and second basis as eigenvectors. These are symmetric, and generated the full matrix algebra.

The case of hermitian matrices is completely analogous.

An elementary proof of Burnside's theorem is in E. Rosenthal, A remark on Burnside's theorem on matrix algebras. Linear Algebra and its Applications 63, 1984, 175-177; Sciencedirect link.

Edit: Here is another article with an even easier proof: V. Lomonosov, P. Rosenthal. The simplest proof of Burnside's theorem on matrix algebras. Linear Algebra and its Applications 383, 2004, 45-47; Sciencedirect link.

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    $\begingroup$ if and only they do not have a common... $\endgroup$ – Chris Godsil Sep 1 '12 at 13:02
  • $\begingroup$ To Chris: thanks, I made the correction. $\endgroup$ – Angelo Sep 1 '12 at 13:09
  • $\begingroup$ You Burnside's theorem link is defunct, I am afraid. $\endgroup$ – Duchamp Gérard H. E. May 12 '17 at 3:46
  • $\begingroup$ @DuchampGérardH.E. See my edit above. $\endgroup$ – Vincent May 12 '17 at 7:52
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As an answer, the lower bound is exactly $2$. To this end take a diagonal matrix of order $n$ with all distinct and non zero entries along diagonal and the matrix of all ones of order $n$. In this way the algebra generated by these two symmetric matrices is simple(i.e. irreducible ), Now use Burnside's theorem to deduce that it should be $M_n(C)$. See Laffey, Thomas J., A structure theorem for some matrix algebras, Linear Algebra Appl. 162-164, 205-215 (1992). ZBL0758.16010.

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  • $\begingroup$ couldn't you merge your two answers into a single one? $\endgroup$ – YCor May 13 '17 at 15:06
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Here is a more direct argument, without Burnside's Theorem. As in M. Karimi's answer, let $A$ be diagonal with distinct entries $\lambda_1,\dotsc,\lambda_n$, and let $B$ be the matrix with all entries equal to $1$. Note that $$ (A^iBA^j)_{st} = \lambda_s^i\lambda_t^j $$

Now let $M$ be an arbitrary $n\times n$ matrix. By Lagrange interpolation, there is a unique polynomial $$ p(x,y) = \sum_{i,j=0}^{n-1} a_{ij}x^iy^j $$ with $p(\lambda_s,\lambda_t)=M_{st}$ for all $s$ and $t$. It is now easy to check that $$ M = \sum_{i,j=0}^{n-1} a_{ij}A^iBA^j. $$

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If the dimension of space is at least 4, then We can always construct two symmetric matrices with all distinct eigenvalues and no eigenvector in common still they have non trivial invariant subspace in common. So they can not generate simple matrix algebra.

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  • $\begingroup$ I believe the claim of Angelo was not that ANY two symmetric matrices with no common eigenvector will generate the whole matrix algebra but rather that you can always FIND two matrices that do. $\endgroup$ – Vincent May 12 '17 at 8:19
  • $\begingroup$ (Clarification: I assumed your post was a reaction to Angelo's answer, written as an answer rather than a comment because of not yet having the right to comment. Please correct me if this interpretation is wrong.) $\endgroup$ – Vincent May 12 '17 at 8:19
  • $\begingroup$ Perhaps it would have been clearer if in the sentence 'So you can take two orthonormal bases of $\mathbb{R}^n$, such that no non-trivial subspace of $\mathbb{C}^n$ is generated by a subset of the first and a subset of the second' from Angelo's answer he had specified that the two subsets together have cardinality $n$ $\endgroup$ – Vincent May 12 '17 at 8:23
  • $\begingroup$ @Vincent Actually, as you mentioned I was not able to add comment. I am ware that Angelo did not mean ANY two symmetric matrices with no common eigenvector will generate the whole matrix algebra. However, the way of constructing such pair of symmetric matrices do not guarantee that they do not admit a non trivial invariant subspace in common. $\endgroup$ – M. Karimi May 13 '17 at 13:51

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