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Good morning. I am a little curious about the motivation for a simple example from linear algebra as well as critique on the example I came up with.

Question: Given a subspace of $M$ commuting $4 \times4$ matrices with complex entries give an example that shows $M$ has five linearly independent elements.

First here is the example I came up with is there a better one? $$ \left( \begin{array} 01 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array} 00 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array} 00 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array} 00 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array} 00 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right) $$

I think the dimension of $M$ is bounded by 4 when there exists a matrix in $M$ with two distinct characteristic values but is that the only reason to construct such an example from a teaching perspective?

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If $M$ has at least five linearly independent elements then its dimension is $\ge 5$ and $\le 16$, regardless of how many eigenvalues any matrix in it has. –  anon Jul 26 '11 at 14:21
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Don't look commuting to me. Look for example at last two. –  André Nicolas Jul 26 '11 at 14:21

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up vote 6 down vote accepted

It is reasonably easy to find a set of n linearly independent commuting n × n matrices: one takes a basis of the space of all diagonal matrices. However, a matrix commuting with all n of those matrices must be diagonal, and so must be linearly dependent. In other words, the space of diagonal matrices is a maximal commuting subspace of the space of matrices.

In your example, your first four matrices generate a maximal commuting subspace: there is no fifth matrix that is both linearly independent of and commuting with your first four matrices. You can write down a generic 4×4 matrix, and calculate what it means for it to commute with each of your first three matrices: you'll find all off-diagonal entries have to be 0.

One might be led to believe that 4 is the maximum dimension, but it is not, and this exercise is asking you to find the surprising example.

$\color{gray}{\textrm{Hover for answer:}}$ For any values of a, b, c, d, e, all of the matrices $$\begin{bmatrix} e & . & a & b \\ . & e & c & d \\ . & . & e & . \\ . & . & . & e \end{bmatrix}$$ commute, where . means 0. Taking each variable to be 1 with the others 0 gives five linearly independent commuting 4×4 matrices. Similarly you can find seven linearly independent commuting 5×5s.

The maximal commuting subspaces of matrices of maximal dimension were classified by Schur (1905). Various proofs have been given, for instance Jacobson (1944) and Mirzakhani (1998).

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@anon: I added the reference. –  Jack Schmidt Jul 26 '11 at 16:38
    
+1: A delightful example!! –  Jyrki Lahtonen Jul 26 '11 at 17:42
    
@Jack: Thanks.$\text{}$ –  anon Jul 28 '11 at 8:52

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