Example of a space of commuting $4 \times 4$ matrices with 5 linearly indpendent elements 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?
 A: 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).


*

*Schur, J.
Zur Theorie der vertauschbaren Matrizen.
J. für M. 130, 66-76 (1905).
JFM36.0140.01
GDZ:archival copy.

*Jacobson, N.
Schur's theorems on commutative matrices.
Bull. Amer. Math. Soc. 50, (1944). 431–436.
MR10540
DOI:10.1090/S0002-9904-1944-08169-X

*Mirzakhani, M.
A simple proof of a theorem of Schur.
Amer. Math. Monthly 105 (1998), no. 3, 260–262.
MR1615548
DOI:10.2307/2589084
