Thread: Generators of matrix algebra

1. Generators of matrix algebra

Let $\displaystyle M_n(\mathbb{C})$ be the space of all $\displaystyle n\times n$ matrices with complex entries.

Consider the matrices
$\displaystyle u=\left(\begin{array}{cccc}1 &0 &\cdots &0\\0 &q &\cdots &\vdots\\0 &0 &\ddots &\vdots\\0 &0 &\cdots &q^{n-1}\end{array}\right)$ and $\displaystyle v=\left(\begin{array}{ccccc} 0 &1 &0 &\cdots &0\\0 &0 &1 &\cdots &0\\0 &0 &\vdots &\ddots &0\\\vdots &\vdots &\cdots &\cdots &1\\1 &\cdots &0 &\cdots &0\end{array}\right)$

Where $\displaystyle u^n=v^n=1$ and $\displaystyle q^n=1$.

Show that the matrices $\displaystyle u \text{ and } v$ generate the whole of $\displaystyle M_n(\mathbb{C})$.

The two by two case is simple, since you can easily find four linearly independant matrices which you can use as the basis. However, I am convinced there has to be some easier way? For the n by n case we would then need to find $\displaystyle n^2$ linearly independant elements which is a little cumbersome.

2. Re: Generators of matrix algebra

You can certainly do it fairly efficiently if you are prepared to throw a bit of heavy machinery at it. Here is an outline.

Step 1. The algebra A generated by u and v is selfadjoint, so by the double commutant theorem it will suffice to prove that the commutant of A consists only of scalars.

Step 2. The subalgebra of A generated by u comprises all the diagonal matrices.

Step 3. A matrix that commutes with all the diagonal matrices is itself diagonal. So the commutant of A consists of diagonal matrices.

Step 4. A diagonal matrix that commutes with v must be a scalar.

Edit. Step 2 assumes that q is a primitive n'th root of unity. If not, for example of q=1, then the result is false.

3. Re: Generators of matrix algebra

This is such a clever way to go about the problem!

4. Re: Generators of matrix algebra

Originally Posted by Mauritzvdworm
This is such a clever way to go about the problem!
No wonder, Opalg is a clever man.