# Left ideals in simple rings

1. Sep 29, 2010

### gerben

I do not see why this is the case. Take the ring of 3 by 3 matrices over the real numbers and the left ideal, J, generated by:

$$\begin{pmatrix} 0 &1 &1\\ 0 &0 &0\\ 0 &0 &0 \end{pmatrix}$$

then J is not equal to S = {M ∈ M(3,ℝ) | The 1st column of M has zero entries},
since for example the following matrix is in S but not in J:

$$\begin{pmatrix} 0 &1 &0\\ 0 &0 &1\\ 0 &0 &0 \end{pmatrix}$$

Can anybody help me understand what the wikipedia page is trying to say, or where I am seeing things wrong?

Last edited by a moderator: May 5, 2017
2. Sep 30, 2010

### gerben

Can somebody shed some light on what is so important about these zero columns? I would think that the number of independent columns is important. I just do not see why zero columns are necessary at all.

Isn't the ideal generated by the following matrix, without any zero columns, also a nontrivial left ideal:

$$\begin{pmatrix} 1 &1 \\ 0 &0 \\ \end{pmatrix}$$

3. Oct 3, 2010

I think you can show that for any left ideal I there is a unique vector subspace L of $$\mathbf{R}^n$$ such that $$I=\{A\in I:\,AL=0\}$$. Then, for any k-dimensional subspace of $$\mathbf{R}^n$$ you can choose a basis such that the first k vectors are in L. This will put your ideal into a canonical form that you are looking form .

$$A=\begin{pmatrix} 1 &1 \\ 0 &0 \\ \end{pmatrix}$$

$$SAS^{-1}=\begin{pmatrix} 1 &0 \\ 1 &0 \\ \end{pmatrix}$$
where
$$S=\begin{pmatrix} 1 &1 \\ 1 &-1 \\ \end{pmatrix}$$

But I am not an expert.

Last edited: Oct 3, 2010
4. Oct 3, 2010

### gerben

Yes, I see that there is a subspace of $$\mathbf{R}^n$$ that is contained in the kernel of every M ∈ I. This subspace is the orthogonal complement of the row space of matrix A ∈ I that has maximum number of independent rows.

Ah yes I guess that was the idea: on an appropriate basis there will be zero columns.

Thank you very much.

Last edited: Oct 3, 2010