How do I see that every left ideal of a square matrix ring over a field is principal?

As the question title suggests, how do I see that every left ideal of a square matrix ring over a field is principal?

Solution 1:

Let $I$ be a left ideal. The first step is that if a matrix is in the ideal then all the rows can be obtained separately: Say $$\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} \in I$$

then

$$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}1&2&3\\0&0&0\\0&0&0\end{pmatrix} \in I$$ and

$$\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}4&5&6\\0&0&0\\0&0&0\end{pmatrix} \in I$$ $$\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}7&8&9\\0&0&0\\0&0&0\end{pmatrix} \in I$$

Further we can reconstitute the original matrix from the above three by permuting the rows and adding. For example

$$\begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix} \begin{pmatrix}7&8&9\\0&0&0\\0&0&0\\\end{pmatrix} =\begin{pmatrix}0&0&0\\0&0&0\\7&8&9\end{pmatrix} \in I$$

Thus it suffices to consider matrices of the form, $$\begin{pmatrix}*&*&*\\0&0&0\\0&0&0\\\end{pmatrix}$$

There may be many such matrices in $I$ but since this is basically an $n$ vector there are at most $n$ linearly independent such such vectors, thus having chosen a basis, we can amalgamate this basis into one matrix. For example say we have $$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\\\end{pmatrix},\begin{pmatrix}0&2&1\\0&0&0\\0&0&0\\\end{pmatrix}\in I.$$ The we have

$$\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}0&2&1\\0&0&0\\0&0&0\\\end{pmatrix} =\begin{pmatrix}0&0&0\\0&2&1\\0&0&0\\\end{pmatrix} \in I.$$

So in this example,

$$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\\\end{pmatrix}+\begin{pmatrix}0&0&0\\0&2&1\\0&0&0\\\end{pmatrix}=\begin{pmatrix}1&0&0\\0&2&1\\0&0&0\\\end{pmatrix}\in I$$ will generate the ideal.