Representing a matrix in its "natural" space

Question

Hi there,

Any n*m matrix defines two vector spaces. The row space and the column space, and these both have the same dimensionality r, which is the rank of the matrix. Is there a way to represent that matrix in as an r*r matrix, such that both itself and its transpose becomes full-rank matrices?

Thank you.

Answer 1

You can do so, but you will have to make some choices and lose some information along the way.

Say you have an m*n matrix A of rank r. Then there are r linearly independent columns in A. Removing the rest of the columns from A, you get an m*r matrix B that still has rank r. There are also r linearly independent rows in B. Removing the rest of the rows from B, you get an r*r matrix C with full rank r. Whether you delete rows or columns first, the final answer will be the same. However, the choices of rows/columns to delete is not unique (and you lose the information contained in these rows and columns).

So what is happening here on the level of linear transformations? The matrix A represents a linear transformation A:Rⁿ→R^m. Removing columns and rows is the same as writing a sequence of maps R^r→Rⁿ→R^m→R^r. Here, the map R^r→Rⁿ is the inclusion of a subspace where certain coordinates are forced to be zero. The map R^m→R^r is an orthogonal projection, given by setting certain coordinates to zero. When we delete columns, we go from the map A:Rⁿ→R^m to the map B:Rⁿ→R^r. When we delete rows, we go from the map B:Rⁿ→R^r to the map C:R^r→R^r. This final map is an invertible matrix/transformation.