MHB Proving Singular Matrix and Non-Zero Solutions: A Tutorial

[Poirot1]

How would I prove that if A is singular, then Av=0 has a non-zero solution?.

 

[Jameson]

How would I prove that if A is singular, then Av=0 has a non-zero solution?.

If A is singular then it isn't invertible, so by the invertible matrix theorem the columns of A are not linearly independent.

 

[Poirot1]

If A is singular then it isn't invertible, so by the invertible matrix theorem the columns of A are not linearly independent.

How can I prove that the columns of an invertible matrix are linearly independent (from 'first principles')?

Thanks

 

[Jameson]

I need to know what you've covered and what tools are available. The proof of the invertible matrix theorem is widely available all over Google so I suggest skimming through some of those proofs and then posting any followup ideas or questions.

Many of these proofs also work by proving a couple of statements and then using that to imply the other statements. Any true statement of the IMT implies all of the others so there are lots of ways to go between these ideas.

Here is an example of an answer to your question:

"Assume that for the matrix A, Row i = Row j. By interchanging these two rows, the determinant changes sign (by Property 2). However, since these two rows are the same, interchanging them obviously leaves the matrix and, therefore, the determinant unchanged. Since 0 is the only number which equals its own opposite, det A = 0"

This uses the property that switching two rows of a matrix will reverse the sign of the determinant.

 

[Poirot1]

I'm not quite sure how your answer pertains to my question. I see on wikipedia there is a list of equivalent statements which comprise the invertblie matrix theorem. I suppose what I want is to prove these in a non-circular manner, i.e. without invoking the invertible matrix theorem.

 

[Jameson]

The definition of a singular matrix A, as far as I know, is a square matrix that does not have an inverse. This occurs iff when det(A) =0. That's my reasoning for starting with the determinant.

Anyway, that's all I have to offer since I don't know the way you want to approach it but I know that a handful of members here are very knowledgeable of linear algebra so hopefully one of them can comment further.