Proving Singular Matrix and Non-Zero Solutions: A Tutorial

In summary, if A is singular, then det(A) cannot be zero and so the columns of A are not linearly independent.
  • #1
Poirot1
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How would I prove that if A is singular, then Av=0 has a non-zero solution?.
 
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  • #2
Poirot said:
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.
 
  • #3
Jameson said:
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

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

What is a singular matrix?

A singular matrix is a square matrix that does not have an inverse. This means that it cannot be multiplied with another matrix to produce the identity matrix. In other words, a singular matrix is a matrix that cannot be inverted.

How can I determine if a matrix is singular?

To determine if a matrix is singular, you can calculate its determinant. If the determinant is equal to zero, then the matrix is singular. You can also check if the matrix has linearly dependent rows or columns, as this also indicates singularity.

What is the significance of proving a matrix is singular?

Proving that a matrix is singular is important in many areas of mathematics and science. In linear algebra, singular matrices have special properties that can help with solving systems of equations and understanding linear transformations. In statistics, singular matrices are used in multivariate analysis to identify relationships between variables.

Can a singular matrix have a non-zero solution?

Yes, a singular matrix can have a non-zero solution. This means that there exists a non-zero vector that, when multiplied by the singular matrix, results in the zero vector. However, a singular matrix cannot have a unique solution.

How can I prove that a matrix has a non-zero solution?

To prove that a matrix has a non-zero solution, you can use the properties of singular matrices. For example, you can show that the determinant of the matrix is equal to zero, or that the matrix has linearly dependent rows or columns. These properties indicate that there exists a non-zero solution for the matrix.

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