Finding a solution vector for a singular matrix that equals zero vector

Question

(Sorry if title was confusing)

So I am doing some linear analysis work, primarily with Gaussian Elimination, and am having a hard time figuring out how to solve this particular problem. I have done some research online but can't seem to find an example similar to this with a singular matrix.

So I have matrix A:

$ \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \\ 1 & 2 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ \end{bmatrix}$

with a solution vector b: $ \begin{bmatrix} 0 \\ 1 \\ 2 \\ 4 \end{bmatrix}$

My job is to perform gaussian elimination to find L, U, and p. Then I am to solve: if A is not singular solve the linear system Ax=b if A is singular, find a nonzero vector Ax = $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

After performing the elimination steps I am left with these:

$ L = \begin{bmatrix} 1 & 0 & 0 & 0 \\ \frac{1}{2} & 1 & 0 & 0 \\ 0 & \frac{2}{3} & 1 & 0 \\ \frac{1}{2} & \frac{-1}{3} & 1 & 1 \end{bmatrix}$

$ U = \begin{bmatrix} 2 & 1 & 0 & 1 \\ 0 & \frac{3}{2} & 1 & \frac{-1}{2} \\ 0 & 0 & \frac{4}{3} & \frac{2}{3} \\ 0 & 0 & 0 & \frac{1}{2} \end{bmatrix}$

$ p = \begin{bmatrix} 4 \\ 0 \\ 1 \\ 3 \end{bmatrix}$

I know that matrix A is singular so I need to solve for the zero vector but I am not sure how to do that. I have tried to perform elimination on the p vector using row subtraction but i do not see a way to cancel all of them to 0 and solve it without resulting in a zero vector.

Answer 1

To help you on your way, what you are trying to do is find a (non-zero) vector in the null space of your matrix i.e. the sub-space of all vectors mapped to the zero vector.

If you find a basis for your null space then you know all vectors in that null space.

There are plenty of online resources describing null space and they are easy to find.