# Existent solution to the linear system Ax=b

1. Jun 21, 2012

### onako

Suppose that the linear system Ax=b is given for some symmetric A, and it is known that vector c spans the null-space of A.

How could one formally show that if b is not orthogonal to c, the solution to the system Ax=b does not exist?
To remind you, the null-space of A contains all vectors u for which Au=0.

2. Jun 21, 2012

### micromass

Staff Emeritus

$$(Ax)^T c$$

where x is an arbitrary vector and c spans the null-space.

3. Jun 22, 2012

### onako

That would imply x^T Ac=x0=0. How do you relate this to the above problem?

4. Jun 22, 2012

### chiro

Hey onako.

I'm assuming A is nxn (since you said it is symmetric). From this if c spans the null-space it must be an n-dimensional column vector.

From this you can use the decomposition argument that a basis can be broken into something and its perpendicular element (some books write it as v_perp + v = basis). Your zero vector c is perpendicular to b if you wish to have full rank.

If this is not the case, then you can show that you don't have full rank and that a solution should not exist. For specifics you should look at rank nullity, and for the v_perp + v = basis thingy, this is just a result of core linear algebra with spanning, dimension, and orthogonality.

5. Jun 22, 2012

### micromass

Staff Emeritus
Doesn't this imply that Ax is perpendicular to c?