## Existent solution to the linear system Ax=b

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.
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 Mentor Blog Entries: 8 Think about $$(Ax)^T c$$ where x is an arbitrary vector and c spans the null-space.
 That would imply x^T Ac=x0=0. How do you relate this to the above problem?

## Existent solution to the linear system Ax=b

 Quote by 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.
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.

Mentor
Blog Entries: 8
 Quote by onako That would imply x^T Ac=x0=0. How do you relate this to the above problem?
Doesn't this imply that Ax is perpendicular to c?

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