# Is the null space of matrix $A$ same as null space of matrix $S_A$ if $S_A=\operatorname{rref}(A)$?

If $$A\in \Bbb M_{m\times n}$$ and if $$S_A=\operatorname{rref}(A)$$, then $$\operatorname{Ker}(A)=\operatorname{Ker}(S_A)$$. True or not?

I know that from RREF we can find wich columns are linearly independent, so that form helps a lot to find out connections between columns of matrix $$A$$. Knowing all this, can I say that they will have the same null spaces?

• What is rref? row-reduced echelon form? Thus obtained from $A$ by only row operations? – Hagen von Eitzen Feb 10 at 17:43

$$S_A$$ is obtained from $$A$$ by elementary row operations, i.e., by left multiplying $$A$$ with an invertible matrix: $$S_A=BA$$ where $$B$$ is invertible. Then $$v\in\ker A\implies Av=0\implies S_Av=BAv=B0=0\implies v\in \ker S_A$$ and $$v\in\ker S_A\implies S_Av=0\implies Av=B^{-1}S_Av=B^{-1}0=0\implies v\in \ker A.$$