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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?

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  • $\begingroup$ What is rref? row-reduced echelon form? Thus obtained from $A$ by only row operations? $\endgroup$ – Hagen von Eitzen Feb 10 at 17:43
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$S_A$ 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 $$ v\in\ker S_A\implies S_Av=0\implies Av=B^{-1}S_Av=B^{-1}0=0\implies v\in \ker A.$$

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The null space of a matrix is the orthogonal complement of its row space. Elementary row operations reversibly (that’s very important) replace rows of the matrix with linear combinations of rows of the matrix, so they don’t change the row space. Therefore...

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