# Basis for the Null Space

I was just watching Strang's test review for Unit 1 and he made a comment about the null space in question 4. I want to check if it's generally true.

Given a matrix $A_{m\times n}$ over a field and it's reduce row echelon form, $R_{m\times n}$ such that we have

$\begin{bmatrix} I_{r\times r} & F_{r\times n-r} \\ 0 & 0 \end{bmatrix}$

Where $r$ is rank(A), then do the non-zero rows of

$\begin{bmatrix} I_{r\times r} & -F_{r\times n-r} \\ 0 & 0 \end{bmatrix}$

always form a basis for the null space?

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No; consider $F_{r\times n-r}=0$.
So if I make the condition that none of $F$'s rows are $0$ then is it true? – Robert S. Barnes Oct 24 '12 at 9:52
@Robert: My advice would be to develop a greater taste for exploratory experimentation. How about $(1,1/2)$? – joriki Oct 24 '12 at 9:56
You mean $(1,1/2)$ as a column of $F$? – Robert S. Barnes Oct 24 '12 at 9:59
@Robert: No, as $A_{1\times2}$ and $R_{1\times2}$. – joriki Oct 24 '12 at 9:59
I'm not exactly sure what you mean there, but I've been experimenting with the case of $I_{2\times 2}$ and $F_{2\times 4}$ and it only seems to work when there is at least one entry in each row and column of $F$. I'm not sure of course if that would carry over to larger matrices. – Robert S. Barnes Oct 24 '12 at 10:09