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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 $F_{r\times n-r}=0$.

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    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.2012-10-24