Proving that proposed vectors actually form a basis for Null space

Question

We have the matrix

A = \begin{bmatrix}1&4&5&6&9\\3&-2&1&4&-1\\-1&0&-1&-2&-1\\2&3&5&7&8\end{bmatrix} and I am asked to find a basis for the Null space of A, as well as verify that my proposed vectors actually form a basis for the Null space. I got the RREF of matrix A and ended up with the following result for my basis:

B = {v = (-1, -1, 1, 0, 0), u = (-2, -1, 0, 1, 0), w = (-1, -2, 0, 0, 1)}

My problem arises in trying to "prove" that these vectors actually form the basis. From my understanding, a subset of a vector space is a basis if the vectors in it are linearly independent and the span of the vectors equals the vector space. Now, when trying to prove that my vectors are linearly independent, I am faced with a problem, as I obtain the following matrix (which will inevitably have a free variable, making the columns linearly dependent):

\begin{bmatrix}-1&-2&-1&0\\-1&-1&-2&0\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}

I'm having a hard time understanding where I went wrong. Any help would be appreciated.

Answer 1

I believe you are trying to show that

$$av+bc+cw=0$$

implies that $a=b=c=0$ and hence the system is linearly independent.

And in doing so, you form the augmented system.

The last column is the augmented column and it doesn't introduce a free variable.

You can conclude that $a=0$ from the third row, conclude that $b=0$ from the fourth row and conclude that $c=0$ from the fifth row.

Remark:

Since the first two rows of $A$ are not multiple of each other, the rank of $A$ is at least $2$, hence by rank-nullity theorem, the nullity is at most $3$. We have found $3$ independent elements of the nullspace of $A$, hence we have found a basis.

Answer 2

Notice that your matrix$$\begin{bmatrix}1&4&5&6&9\\3&-2&1&4&-1\\-1&0&-1&-2&-1\\2&3&5&7&8\end{bmatrix}$$ has rank $2$ and nullity $3$

Thus if you show your three vectors $$v = (-1, -1, 1, 0, 0), u = (-2, -1, 0, 1, 0), w = (-1, -2, 0, 0, 1)$$ in the null space are linearly independent, then you have a basis for your null space.

You may verify the linear independence of these vectors by row reduced echelon form of the matrix whose rows are your$ v$,$u$, and $ w.$ Thus B={$v,u,w$} is indeed a basis for your null space.