# Finding a basis for the nullspace

$A$=\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}

I need to find the null space for this matrix. After performing row operations $R_3 + 3R_1$, then $R_{new row 3}$ + $R_2$, I got the reduced row echelon form. (the variable I chose were $x$, $y$, $z$, $w$). I identified the pivot variables as $x$ and $y$, after expressing both of these in terms of the free variables $z$ and $w$, I got $$\begin{bmatrix} x\\ y\\ z\\ w\\ \end{bmatrix} = z \begin{bmatrix} 23/2\\ 4\\ 1\\ 0\\ \end{bmatrix} + w\begin{bmatrix} -11/2\\ -2\\ 0\\ 1\\ \end{bmatrix}$$, and then chose both the column vectors alongside $z$ and $w$ as the basis for the nullspace. However the answer given is $$\begin{bmatrix} 23\\ 8\\ 2\\ 0\\ \end{bmatrix}\begin{bmatrix} 9\\ 4\\ 0\\ 2\\ \end{bmatrix}$$. What am I doing wrong?

• Could you show your reduced row echelon form? Two row operations don't sound like enough to get to a reduced form for a matrix with that many nonzero entries. – Henning Makholm May 31 '15 at 10:26
• The rank of the matrix is $3$, so the null-space has dimension $1$. The first vector you mentioned is member of the null-space. – Peter May 31 '15 at 10:27
• If you multiply $A$ with the second vector, you see that it does not belong to the null-space. – Peter May 31 '15 at 10:28
• The given answer is also incorrect. – Peter May 31 '15 at 10:30
• Are all the signs of the last column of your matrix right? If the column was $[-1,-2,1,0]$ instead, the matrix would have rank 2, and $[9,4,0,2]$ would be in the null space. – Henning Makholm May 31 '15 at 10:30

$$\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}$$ Row reduced form of matrix is $$\begin{bmatrix}1 & 0 & -\frac{23}{2} & 0\\ 0 & 1 & -4 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &0 &0\end{bmatrix}$$ So basis is $$\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix}$$
If matrix is $$\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & -2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}$$ Row reduced form of matrix is $$\begin{bmatrix}1 & 0 & -\frac{23}{2} & -\frac{9}{2}\\ 0 & 1 & -4 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 &0 &0\end{bmatrix}$$ So basis is $$\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix} \begin{bmatrix}9 \\ 4 \\ 0 \\ 2\end{bmatrix}$$
• No I don't think you understand my question. I'm saying if we have $v_1$, $v_2$ as basis for a subspace, then is $kV_1$,$bV_2$ also a basis for that subspace? (b and k are scalars) – user140161 May 31 '15 at 11:04
• A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every other vector in the vector space is linearly dependent on these vectors . If you multiply that vector with any scalar , say $kv_1=kv_1+0v_2$ so it is dependent on $v_1$ and $v_2$ therefore $kv_1$ cannot be a basis. Similarly $kv_2$. – vidhan May 31 '15 at 11:09