Null Space of A: Find Rank & Dim.

[karush]

Let
$$\left[\begin{array}{rrrrrrr}
1 & 0 & -1 & 0 & 1 & 0 & 3\\
0 & 1 & 0 & 0 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 4 & 0 & 2\\
0 & 0 & 0 & 0 & 0 & 1 & 3
\end{array}\right]$$
Find a basis for the null space of A, the dimension of the null space of A, and the rank of A.ok following an book example I did this $Ax=b$
$$\left[ \begin{array}{ccccccc}
1 & 0 & -1 & 0 & 1 & 0 & 3 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 4 & 0 & 2 \\
0 & 0 & 0 & 0 & 0 & 1 & 3
\end{array} \right]
\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \\ x_{7}
\end{array} \right]
=\left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0
\end{array} \right]$$
which would result in
$$\begin{array}{rrrrrrr}
x_1 & &-x_3 & &x_5 & &3x_7=0 \\
&x_2 & & &x_5 & &x_7 =0\\
& & & x_4 & 4x_5 & &2x_7=0 \\
& & & & & x_6 &3x_7=0
\end{array}$$ hopefully so far !

 

[Olinguito]

It is clear (by inspection, working from the last equation upwards) that you can let $x_7,x_5,x_1$ be arbitrary, then $x_2,x_3,x_4,x_6$ can be expressed in terms of them as:
$$x_6\ =\ -3x_7 \\ x_4\ =\ -4x_5-2x_7 \\ x_2\ =\ -x_5-x_7 \\ x_3\ =\ x_1+x_5+3x_7.$$
Hence:
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \\ x_7\end{pmatrix}\ =\ x_1\underbrace{\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}}_{\mathbf v_1} + x_5\underbrace{\begin{pmatrix} 0 \\ -1 \\ 1 \\ -4 \\ 1 \\ 0 \\ 0\end{pmatrix}}_{\mathbf v_2} + x_7\underbrace{\begin{pmatrix} 0 \\ -1 \\ 3 \\ -2 \\ 0 \\ -3 \\ 1\end{pmatrix}}_{\mathbf v_3}.$$
Verify that $\mathbf v_1,\mathbf v_2,\mathbf v_3$ are linearly independent and thus form a basis for the nullspace of $A$; hence $A$ has nullity $3$. The rank of $A$ can then be found from the formula $r(A) = \dim(A)-n(A)$.

 

[HOI]

You have 4 equations in 7 unknowns so the null space is 7- 4= 3 dimensional. You will need 3 basis vectors.

 

[karush]

Ok that was very helpful

Not sure if I would know the linear independence of these

 

[HOI]

You prove those vectors are independent using the definitions of "independent" or "dependent".
Suppose the three vectors given by Olinguito were NOT independent. Then there would exist numbers, a, b, and c, not all zero, such that
a\begin{pmatrix}1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0\end{pmatrix}+b\begin{pmatrix}0 \\ -1 \\ 1 \\ -4 \\ 1 \\ 0 \\ 0 \end{pmatrix}+c\begin{pmatrix} 0 \\ -1 \\ 3 \\ -2 \\ 0 \\ -3 \\ 1 \end{pmatrix}=\begin{pmatrix}a \\ -b- c \\ a+ b+ 3c \\ -4b- 2c \\ b \\ -3c \\ c\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.

That gives 7 equations: a= 0, -b- c= 0, a+ b+ 3c= 0, -4b- 2c= 0, b= 0, -3c= 0, c= 0. Clearly the only a, b, and c that satisfy all 7 equations are a= b= c= 0. Therefore the three vectors are independent.