# Thread: Help figuring out how to find this vector

1. ## Help figuring out how to find this vector

I have a matrix $A$

$2 4 1 1$
$3 4 0 2$
$1 2 1 2$

Now im trying to find a basis for the nullspace(A)

After applying GJ to A i get

$1 0 0 4$
$0 1 0 0$
$0 0 1 3$

Now how would i go from this matrix to finding the single basis vector

b4 =

1
0
2
1

I don't understand how they went from the matrix to this vector. Any help would be greatly appreciated

2. Originally Posted by BigBoss22
I have a matrix $A$

$2 4 1 1$
$3 4 0 2$
$1 2 1 2$

Now im trying to find a basis for the nullspace(A)

After applying GJ to A i get

$1 0 0 4$
$0 1 0 0$
$0 0 1 3$

Now how would i go from this matrix to finding the single basis vector

b4 =

1
0
2
1

I don't understand how they went from the matrix to this vector. Any help would be greatly appreciated

I really don't understand what you did: first, I interchanged the 3rd and 1st rows in the original matrix and brought it to echelon form and got:

$\left(\begin{array}{cccc}1&2&1&2\\0&2&3&4\\0&0&1&3 \end{array}\right)$

Using this matrix a coefficient matrix for a homogeneous system (i.e., equating every row to zero), the general solution is

$\left(\begin{array}{c}\!\!\!\!-8w\\5w\\\!\!\!\!-6w\\2w\end{array}\!\!\right)\,,\,\;with\;\;w\in \mathbb{F}=\mbox{ the field of de{}finition}$

You can now choose any non-zero value for w and you get a basis for the null space of A. The vector you mention above is NOT in the null space of A, as you can easily check.

Tonio

3. Ok i'll give you the full example from the notes that i'm looking at.

Let W be a subspace of Kn and
v is also contained in Kn

Need to determine if v is contained in W

and v =
3
1
2
3

W = rowspace(A)

A =

2411
3402
1212

over $Z5$

According to the notes we will find another matrix B where:
W = nullspace(B)

Then after GJ A =

1004
0100
0013

And then aparantlly

From this information you can get a single bias vector
$b4$=
1
0
2
1

But i dont see how??

4. Originally Posted by BigBoss22
I have a matrix $A$

$2 4 1 1$
$3 4 0 2$
$1 2 1 2$

Now im trying to find a basis for the nullspace(A)

After applying GJ to A i get

$1 0 0 4$
$0 1 0 0$
$0 0 1 3$

Now how would i go from this matrix to finding the single basis vector

b4 =

1
0
2
1

I don't understand how they went from the matrix to this vector. Any help would be greatly appreciated
$rref(A)=\begin{bmatrix} 1&0&0&4\\0&-2&0&5\\0&0&1&3 \end{bmatrix},\ker(A)=span(\begin{bmatrix} -4\\ \frac{5}{2} \\-3 \\1\end{bmatrix})=span(\begin{bmatrix} -8\\5 \\-6 \\2\end{bmatrix})$