# Help figuring out how to find this vector

• Nov 2nd 2009, 01:38 PM
BigBoss22
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
• Nov 2nd 2009, 02:42 PM
tonio
Quote:

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
• Nov 2nd 2009, 04:18 PM
BigBoss22
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??
• Nov 2nd 2009, 08:05 PM
math2009
Quote:

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

$rref(A)=\begin{bmatrix} 1&0&0&4\\0&-2&0&5\\0&0&1&3 \end{bmatrix},\ker(A)=span(\begin{bmatrix} -4\\ \frac{5}{2} \\-3 \\1\end{bmatrix})=span(\begin{bmatrix} -8\\5 \\-6 \\2\end{bmatrix})$