1. ## Dealing with

Let $\displaystyle W$be a subspace of $\displaystyle Kn$ and

$\displaystyle v$is also contained in $\displaystyle Kn$

Need to determine if $\displaystyle v$ is contained in $\displaystyle W$
$\displaystyle v$ =

3
1
2
3

W = rowspace(A)

$\displaystyle A =$

2411
3402
1212

over

According to the notes we will find another matrix B where:

$\displaystyle W = nullspace(B)$
Then after GJ $\displaystyle A =$
1004
0100
0013

And then aparantlly

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

But i dont see how??

Any help would be great

2. I wouldn't worry about finding this "B". The definition of "row space of A" is the space spanned by the rows of A. In order to prove that < 3 1 2 3> is in that rwo space, you must show that it can be written as a linear combination of rows of A: that is, find a, b, c such that a<2 4 1 1>+ b<3 4 0 2>+ c<1 2 1 2>= <3 1 2 3>. That is the same as showing that it is possible to solve 2a+ 3b+ c= 3, 4a+ 4b+ 2c= 1, a+ c= 2, and a+ 2b+ 2c= 3.