# Thread: Bases over vector spaces.

1. ## Bases over vector spaces.

http://img541.imageshack.us/img541/5...2nov211904.gif
i have this mental block, it's probably really simple but i'm just not getting it
is this correct?

for Q, e1=(-1 0 1 0) and e2=(0 -1 0 1)?
and for F_3, e1=(1 1 0 0), e2=(2 0 1 0) and e3=(1 0 0 1)
?

thank you

2. Actually, you have done extremely well!

When I did the problem, I got (1, 0, -1, 0) and (0, 1, 0, -1) but since those are just multiples of your vectors, they are both bases. Yes, your answer is correct.

I assume that your $F_3$ is the field of integers modulo 3 so that 2= -1 and the two equations we get from applying the matrix $\begin{pmatrix}1 & 2 & 1 & 2 \\ 1 & -1 & 1 & -1\end{pmatrix}$ to any member, $\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}$, of the kernel, x+ 2y+ z+ 2w= 0 and x- y+ z- w= 0, are exactly the same. Since there is only one condition, the dimension of the kernel is 4- 1= 3 rather than 4- 2= 2 as before. We can solve the second equation for x: x= y- z+ w and so $\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}= \begin{pmatrix}y- z+ w \\ y \\ z \\ w\end{pmatrix}$ $\begin{pmatrix}y \\ y \\ 0 \\ 0\end{pmatrix}+ \begin{pmatrix}-z \\ 0 \\ z \\ 0\end{pmatrix}+ \begin{pmatrix} w \\ 0 \\ 0 \\ w\end{pmatrix}= y\begin{pmatrix}1 \\ 1 \\ 0 \\ 0\end{pmatrix}+ z\begin{pmatrix}-1 \\ 0 \\ 1 \\ 0\end{pmatrix}+ w\begin{pmatrix}1 \\ 0 \\ 0 \\ 1\end{pmatrix}$.

Since, in $F_3$, 2= -1, that is exactly what you have.