I've to find the basis of $W=\operatorname{span}\{v_1,v_2,v_3,v_4,v_5\} \subseteq\mathbb{F}_{2} ^{5}$
$v_{1}=(1,0,1,0,1)$ , $v_{2}=(0,0,1,0,1)$ , $v_{3}=(0,0,0,1,0)$ , $v_{4}=(1,0,1,1,1)$ , $v_{5}=(1,0,0,1,0)$
$\mathbb{F}_{2}$ is a Field with $2$ Elements
so now first I've wrote a linear combination for the span:
$a \cdot1$ + $b \cdot0$ + $c \cdot0$ + $d \cdot1$ + $e \cdot1$
$a \cdot0$ + $b \cdot0$ + $c \cdot0$ + $d \cdot0$ + $e \cdot0$
$a \cdot1$ + $b \cdot1$ + $c \cdot0$ + $d \cdot1$ + $e \cdot0$
$a \cdot0$ + $b \cdot0$ + $c \cdot1$ + $d \cdot1$ + $e \cdot1$
$a \cdot1$ + $b \cdot1$ + $c \cdot0$ + $d \cdot1$ + $e \cdot0$
so i got:
$a+d+e=w_{1}$
$0=w_{2}$
$a+b+d=w_{3}$
$c+d+e=w_{4}$
$a+b+d=w_{5}$
am i now on the right Way? or completly wrong? Am i right that i have to show the Basis with the linear combination?
Consider the matrix$$\begin{bmatrix}1&0&1&0&1\\0&0&1&0&1\\0&0&0&1&0\\1&0&1&1&1\\1&0&0&1&0\end{bmatrix}.$$Its rows are the vectors that you provided. If the determinant was non null, that would be great, but, alas, it is $0$. So, add the first row to the fourth and the fifth rows, thereby getting:$$\begin{bmatrix}1&0&1&0&1\\0&0&1&0&1\\0&0&0&1&0\\0&0&0&1&0\\0&0&1&1&1\end{bmatrix}.$$Now, you add the second row to the fifth one, and you get:$$\begin{bmatrix}1&0&1&0&1\\0&0&1&0&1\\0&0&0&1&0\\0&0&0&1&0\\0&0&0&1&0\end{bmatrix}.$$Now, you can stop, since the third, and fourth and the fifth rows are all equal. Thereby, a basis of your space is$$\bigl\{(1,0,1,0,1),(0,0,1,0,1),(0,0,0,1,0)\bigr\}.$$