If we want to check if the following set of vector span $R^3$ (1,0,0) (0,1,-1) (0,4,-3) (0,2,0) then we forme an augmented matrix formed by the vectors which form the columns of the augmented matrix and add and extra vector u=$(u_1,u_2,u_3)$.
Solving this augmented matrix give me the following solution
$C_1=u_1;
C_2=6c_4-3u_2-4u_3;
C_3=-2c_4+u_2+u_3$
Therefore we got many solutions to the augmented matrix ,and that is expected given we got more unknows than equations.
However what does this mean for the span?
I know that this actually span $R^3$ but how do i conclued that?
Edit: forming an equation to see that if every vector from $R^3$ can be written as a combination of the of vector above give the following equation
$c_1(1,0,0)+c_2(0,1,-1)+c_3(0,4,-3)+c_4(0,2,0)= (u_1,u_2,u_3)$
Making the left side as an unique vector ,and then relating the components of each side then forme a coefficient matrix of it and then do following type $\boldsymbol Ac=u$.
The augmented matix is the following matix
$$\left[\begin{array}{r,r,r,r|r}1&0&0&0&u_1\\0&1&4&2&u_2\\0&-1&-3&0&u_3\end{array}\right]$$
Now solving this matrix to reduce row-echelon give the result above i add also here
$C_1=u_1;
C_2=6c_4-3u_2-4u_3;
C_3=-2c_4+u_2+u_3$
Getting to this point i know.so what implication does it have on the span,how do we imply that the set of vectors span$R^3$?
Best Answer
What you would do is to take the determinant of a $3\times 3$ matrix whose columns are the vectors that you conjecture to be independent (thus they span $\mathbb R^3$). Notice that since you have $4$ of these vectors you select three. The procedure is to take the determinant. If it is different from $0$ then the $3$ vectors you selected are independent and they span $\mathbb R^3$. If the $\det$ is $0$ select other $3$ vectors... If all of the combinations of vectors that form a $3\times 3$ matrix have determinant $0$, then the vectors are dependent and they don't span $\mathbb R^3$. If any of these matrices formed by putting $3$ vectors as its columns has determinant different from $0$, then the other vector can be written as a combination of the three independent vectors you've found.
For instance:
$$\det\left[\begin{array}{r,r,r}1&0&0\\0&1&4\\0&-1&-3\end{array}\right]=-3+4=1\neq0.$$
Therefore the first three vectors you mentioned are independent and span $\mathbb R^3$.
Moreover, we know that if there are constants $k_1,k_2,k_3,k_4$, at least one of them different from $0$, such that for every $c_1,c_2,c_3,c_4\in\mathbb R$ (these come from the vector solution $\mathbf c$) we have that (if we name the columns $C_1,C_2,C_3,C_4$) $k_1C_1+k_2C_2+k_3C_3+k_4C_4=\boldsymbol0$, then the columns are dependent.
So let's assume that the columns are dependent and we will try to reach a contradiction. If $k_1\neq 0$ the contradiction is trivial since the first column of your matrix is $\left[\begin{array}{c}1\\0\\0\end{array}\right]\Longrightarrow k_1c_1\left[\begin{array}{c}1\\0\\0\end{array}\right]+\boldsymbol \xi\neq \left[\begin{array}{c}0\\0\\0\end{array}\right]\text{for some } c_1\in\mathbb R.$ The same happens if we select $k_4\neq0$. So the thing turns interesting when $k_1,k_4=0$. Then we are left with the two columns of the middle. Of course if additionally $k_2=0$ or $k_3=0$ then the contradiction is trivial as well. But if $k_2,k_3\neq 0$, then we have the system: \begin{align} c_2+4c_3=u_2, && -c_2-3c_3=u_3. \end{align} $\text{That yields } c_3=u_2+u_3.$ Therefore $c_3$ does not depend on $c_2$ and it is impossible that $k_2(c_2+4c_3)+k_3(-c_2-3c_3)=0,\,\forall c_2,c_3\in\mathbb R$.