I'm asked the following.
Let $V$ be a vector space, and suppose $v_1,v_2,v_3$ span $V$. If $u_1=v_1+v_2$, $u_2=v_1-v_2$, $u_3=2v_1+3v_2-v_3$, show that $u_1,u_2,u_3$ span $V$.
I have three ideas of going about this.
For $u_1,u_2,u_3$ to span $V$, I should be able to write this in matrix form and obtain a consistent solution. Namely, I should be able to write:
$$\begin{pmatrix}
1 & 1 & 0 \\
1 & -1 & 0\\
2 & 3 & -1 \\
\end{pmatrix}$$
If I row reduce this and obtain a consistent solution (Without a row of $0$'s), I can say that this spans $V$.
My second idea was that, would I also be correct in saying that if I can write $u_3$ in terms of $u_1$ and $u_2$ that this also spans $V$ since I can express $u_3$ as a linear combination of $u_1$ and $u_2$?
Finally, could I simply say that if the determinant of $$\begin{pmatrix}
1 & 1 & 0 \\
1 & -1 & 0\\
2 & 3 & -1 \\
\end{pmatrix}$$
is non zero, that also spans $V$?
I think my 1st and 3rd idea is right but I am unsure if my 2nd methods works to show the set of vectors span $V$. Can someone explain how linear combinations helps in determining if a set of vectors span a space? I know the span is the set of all possible linear combinations but I'm not sure how linear combinations can actually be used here for showing that the three vectors belong/don't belong to the span.
Best Answer
$v_1=\frac 1 2 u_1+\frac 1 2 u_2, v_2=\frac 1 2 u_1-\frac 1 2 u_2$ and $v_3=2v_1+3v_2 -u_3=\frac 5 2 u_1-\frac1 2 u_2-u_3$. Since any vector in $V$ is a linear combination of $v_1,v_2,v_3$ it follow from these equations that any vector in $V$ is also a linear combination of $u_1,u_2,u_3$.