Proof of linear independence How would I be able to prove this question:
Let $x$ and $y$ be linearly independent elements of a vector space $V$. Show that $u = ax+by$ and $v=cx+dy$ are linearly independent if and only if $ad-bc$ does not equal $0$. 
I know that $u$ and $v$ are linear combinations of $V$ which will make them span $V$. Also, if the determinant is equal to $0$ then it will be a singular matrix and if it is singular then it will have free variables which will make it dependent, but how can I show this mathematically? 
 A: You don’t have any reason to think that $x$ and $y$ span $V$: all you know is that they are linearly independent. The dimension of $V$ might well be greater than $2$, in which case no two-element subset of $V$ will span $V$, though many will be linearly independent.
You have two things to show:


*

*If $ad-bc\ne0$, then $u$ and $v$ are linearly independent.  

*If $u$ and $v$ are linearly independent, then $ad-bc\ne0$.


It’s probably easiest to prove (1) by proving the contrapositive: if $u$ and $v$ are linearly dependent, then $ad-bc=0$. That’s because the assumption of linear dependence gives you something very concrete to work with: if $u$ and $v$ are linearly dependent, there are scalars $\alpha$ and $\beta$, at least one of which is non-zero, such that $\alpha u+\beta v=0$. Now write this out in terms of $x$ and $y$: 
$$\alpha(ax+by)+\beta(cx+dy)=0\;.$$
Collect the $x$ and $y$ terms on the lefthand side:
$$(\alpha a+\beta c)x+(\alpha b+\beta d)y=0\;.$$
By hypothesis $x$ and $y$ are linearly independent, so 
$$\left\{\begin{align*}
&\alpha a+\beta c=0\\
&\alpha b+\beta d=0\;.
\end{align*}\right.$$
This says that $\begin{bmatrix}\alpha\\\beta\end{bmatrix}$ is a non-zero solution to the equation
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}z=0\;.$$
what does that tell you about $\det\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and hence about $ad-bc$?
To prove (2), again go for the contrapositive: if $ad-bc=0$, then $u$ and $v$ are linearly dependent. You can pretty much just reverse the reasoning in the argument that I outlined for (1).
A: For one direction, suppose $ad-bc \ne 0$, and suppose $\lambda, \mu \in \mathbb{R}$ satisfy
$$\lambda u + \mu v = 0$$
Show that this implies that $\lambda = \mu = 0$, and hence that $u$ and $v$ are linearly independent.
For the other direction, suppose $ad-bc = 0$ and find nonzero $\lambda, \mu$ satisfying the above equation.
Post in the comments if you need more help.
