For linearly independent vectors {u, v}, {$ au+bv, cu+dv$} is linearly independent iff $\begin{pmatrix}a & b\\\ c & d\end{pmatrix}$ is invertible.

Question

Given $\{u, v\}$ is linearly independent vectors, I want to prove that $\{au+bv, cu+dv\}$ is linearly independent if and only if $M = \begin{pmatrix}a & b\\\ c & d\end{pmatrix}$ is invertible.

I started from assuming the two linear combinations are linearly independent, however I do not know how to show that $\det(M) > 0$. Should I use other theorem to show it is invertible?

Answer 1

Consider the equation $t(au+bv)+s(cu+dv)=0$. By linear independence of $u$ and $v$ this is equivalent to the system of equations

$ta+sc=0$

$tb+sd=0$

This system has a non-trivial solution iff the coefficient matrix is non -singular which means $\det(A) = 0$.

Hence $det(A) = 0$ iff there exist $t$ and $s$ not both $0$ satisfying these equations iff $au+bv$ and $cu+dv$ are linearly dependent.