Linear Independence proof Looking to get some help with this practice proof. If we let $u$ and $v$ be linearly independent vectors. Show that $2u + 3v$ and $u+v$ are linearly independent.
 A: HINT: Just apply the definition: assume that $a,b\in\Bbb R$ are such that
$$a(2u+3v)+b(u+v)=\mathbf{0}\;,\tag{1}$$
and show that this forces $a$ and $b$ both to be $0$. A little algebra immediately reduces $(1)$ to
$$(2a+b)u+(3a+b)v=\mathbf{0}\;;$$
now use the hypothesis that $u$ and $v$ are linearly independent to conclude something about $2a+b$ and $3a+b$, and use that conclusion so show that $a=b=0$.
A: You have:
$$
a(2u+3v)+b(u+v)=0 \quad \iff \quad u(2a+b)+v(3a+b)=0
$$
and, since $u$ and $v$ are linearly independent this is true only if
$$
\begin{cases}
2a+b=0\\
3a+b=0
\end{cases}
$$
but this gives $a=b=0$
A: A set of vectors vectors $\{a_1, \ldots, a_k\}$ is linearly independent if it is not linearly dependent, i.e.,
$$
\beta_1 a_1 + \ldots + \beta_k a_k = 0
$$
holds only when $\beta_1 = \cdots = \beta_k = 0$. 
So, let's try to use this definition to prove linearly independence. Suppose we have 
a set of coefficients $\beta_1, \beta_2$ so that
$$
\beta_1 (2u + 3v) + \beta_2 (u + v) = 0.
$$
If we expand this out, we have 
$$
(2 \beta_1  + \beta_2) u + (3 \beta_1 + \beta_2) v = 0.
$$
Now, because $u$ and $v$ are linearly independent 
$$
2 \beta_1 + \beta_2 = 3 \beta_1 + \beta_2 = 0 .
$$
From this equation, we can deduce $\beta_1 = \beta_2 = 0$. 
