Let $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ and $\mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \end{bmatrix}$ be two linearly independent vectors in $\mathbb{R}^2$.
Let $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ be an arbitrary vector in $\mathbb{R}^2$.
We want to show that there exist real numbers $a$ and $b$ such that:
$a\mathbf{v} + b\mathbf{w} = \mathbf{x}$
$a \begin{bmatrix}v_1\\v_2\end{bmatrix} + b \begin{bmatrix}w_1\\w_2\end{bmatrix} = \begin{bmatrix}x_1\\x_2\end{bmatrix}$
This is equivalent to solving the system of linear equations:
$av_1 + bw_1 = x_1$
$av_2 + bw_2 = x_2$
In matrix form, this is:
$\begin{bmatrix}
v_1 & w_1 \\
v_2 & w_2
\end{bmatrix}
\begin{bmatrix}
a \\
b
\end{bmatrix} =
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$
Let $A = \begin{bmatrix} v_1 & w_1 \\ v_2 & w_2 \end{bmatrix}$. We need to show that this system always has a unique solution.
Since $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, $\det(A) = v_1w_2 - v_2w_1 \neq 0$.
When $\det(A) \neq 0$, the matrix $A$ is invertible, which means the system has a unique solution for any arbitrary vector $\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$.
The solution is given by:
$\begin{bmatrix}
a \\
b
\end{bmatrix} = A^{-1}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$
Therefore, for any vector $\mathbf{x}$ in $\mathbb{R}^2$, we can find unique $a$ and $b$ such that $\mathbf{x} = a\mathbf{v} + b\mathbf{w}$.
This proves that any vector in $\mathbb{R}^2$ can be expressed as a unique linear combination of two linearly independent vectors in $\mathbb{R}^2$.
