Q:

Understanding the result of applying a linear transformation to a basis

There are two related concepts I do not grasp.

A linear transformation is determined by its action on a basis.
The action on a basis can be arbitrary.

Can someone give insight into this?

A:

Let us consider linear transformations from $\mathbb{R}^2$ to $\mathbb{R}^2$ as an example.
The standard basis of $\mathbb{R}^2$ is $(1,0), (0,1)$. We first arbitrarily pick an image for $(1,0)$. We have to make two choices: the $x-$coordinate of its image, and the $y-$coordinate of its image:
$$L (1,0) = (x_1, y_1).$$
Next we arbitrarily (this does not depend on or is in any way restricted by our previous choices!) pick an image for $(0,1)$. Again we make two choices.
$$L (1,0) = (x_2, y_2).$$
We have now made four choices which will completely determine our linear transformation. This is exactly the number of elements in your matrix, and this is not coincidental. We can now calculate the image of an arbitrary vector. Why? Let $(a,b)$ be an arbitrary vector. By linearity:
$$L(a,b) = a \cdot L(1,0) + b \cdot L(0,1) = a \cdot (x_1, y_1) + b \cdot (x_2,y_2).$$

