# The components of basis vectors in linear algebra under a change of basis

If I choose a non-orthonormal basis for $$\Bbb{R}^2$$, I have to label the basis vectors by their components. But doing this requires me to specify the components of one of the basis vectors relative to that of the other one. For instance if I take the standard $$(0,1)$$, $$(1,0)$$ basis vectors and double the length of one of them, is there not ambiguity in the "unit" length? For instance the basis is now $$(0,2)$$, $$(1,0)$$, but could I equally define them as $$(0,1)$$, $$(0.5,0)$$?

My guess at the resolution is that when you move into a new basis, you redefine the "unit" component length to be $$1$$ in each direction. So that the vector $$(1,1)$$ corresponds to different vectors in two different bases. And my mistake comes from talking about the new basis relative to the old basis.

Your guess is correct. The notation $$(x, y)$$ is meaningless without reference to a chosen basis.
If you choose a basis $$v_1, v_2$$ of $$\mathbb{R}^2$$, then $$(1, 2)$$ denotes the vector $$v_1+2v_2$$.
If you choose a basis $$u_1, u_2$$ of $$\mathbb{R}^2$$, then $$(1, 2)$$ denotes the vector $$u_1+2u_2$$. This can be a completely different vector from $$v_1+2v_2$$.
If I choose a non-orthonormal basis for $$\mathbb{R}^2$$, I have to label the basis vectors by their components. But doing this requires me to specify the components of one of the basis vectors relative to that of the other one.
This is not true. You are specifying the components relative to the standard basis. Therefore $$(0,2), (1, 0)$$ is not the same basis as $$(0, 1), \left (\frac 12, 0 \right)$$.