# Alternate Definition of the Change of Basis Matrix

Let $u = (u_1, \dots, u_n)$ and $v = (v_1, \dots, v_n)$ be bases for the vector space $V$. According to the definition supplied by Arturo in the answer to this question, the change of basis matrix from $u$ to $v$ is the matrix $P$ such that for any vector $a \in V$ $$P[a]_u = [a]_v$$ where $[a]_u$ is the representation of $a$ with respect to to the basis $u$ and $[a]_v$ is the representation of $a$ with respect to the bases $v$. On page 251 of Birkhoff & Mac Lanes Algebra (3rd ed), the change so basis matrix from $u$ to $v$ is defined as the matrix $P$ whose components are given by $$P^i_j = v^i(u_j)$$ where $(v^1, \dots, v^n)$ denotes the basis dual to $(v_1, \dots, v_n)$.

Now, I am trying to demonstrate the equivalence of these two definitions.

I'm aware of the various properties of dual bases such as $$v^i(v_j) = \delta^i_j$$ which implies that $v^i(a)$ represents the $i^{th}$ component of $a$ which implies that $a$ can be expressed as

$$a = v^1(v)v_1 + \cdots + v^n(v)v_n$$

I've tried various approaches to this problem using these facts and the algebra always ends up a mess with more equations than unknowns, or I end up with rank two tensors, etc, but to no avail. My basic strategy has been to expand the LHS of the expression in the original COB definition and then solve for the components of the matrix $P$ to show that they are equal to $v^i(u_j)$.

I'm thinking it might be easier to compute the inverse first but this, of course, would require knowing that $P$ is invertible (which isn't a fact yet established in the development I'm following)

What is the best way to proceed with this?

## Update

Following azarel's hint, I believe I have a handle on this now. Consider, for example, the case of a 2-dimensional vector space and let $(x^1, x^2)$ denote the coordinates of $a$ relative to $u$ and $(y^1, y^2)$ denote the coordinates of $a$ relative to $v$. If $a$ is a unit vector relative to $u$ then the coordinates of $a$ are either $(1, 0)$ or $(0, 1)$. In the first instance, the change of basis equation implies that $P^1_1 = y^1$ and $P^2_1 = y^2$. But, using the summation convention, $$y^1 = v^1(y^jv_j) = v^1(x^ju_j) = x^jv^1(u_j) = v^1(u_1)$$ since $x^1 = 1$ and $x^2 = 0$.

Similarly, $$y^1 = v^2(u_1)$$ Therefore, $P^1_1 = v^1(u_2)$ and $P^2_1 = v^2(u_1)$. The scalars $P^1_2$ and $P^2_2$ can be obtained in an analagous fashion by applying $P$ to the $(0,1)$ unit vector. This verifies that in the case of $n=2$, $P^i_j = v^i(u_j)$. I believe I can now turn this into a general argument that proves the claim.

-
There are quite a number of questions about change of basis matrices already -- just wondering whether you've looked at those (I haven't so they may not help). Some of them appear under "Related" to the right. –  joriki Jan 18 '12 at 17:22
@Jorki I have looked, but haven't been able to locate a question that addressed the issue at hand. –  ItsNotObvious Jan 18 '12 at 17:29
You should be careful in that there are two different conventions; in one of them, "change of basis from $u$ to $v$" is the one you describe; but there are plenty of books that define the "change of basis matrix from $u$ to $v$" to be the matrix $P$ such that $P[x]_v = [x]_u$. –  Arturo Magidin Jan 18 '12 at 19:42
@ArturoMagidin Yes, I am aware of this unfortunate circumstance and it has caused me no small amount of grief. I think though the definition you provided is most common. –  ItsNotObvious Jan 18 '12 at 20:25

Hint: A linear map is determined by its action on the basis. Try applying the matrix $P$ to the $u_i$'s vectors.