Change of Basis
Table of Contents
1 Introduction
Suppose for some finite dimensional K-space \( V \) we are given two bases:
\[ B = \{v_1, ..., v_n \} \] \[ C = \{u_1, ..., u_n \} \]
And we have some \( w \in V \) whose column vector WRT \( B \) is:
\[ [w]_B = c_1 \]
Then there is an easy way to obtain \( [w]_C \) from \( [w]_B \), define the following matrix:
\[ P_{B \to C} = ( [v_1]_C \ [v_2]_C \ ... \ [v_n]_C ) \]
Ie the columns of \( P_{B \to C} \) are given by the coordinate vectors of the elements in the basis \( B \) WRT \( C \).
We claim \( \forall w \in V \):
\[ P_{B \to C} \ [w]_C = [w]_B \]
2 Proof
Clearly \( P_{B \to C} \) is an \( n * n \) matrix, so the multiplication is defined. Consider:
\[ P_{B \to C} \ [v_i]_B \]
Where \( [v_i]_B \) is the coordinate vector of the ith vector in the basis \( B \) WRT to \( B \), and is thus 1 in the ith row and 0 everywhere else.
We can see this multiplication will extract the ith column of \( P_{B \to C} \), and thus by the definition of \( P_{B \to C} \):
\[ P_{B \to C} \ [v_i]_B = [v_i]_C \]
Hence our claim holds for all basis vectors in \( B \). \( B \) spans \( V \), hence we can write any \( w \in V \) as:
\[ w = a_1v_1 + ... + a_nv_n \]
For some \( a_1, ... a_n \in K \). So:
\begin{align} [w]_C &= [a_1v_1 + ... + a_nv_n]_C \\ &= a_1[v_1]_C + ... + a_n[v_n]_C \\ &= P_{B \to C} \ a_1[v_1]_B + ... + P_{B \to C} \ a_n[v_n]_B \\ &= P_{B \to C} (a_1[v_1]_B + ... + a_n[v_n]_B) \\ &= P_{B \to C} ([a_1v_1 + ... + a_nv_n]_B) \\ &= P_{B \to C} [w]_B \end{align}