Matrix change of basis, is $P_{D,E}$ = $P^{-1}_{E,D}$? I have a transition matrix $P_{E,D}$ from a basis $D=(d_{1},d_{2},d_{3})$ to a basis $E=(e_1,e_2,e_3)$. I need to find $P_{D,E}$ and wondering if $P_{D,E}$ is equal to $P^{-1}_{E,D}$. I know that when the standard basis is used this is true but I'm not sure for two general bases.
 A: $P_{C,B}B=C\implies P^{-1}_{C,B}C=B$
A: Assume the standard basis is $S$.
Then 

$$P_{E,D}=P_{S,D}P_{E,S}$$
  $$P_{D,E}=P_{S,E}P_{D,S}$$

Since you know that for standard basis the result is true, so we conclude that:

$$P_{S,D}=P^{-1}_{D,S}$$
  $$P_{S,E}=P^{-1}_{E,S}$$

Try to finish the proof yourself. (hidden until hover over)

$$P_{E,D}=P_{S,D}P_{E,S}=P_{D,S}^{-1}P_{S,E}^{-1}=(P_{S,E}P_{D,S})^{-1}=P_{D,E}^{-1}$$

A: Consider the matrix which have vectors $d_i$ as columns
$$M_D= [d_1 \quad d_2 \quad d_3]$$
then if you consider a vector $v_D$ expressed in the basis $D$ the product
$$v_S=M_D\cdot v_D$$
gives the component of vector v in the standard basis.
So for basis $E$ we have that
$$M_E= [e_1 \quad e_2 \quad e_3]$$
then if you consider a vector $v_E$ expressed in the basis $E$ the product
$$v_S=M_E\cdot v_E$$
gives the component of vector v in the standard basis.
Thus
$$M_E\cdot v_E=M_D\cdot v_D\implies v_E=M_E^{-1}M_D\cdot v_D$$
$$P_{E,D}=M_E^{-1}M_D$$
and of course
$$P_{D,E}=M_D^{-1}M_E=P_{E,D}^{-1}$$
