Matrix change of basis, is $P_{D,E}$ = $P^{-1}_{E,D}$?

Question

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.

Answer 1

$P_{C,B}B=C\implies P^{-1}_{C,B}C=B$

Answer 2

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)

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

Answer 3

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

$$P_{D,E}=M_D^{-1}M_E=P_{E,D}^{-1}$$