I'm having trouble understanding how to solve the following exercise (or rather, what is it asking for):
Find the change of basis matrix for the following basis B and D for $\mathbb{R}^2$.
$B = \{(1 \ \ 2),(3 \ \ 4)\}$
$D = \{(1 \ \ 4),(2 \ \ 3)\}$
So far I've computed the coordinates of the base vectors in D with respect to base B, obtaining the matrix $$ S = \begin{matrix} 4 & 1/2 \\ -1 & 1/2 \\ \end{matrix} $$ Is it the change of base matrix? Or should I compute its inverse, $S^{-1}$ ?
Thanks
Consider this. Matrix multiplication, i.e. forming $A*x$ is the same as building the vector which is of the form $x_1 a_1 + ... x_n a_n$, i.e. forming linear combinations of the columns $a_1,... ,a_n$ of the matrix $A$ (test it by multiplying $A$ with say the second unit vector and you will see that you get exactly the second column of the matrix, i.e. what I call $a_2$.
So if you have a basis, or in other words a quadratic, invertible matrix $A$ you may ask: What are the correct coefficients of $b$ with respect to the basis $a_1,...a_n$, i.e. what is $x$, and you know it has to be $A^{-1} * b$.
So if somebody is telling you, that s/he provides to you the coefficients $c$ in the $B$ system, you have to form the vector $y = B*c$, and from that you get the $D$-coordinates as $c' = D^{-1} * y = (D^{-1}*B) * c$, so the transition from $B$-coordinates $c$ to the $D$-coordinates $c'$ is given by the multiplication from the left with the base-change matrix $H := D^{-1}*B.$