Which one is the correct way to compute the change of basis matrix?

Question

I was looking for a way to compute the change of basis matrix (given the old basis and the new basis), but I found two methods that lead to different results, and I can't understand which one is correct.

Method 1:

$$ B = \left\{ \begin{pmatrix} 1 \\ 2 \end{pmatrix} , \begin{pmatrix} 3 \\ 4 \end{pmatrix} \right\} ~ , ~ D= \left\{ \begin{pmatrix} 1 \\ 4 \end{pmatrix} , \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right\} $$ The vectors in $D$ are expressed as a linear combination of the ones in $B$, and then the coefficients are used to construct the change of basis matrix $S$, i.e.:
$$ \begin{pmatrix}1 \\ 4 \end{pmatrix} = s_{11}\cdot \begin{pmatrix}1 \\ 2 \end{pmatrix} + s_{12} \cdot \begin{pmatrix}3 \\ 4 \end{pmatrix} \ \ \Rightarrow s_{11}=4 \ \ \text{and} \ \ s_{12}=-1 \\ \begin{pmatrix}2 \\ 3 \end{pmatrix} = s_{21}\cdot \begin{pmatrix}1 \\ 2 \end{pmatrix} + s_{22} \cdot \begin{pmatrix}3 \\ 4 \end{pmatrix} \Rightarrow s_{21}=\frac{1}{2} \ \ \text{and} \ \ s_{22}=\frac{1}{2} \\ \Rightarrow S =\begin{pmatrix}4 & -1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} $$

Method 2:

$$\begin{pmatrix}s_{11} & s_{12} \\ s_{21} & s_{22} \end{pmatrix} \cdot \begin{pmatrix}1 & 3 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix}1 & 2 \\ 4 & 3 \end{pmatrix} \\ \Rightarrow s_{11}=0 \ \ ;\ \ s_{12}= \frac{1}{2} \ \ ; \ \ s_{21}= -5 \ \ ; \ \ s_{22}=\frac{9}{2}\\ \Rightarrow S =\begin{pmatrix}0 & \frac{1}{2} \\ -5 & \frac{9}{2} \end{pmatrix} $$

Answer 1

The method 2) is correct. It is the same as to find a matrix $$ S=\begin{pmatrix}s_{11} & s_{12} \\ s_{21} & s_{22} \end{pmatrix} $$ such that $$ S\begin{pmatrix}1 \\2 \end{pmatrix}=\begin{pmatrix}1 \\4 \end{pmatrix} $$ and $$ S\begin{pmatrix}3 \\4 \end{pmatrix}=\begin{pmatrix}2 \\3 \end{pmatrix} $$ that means exactly to transform the vectors of the first basis to the vectors of the second basis.

I don't see how the method 1) comes from...but it is wrong.

Answer 2

In Method 1, what you really did, when looking at the equations you wrote down, was trying to solve the equation $$ \underbrace{ \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix} }_{=: ~A} \begin{pmatrix} s_{11} & s_{21} \\ s_{12} & s_{22} \end{pmatrix} = \underbrace{ \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} }_{=: ~B} $$ So $\color{blue}{ \text{you were trying to solve } AS = B }$ for $S$.

If you compare this to your matrix $S$, you will see that I exchanged $s_{12}$ and $s_{21}$ to correct a mistake in your calculation.

In this case your result will be $$ S = \begin{pmatrix} 4 & \frac{1}{2} \\ -1 & \frac{1}{2} \end{pmatrix} $$ In Method 2 $\color{blue}{ \text{you were trying to solve } SA = B }$ , which explains why you got two different results: you were solving two different matrix equations.

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Which method is the correct way to calculate the change of basis?

Let's say you have a vector $\boldsymbol{\nu}$ written in the basis $\mathcal{A} = \{\boldsymbol{a}_1, \boldsymbol{a}_2 \}$. This means you have coordinates $\nu_1, \nu_2$ such that, if we also define the matrix $A = (\boldsymbol{a}_1, \boldsymbol{a}_2)$, you can write $$ \boldsymbol{\nu} = \nu_1 \boldsymbol{a}_1 + \nu_2 \boldsymbol{a}_2 = A \begin{pmatrix} \nu_1 \\ \nu_2 \end{pmatrix} $$ Now you want to know how it looks in $\mathcal{B} = \{\boldsymbol{b}_1, \boldsymbol{b}_2 \}$. This means you want to know the coordinates $\nu'_1, \nu'_2$ which lead to the same vector, no matter what basis you use : $$ \boldsymbol{\nu} = \nu'_1 \boldsymbol{b}_1 + \nu'_2 \boldsymbol{b}_2 = B \begin{pmatrix} \nu'_1 \\ \nu'_2 \end{pmatrix} = A \begin{pmatrix} \nu_1 \\ \nu_2 \end{pmatrix} $$ You continue by figuring out how the basis $\mathcal{A}$ is written in $\mathcal{B}$, solving the equations $\boldsymbol{a}_j = s_{j1}\boldsymbol{b}_1 + s_{j2} \boldsymbol{b}_2 ~$. This can be written as one matrix equation $$ A = (\boldsymbol{a}_1, \boldsymbol{a}_2) = B (\boldsymbol{s}_1, \boldsymbol{s}_2) = BS $$ Which leads to $$ \boldsymbol{\nu} = A \begin{pmatrix} \nu_1 \\ \nu_2 \end{pmatrix} = BS \begin{pmatrix} \nu_1 \\ \nu_2 \end{pmatrix} $$ Comparing the equations we get $$ \begin{pmatrix} \nu'_1 \\ \nu'_2 \end{pmatrix} = S \begin{pmatrix} \nu_1 \\ \nu_2 \end{pmatrix} ~~~~~ \text{and} ~~~~~ S = (B^{-1})A $$ where $S$ is the change of basis matrix you are looking for: given the coordinates in $\mathcal{A}$, you get the ones in $\mathcal{B}$. As you can see, $S$ can be calculated from $A$ and $B$, for example using Gaussian elimination.

In the methods you were using, you calculated $S = (A^{-1})B$ (method 1) and $S = B(A^{-1})$ (method 2). So none of the two yields the result you wanted. Method 1 is in some sense the closest, since you calculated the inverse of the matrix you actually wanted.