Computing the change of coordinate matrix from one basis to another

Question

Given the bases a = {(0,2),(2,1)} and b = {(1,0),(1,1)} compute the change of coordinate matrix from basis a to b.
Then, given the coordinates of z with respect to the basis a as (2,2), use the previous question to compute the coordinates of z with respect to the basis b.

The way I understood the first part was that I have to multiply the vectors of b by the coordinates of the vectors of a to compute the change of coordinate matrix from a to b. This gives me the following matrix: \begin{bmatrix}2&3\\2&1\end{bmatrix}

For the second part, I then have to take the inverse of the matrix I got from above and then multiply it by the coordinates of z to get the coordinates of z with respect to the basis b. The inverse of the matrix is: \begin{bmatrix}-1/4&3/4\\1/2&-1/2\end{bmatrix} which I then multiply by (2,2) to get the coordinates of z with respect to basis b

I am not sure that is correct however.

Answer 1

What you can do is use the changes of coordinates between each basis and the standard basis. So, you want: $A=\begin{pmatrix}0&2\\2&1\end{pmatrix}$ and $B=\begin{pmatrix}1&1\\0&1\end{pmatrix}$.

Then what is normally called the change of basis matrix from $a$ to $b$ would be the matrix that takes vectors written in terms of $b$ and returns them written in terms of $a$. So we get the matrix $C$, where $C=A^{-1}B$.

For the second part apply $C^{-1}$ to $z$.

Answer 2

You need to change the coordinates from the basis $a$ to the canonical basis (multiplying by $A$), then change from the canonical basis to the base $b$ (multiplying by $B^{-1}$ on the left). Finally you apply to $z$,

$$B^{-1}Az.$$

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As a shortcut, you can compute $Az$ and solve the system

$$Bx=Az,$$ this spares computation (but does not give you the matrix).

Answer 3

Let $\mathcal{B} = \{b_1, b_2\}$ and $\mathcal{B}^{\prime} = \{b_1^{\prime}, b_2^{\prime}\}$ be two bases.

If $v$ is a vector expressed with respect to $\mathcal{B}$, then $$v = \begin{bmatrix} b_{1} & b_{2}\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$ for some scalars $c_1$ and $c_2$. Since $\mathcal{B}^{\prime}$ is a basis, the elements of $\mathcal{B}$ can be expressed with respect to $\mathcal{B}^{\prime}$. Hence, \begin{align} v &= \begin{bmatrix} (d_1b_1^{\prime} + d_2b_2^{\prime}) & (d_3b_1^{\prime} + d_4b_2^{\prime}) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \\ &= \begin{bmatrix} b_{1}^{\prime} & b_{2}^{\prime}\end{bmatrix} \begin{bmatrix} d_1 & d_3 \\ d_2 & d_4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \end{align} for some scalars $d_1, d_2, d_3$, and $d_4$.

The vector $\begin{bmatrix} d_1 & d_3 \\ d_2 & d_4 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$ is the coordinate vector of $v$ with respect to $\mathcal{B}^{\prime}$.

In your case, $\mathcal{B} = a$, $\mathcal{B}^{\prime} = b$, and $\begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = z$. Can you determine $\begin{bmatrix} d_1 & d_3 \\ d_2 & d_4 \end{bmatrix}$?