Change of coordinates between bases

Question

I have a subspace of $\Bbb R^4$ of dimension 2, with two bases of the same subspace. A = {$\vec a_1, \vec a_2$} and B = {$\vec b_1, \vec b_2$}.

How can I find the matrix C for the change of basis from B to A, or said in other words how can I find the matrix C that when I multiply a vector $\vec v$ represented in B and A coordinates, the result is the same vector represented in A coordinates, $ C·\begin{bmatrix} \vec v \end{bmatrix}_B = \begin{bmatrix} \vec v \end{bmatrix}_A$.

I see that the matrix C should be a 2x2 matrix, and using the formulas for passing from B coordinates to standard coordinates I can construct an equation, something like:

$$\begin{bmatrix} b_{1x} & b_{2x} \\ b_{1y} & b_{2y} \\ b_{1z} & b_{2z} \\ b_{1t} & b_{2t}\end{bmatrix} · \begin{bmatrix} c_{1b} \\ c_{2b}\end{bmatrix} = \vec v$$

$$\begin{bmatrix} a_{1x} & a_{2x} \\ a_{1y} & a_{2y} \\ a_{1z} & a_{2z} \\ a_{1t} & a_{2t}\end{bmatrix} · \begin{bmatrix} c_{1a} \\ c_{2a}\end{bmatrix} = \vec v$$

$$\begin{bmatrix} b_{1x} & b_{2x} \\ b_{1y} & b_{2y} \\ b_{1z} & b_{2z} \\ b_{1t} & b_{2t}\end{bmatrix} · \begin{bmatrix} c_{1b} \\ c_{2b}\end{bmatrix} = \begin{bmatrix} a_{1x} & a_{2x} \\ a_{1y} & a_{2y} \\ a_{1z} & a_{2z} \\ a_{1t} & a_{2t}\end{bmatrix} · \begin{bmatrix} c_{1a} \\ c_{2a}\end{bmatrix}$$

But I don't see how can I obtain matrix C from here.

Answer 1

Suppose that $\vec{b_1}=\alpha_{11}\vec{a_1}+\alpha_{21}\vec{a2}$ and that $\vec{b_2}=\alpha_{12}\vec{a_1}+\alpha_{22}\vec{a2}$. Then $C=\left(\begin{smallmatrix}\alpha_{11}&\alpha_{12}\\\alpha_{21}&\alpha_{22}\end{smallmatrix}\right)$.

Answer 2

HINT

To find the change of basis we need to complete the two basis with the vectors $c_3$ and $c_4$ and use the standard basis, notably

• consider the matrices $M_A=[\vec a_1 \quad \vec a_2 \quad \vec c_3 \quad \vec c_4]$ and $M_B=[\vec b_1\quad \vec b_2 \quad \vec c_3\quad \vec c_4]$
• consider a vector $\vec v$ and indicate with $\vec v_A$ its representation in the basis A, then the representation of $\vec v$ in the standard basis is $\vec v_S=M_A\vec v_a$
• for B we obtain $\vec v_S=M_B\vec v_B$ and then

$$M_A\vec v_A=M_B\vec v_B\implies \vec v_A=M_A^{-1}M_B\vec v_B \quad C=M_A^{-1}M_B$$