The matrix of basis change from basis $B = \{b_1,b_2\}$ to the basis $C = \{(1,1),(0,2)\}$ is
$$ M = \begin{bmatrix} 1 & 0 \\2&3\end{bmatrix}$$
Find basis $B$.
Well, here's what I did:
Since $M$ changes from $B$ to $C$, then it Works like this:
$${\begin{bmatrix}& \\\\\\\end{bmatrix}}_C = \begin{bmatrix} 1 & 0 \\2&3\end{bmatrix}{\begin{bmatrix}a_1\\a_2\end{bmatrix}}_B$$
In other words, a vector with coordinates in $C$ is written as the matrix $M$ multiplied by the vector written in $B$.
$${\begin{bmatrix}& \\\\\\\end{bmatrix}}_C = a_1\begin{bmatrix} 1 \\2\end{bmatrix}+a_2\begin{bmatrix} 0 \\3\end{bmatrix}$$
where $\begin{bmatrix} 1 \\2\end{bmatrix}$and $\begin{bmatrix} 0 \\3\end{bmatrix}$ are the vectors $b_1,b_2$ written in $C$, so:
$$\vec{b_1} = 1(1,1) + 2(0,2) = (1,5)\\\vec{b_2} = 0(1,1) + 3(0,2) = (0,6)$$
but this is not the answer.
Could somebody tell me what I'm doing wrong? (instead os just solving in another way)
$\endgroup$ 143 Answers
$\begingroup$Here is how I would do it:
- First remember the change of basis matrix $P_\mathcal B^{\mathcal B'}$ from an
old base$\mathcal B$ of a vector space $ E $ to anew base$\mathcal B\,' $ has as column-vectors the coordinates of thenewbasein theoldbase. It is the matrix of theidentity mapfrom $ (E, \mathcal B\,') $ to $ (E, \mathcal B) $. - The change of basis matrix the other way is just the inverse matrix of the previous one: $$P_{\mathcal B'}^{\mathcal B}=\bigl(P_\mathcal B^{\mathcal B'}\bigr)^{-1}$$
- It allows to express the
old coordinates$X$ of a vector from thenew coordinates$X'$: $$X=P_\mathcal B^{\mathcal B'}X'$$ - You can compose the change of basis matrix: $$ P_{\mathcal B}^{\mathcal B''} = P_{\mathcal B}^{\mathcal B’}P_{\mathcal B’}^{\mathcal B''}$$
Denote the canonical base as $\mathrm{Can}$. Determining $\mathcal B$ is the same as determining $P_{\mathrm{Can}}^{\mathcal B}$. By the composition formula, we have: $$P_{\mathrm{Can}}^{\mathcal B}=P_{\mathrm{Can}}^{\mathcal C}P_{\mathcal C}^{\mathcal B}=P_{\mathrm{Can}}^{\mathcal C}\bigl(P_{\mathcal B }^{\mathcal C}\bigr)^{-1}.$$
Now $ P_{\mathrm{Can}}^{\mathcal C}= \begin{bmatrix} 1 & 0 \\1&2\end{bmatrix} $, and we compute with the pivot method that $$\bigl(P_{\mathcal B }^{\mathcal C}\bigr)^{-1}= \begin{bmatrix} 1 & 0 \\2&3\vphantom{\tfrac 13}\end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 \\-\tfrac23& \tfrac 13\end{bmatrix}$$ so that $$P_{\mathrm{Can}}^{\mathcal B}=\begin{bmatrix} 1 & 0 \\-\tfrac 13& \tfrac 23\end{bmatrix}$$
Hence $b_1=\begin{bmatrix} 1 \\-\tfrac 13\end{bmatrix}$, $\quad b_2=\begin{bmatrix} 0 \\\tfrac 23\end{bmatrix}$.
$\endgroup$ 2 $\begingroup$Apply the matrix $M$ to the basis vectors in $B = \{b_1=(a,b),b_2=(c,d)\}$ and note that
$$M b_1^T = (1,1)^T \longrightarrow (1) $$
$$M b_2^T = (0,2)^T \longrightarrow (2) $$
then solve the resulting system in $\left\{ a, b, c, d \right\}$. Note that you can solve $(1)$ for $a$ and $b$ and $(2)$ for $c$ and $d$ separately.
$\endgroup$ 6 $\begingroup$I know it's an old question but I'm gonna post my answer for future readers.
I think you are right as a matter of fact.
Let M be the change of basis matrix from base B to base C.
Let [v]B be the coordinate vector of some vector v with respect to basis B,
let [v]C be the coordinate vector of that same vector v with respect to basis C.
By definition M[v]B = [v]C
Now let B = {b1,...,bn },
the coordinate vector of any bi with respect to B is simply ei,
from the standard basis:
$$\mathbf{[b_i]}_B = \mathbf{e}_i =\begin{bmatrix} 0 \\ .\\.\\.\\1\\.\\.\\.\\0\\\end{bmatrix}$$ where 1 is the i-th coordinate.
Now, as you may know, the product between any matrix A and ei gives back Ai (the i-th column of A),
We can use these facts to arrive to the following conclusion:
M[bi]B = [bi]C = Mi,
hence in your particular case we can infer that $$\begin{bmatrix} 1 \\2\end{bmatrix} = \mathbf{[b_1]}_C $$
$$\begin{bmatrix} 0 \\3\end{bmatrix} = \mathbf{[b_2]}_C $$
Therefore: $$(1)\begin{bmatrix} 1\\1\end{bmatrix} + (2) \begin{bmatrix} 0\\2\end{bmatrix} = \begin{bmatrix} 1\\5\end{bmatrix} = \mathbf {b_1} $$
$$(0)\begin{bmatrix} 1\\1\end{bmatrix} + (3) \begin{bmatrix} 0\\2\end{bmatrix} = \begin{bmatrix} 0\\6\end{bmatrix} = \mathbf {b_2} $$
