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)

  • $\begingroup$ You're wrong: the change of base matrix allows to compute old coordinates as linear functions of the new coordinates. $\endgroup$ – Bernard Jan 4 '15 at 3:56
  • $\begingroup$ @Bernard sorry, I didn't understand. Shouldn't the matrix $M$ multiply a vector with coordinates in $B$ and transform it to $C$? $\endgroup$ – Guerlando OCs Jan 4 '15 at 3:57
  • 1
    $\begingroup$ @Timbuc: Would you care to explain your comment? I understand there is confusion about which direction to call change of basis from B to C, but this is just a matter of nomenclature. I cannot see how the situation would be different if one wrote matrices on the other side (and they would have to be transposed of course). The relation between change of bases is inverse, not transpose. $\endgroup$ – Marc van Leeuwen Jan 12 '15 at 11:37
  • 1
    $\begingroup$ @Timbuc: I don't understand anything of what you are saying. But the following is a fact, independent of how one writes coordinate vectors and matrices. In $\Bbb R^2$, if apart from the standard basis $B$ you consider the ordered basis $C=[(2,0),(0,2)]$, then transforming a vector with coordinates $(x,y)$ w.r.t. $B$ into coordinates w.r.t. $C$ gives $(x/2,y/2)$. Traditionally one calls the matrix $P=2I$ with the coordinates of $C$ the change of basis matrix (from $B$ to $C$). Converting $(x,y)\to(x/2,y/2)$ by left or right matrix multiplication requires the inverse matrix $P^{-1}$. $\endgroup$ – Marc van Leeuwen Jan 12 '15 at 12:45
  • 1
    $\begingroup$ @Timbuc: Yes, I know this and it is exactly what I meant in my first comment. The point is that in the end everything is the same, just written differently. In particular it does not change the fact that given the coordinates of a new basis, one needs to compute the inverse of a matrix in order to transform coordinates to that new basis. The two issues are quite unrelated, so changing conventions on one issue does not solve anything for the other issue. $\endgroup$ – Marc van Leeuwen Jan 12 '15 at 13:03

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 a new base $\mathcal B\,' $ has as column-vectors the coordinates of the newbase in the oldbase. It is the matrix of the identity map from $ (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 the new 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 $b_1=\begin{bmatrix} 1 \\-\tfrac 13\end{bmatrix}$, $\quad b_2=\begin{bmatrix} 0 \\\tfrac 23\end{bmatrix}$.

  • $\begingroup$ Why the hell it's named "from $B$ to $B'$", if when we multiply this matrix by the vector $X'$ we end up in the basis $B$, not $B'$? Why 'from $B$ to $B'$'??? $\endgroup$ – Guerlando OCs Jan 4 '15 at 19:38
  • $\begingroup$ Probably because most of the time, it's the vectors of the basis $\mathcal B'$ which are given. Anyway, if you don't understand that, you'll never understand the formulae for the matrix of a linear or bilinear map on a change of basis. $\endgroup$ – Bernard Jan 4 '15 at 21:56

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.

  • $\begingroup$ I don't even know what your notation means, I'm so lost $\endgroup$ – Guerlando OCs Jan 4 '15 at 4:50
  • $\begingroup$ You do not know what the transpose $u^T$ of a vector $u$! $\endgroup$ – Mhenni Benghorbal Jan 4 '15 at 4:53
  • $\begingroup$ I mean, I don't know why you're multiplying the matrix. I'm lost with the part of 'change from $B$ to $C$' $\endgroup$ – Guerlando OCs Jan 4 '15 at 4:54
  • $\begingroup$ You are asked to find the vectors $b_1$ and $b_2$ so we need to assume them as $b_1=(a,b)$ and $b_2=(c,d)$. Note that the Matrix $M$ maps the vector $b_1=(a,b)$ to the vector $(1,1)$ and same for $b_2=(c,d)$. Now just take your time to work the problem out. $\endgroup$ – Mhenni Benghorbal Jan 4 '15 at 4:59
  • 1
    $\begingroup$ Nice, I understand what you're doing. It's because I was following my book. According to the construction the book made, the matrix first column should be $b_1$ written in terms of $C$. The same for the second column. Then, if I have $b_1$ and $b_2$ written in $C$, I should solve for $b_1$ and $b_2$, rigth? $\endgroup$ – Guerlando OCs Jan 4 '15 at 5:08

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: $$(1)\begin{bmatrix} 1\\1\end{bmatrix} + (2) \begin{bmatrix} 0\\2\end{bmatrix} = \begin{bmatrix} 1\\5\end{bmatrix} = \mathbf {b_1} $$
$$(0)\begin{bmatrix} 1\\1\end{bmatrix} + (3) \begin{bmatrix} 0\\2\end{bmatrix} = \begin{bmatrix} 0\\6\end{bmatrix} = \mathbf {b_2} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.