# change of coordinate matrix with non-standard basis

I want to find the change of coordinate matrix that changes ${\beta^{'}}$ coordinates into ${\beta}$ coordinates.

$${\beta^{'}}=\{(0,10),(5,0)\}$$ $${\beta}=\{(-1,3),(2,-1)\}$$

I received a hint that said to just compute $[I]_{\beta^{'}}^\beta$

I don' have any problems writing a linear transformation as a matrix when the basis are the standard basis but when I try this case I don't get the same answer that's in the solution.

I computed $[I]_{\beta^{'}}^\beta$ by first computing the identity function with respect to $\beta^{'}$ and got

$$I(0,10)=(0,10)$$ $$I(5,0)=(5,0)$$

Then when I try to write these vectors in with respect to $\beta$, similar to how I would with respect to the standard basis I get

$$\begin{bmatrix} 30 & -5 \\ -10 & 10\\ \end{bmatrix}$$

• Are you asking why the suggested method works (i.e. computing $[I]_{\beta'}^{\beta}$) or are you just unable to compute what it is? – user160738 Dec 12 '17 at 14:41
• unable to compute – john fowles Dec 12 '17 at 14:41

To avoid slightly confusing notations, let's denote $x'=(0,10)$ and $y'=(5,0)$ and $x=(-1,3)$,$y=(2,-1)$.

Your aim is, given any element $v=ax'+by'=cx+dy$ (i.e. its coordinate is $(a,b)$ with respect to $\beta'$ and $(c,d)$ with respect to $\beta$) you want some matrix $A$ such that $A(a,b)^{t}=(c,d)^{t}$ ($t$ is transpose here. So these are column vectors)

At this point, you might notice that in fact the whole job is dependent on knowing how to represent $x'$ in terms of $x,y$ and the same for $y'$. Because, then, you can write $ax'+by'$ in terms of $x$ and $y$.

But it's easy to see that $x'=2(2x+y)=4x+2y$ and $y'=x+3y$. This means, in fact, $$cx+dy=(4x+2y)a+(x+3y)b=(4a+b)x+(2a+3b)y$$

Since $x,y$ are members of a basis, it follows that $4a+b=c$ and $2a+3b=d$. So

$$A=\begin{pmatrix} 4 & 1 \\ 2 & 3\end{pmatrix}$$.

Some simple checks can assure you that this works;

This method also works in more general case other than $\mathbb{R}^2$.

Summary: Write the member of the current basis as a linear combination of members of a basis you want to change to. The coefficients of these linear combinations give you the matrix you want.

• I understand this method, but what about using the process of converting the linear transformation $[I]_{\beta^{'}}^{\beta}$ into a matrix? – john fowles Dec 12 '17 at 15:50
• @johnfowles I don't understand what you mean by process of converting linear transformations in to matrices. What I've done is exactly that. In general, if everything is represented by coordinates, then to find matrix representing a linear transformation $T$, you need to compute $T(1,0)^{t}$ and $T(0,1)^{t}$ with respect to the other basis of the space containing the image of $T$. This is because of the linearity of $T$. What I've done is exactly this – user160738 Dec 12 '17 at 16:17
• I seemed to have thought that there were two distinct approaches; but I now see that they're the same. Thanks! – john fowles Dec 12 '17 at 18:34