# Change of coordinate matrices “example”

My book gives an example in its attempt to explain change of coordinate matrices:

In $\Bbb R^2$, let $\beta = \{(1,1),(1,-1)\}$ and $\beta' = \{(2,4),(3,-1)\}$. Since $(2,4) = 3(1,1) - 1(1,-1)$ and $(3,1)=2(1,1)+1(1,-1)$, the matrix that changes $\beta'$-coordinates into $\beta$-coordinates is $$Q = \begin{pmatrix} 3&2\\ -1&1 \end{pmatrix}$$ Thus, for instance, $$[(2,4)]_\beta=Q[(2,4)]_{\beta'} = Q\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 3\\-1 \end{pmatrix}.$$

I follow the original statement, and I understand how Q is found (though I am definitely shaky on doing it myself), but then that "thus for instance" part makes absolutely no sense to me. What just happened there? I'm continuing to read and trying to understand the following examples, but some seem clear as day (like the first part of this) and others completely opaque (as is the latter part).

(Perhaps of note, I know what eigen-vectors/values are, but we have not reached that in the book yet.)

The expression $[(2,4)]_{\beta'}$ means the coordinates of vector $(2,4)$ with respect to the basis $\beta'$. These coordinates are $(1,0)$ since $$(2,4)=1\cdot(2,4)+0\cdot(3,-1).$$ To find the coordinates of the same vector with respect to basis $\beta$ you can use the matrix you found. The calculation shows that these coordinates are $(3,-1)$. Indeed, $$(2,4)=3\cdot(1,1)+(-1)\cdot (1,-1).$$
Recall that if $[(x, y)]_\beta = (a, b)$, then: $$(x, y) = a(1, 1) + b(1, -1)$$ So in particular, we know that: $$[(2, 4)]_\beta = (3, -1) \\ [(3, -1)]_\beta = (2, 1) \\ [(2, 4)]_{\beta'} = (1, 0) \\ [(3, -1)]_{\beta'} = (0, 1) \\$$ The first equal sign is just the definition of a change of basis matrix. After that, we can just substitute a particular vector and convert back to the standard basis and see that everything makes sense: \begin{align*} [(x, y)]_\beta &= Q[(x, y)]_{\beta'} \\ [(2, 4)]_\beta &= Q[(2, 4)]_{\beta'} \\ \begin{pmatrix} 3 \\ -1 \end{pmatrix} &= \begin{pmatrix} 3&2\\ -1&1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{align*}