# Find the change of basis matrix for this non standard basis of $\mathbb{R^2}$

I have a question on a revision problem sheet. Let $$e_1=(1,0)$$ and $$e_2=(0,1)$$ be the standard basis of $$\mathbb{R^2}$$. And let $$e_1'=(1,1)$$ and $$e_2'=(1,-1)$$ be a non-standard basis of $$\mathbb{R^2}$$. Find the change of base matrix that converts from the standard basis $$\{e_1,e_2\}$$ to the non-standard basis $$\{e_1',e_2'\}$$.

My answer was the following matrix: $$\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \quad$$

I got this from finding a matrix that maps each standard basis element to the non standard basis element.

The answers on the problem sheet however say it should be: $$\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \quad$$

Am I wrong?

This made me wonder are basis ordered or unordered sets? Intuitively I would think that the geometrically the order doesn't matter, but the orientation would, so I would assume that order does intact matter?

In fact, it’s neither. You can easily see that neither matrix is correct: applying the change of basis to $$e_1'$$ should produce $$(1,0)^T$$, but the neither matrix gives this result.
You’ve made a fairly common mistake here. The issue isn’t the order of the basis elements, but the direction in which you’re performing the coordinate mapping. Since a linear transformation is determined by its action on the basis vectors, the change-of-basis matrix $$B$$ is the solution to the equation $$\begin{bmatrix}1&0\\0&1\end{bmatrix} = B\begin{bmatrix}1&1\\1&-1\end{bmatrix}.$$ That is, it’s the inverse of the matrix that you constructed, which is easily found to be $$B = \frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}.$$
Another way to look at it is in terms of the inputs and outputs to the transformation represented by the matrix. The product of a matrix and vector is a linear combination of the columns of the matrix. In the matrix that you constructed, those columns are expressed relative to the standard basis, so the product is also expressed relative to the standard basis. For a change of basis, however, you want the product to be expressed relative to the new basis, not the old one. This means that columns of the change-of-basis matrix must also be expressed relative to this new basis, so constructing the matrix by writing down the coordinates of $$e_1'$$ and $$e_2'$$ relative to the standard basis can’t be right.
The matrix you gave converts vectors from the non-standard basis to the standard basis. The inverse of that matrix would answer the question. I get : $$\begin{pmatrix}\dfrac 12&\dfrac12\\\dfrac 12&-\dfrac 12\end{pmatrix}$$.