# How to find the matrix representation of an operator from its action on a basis?

First, I apologize if something is poorly written but English is not my first language.

I know that these exercises have been solved in this question. But I do not agree. Inner product and concrete vectors are used and I think that this question has to be solved only with information we have from the beginning of chapter 2 until the exercise.

So I think we only have to use eq. 2.12 in this way:

if we write $$\left|0\right>=\left|v_{0}\right>$$ and $$\left|1\right>=\left|v_{1}\right>$$ and we use them as input and output basis, we can write (2.12) as $$A\left|v_{j}\right> = \sum_{i} A_{ij} \left|v_{i}\right>$$; so,

$$A\left|v_{0}\right> = A_{00}\left|v_{0}\right> + A_{10}\left|v_{1}\right> = \left|v_{1}\right> \Rightarrow A_{00}=0; A_{10}=1$$

$$A\left|v_{1}\right> = A_{01}\left|v_{0}\right> + A_{11}\left|v_{1}\right> = \left|v_{0}\right> \Rightarrow A_{01}=1; A_{11}=0$$

$$A = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$

As we can see I don't use concrete basis vectors and this works for basis $$\begin{pmatrix}0 \\1 \end{pmatrix}$$ and $$\begin{pmatrix}1 \\0 \end{pmatrix}$$.

But not with basis like $$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\1 \end{pmatrix}$$ and $$\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\-1 \end{pmatrix}$$.

My feeling was that this solution doesn't depend on the basis but it does. So why?

• @Sam Palmer Erm, why not? You can write each element in the first basis as a linear combination of the second basis elements and the other way around. So their spans are the same. Jun 26 '20 at 19:07