I'm solving a basic problem in quantum mechanics to change the matrix A into a representation in the basis consisting of eigenvectors of matrix B. And I've noticed that the new mateix A looks significantly different depending on the order I arrange the eigenvectors in the transformation matrix. Why does this matter?
For context, the matrices are: $A=\begin{bmatrix}1&0&0\\0&0&0\\0&0&-1\end{bmatrix}$
$B=\frac{1}{\sqrt{2}}\begin{bmatrix}0&1&0\\1&0&1\\0&1&0\end{bmatrix}$
The eigenvalues and vectors are $a_1=0, a_2=1,a_3=-1$
$v_1=\frac{1}{\sqrt{2}}\begin{bmatrix}1\\0\\-1\end{bmatrix}$
$v_2=\frac{1}{2}\begin{bmatrix}1\\\sqrt{2}\\1\end{bmatrix}$
$v_3=\frac{1}{2}\begin{bmatrix}1\\-\sqrt{2}\\1\end{bmatrix}$
Now, I calculate the matrix A in the new basis using formula $A^v=U^\dagger AU$. Depending on if $U=\begin{bmatrix}v_1&v_2&v_3\end{bmatrix}$ or if $U=\begin{bmatrix}v_3&v_1&v_2\end{bmatrix}$ (the way my prof has in the solutions), I get completely different $A^v$. Could you please explain why the order of the new basis (I thought it is a set?) matters, or point me to some theorem?
The order of a basis does matter. Bases are better defined as a "tuple" of vectors than as a set.
One explanation for the effect of switching vectors (in an orthonormal basis) is as follows. In general, we have $$ \pmatrix{v_1 & v_2 & v_3} \pmatrix{\lambda_1 \\ & \lambda_2 \\ && \lambda_3}\pmatrix{v_1 & v_2 & v_3}^\dagger = \lambda_1 v_1 v_1^\dagger + \lambda_2 v_2 v_2^\dagger + \lambda_3 v_3 v_3^\dagger. $$ If the eigenvalues $\lambda_1,\lambda_2,\lambda_3$ are distinct, then it is clear that switching the order of the vectors produces a different sum on the right hand side.