Change of basis (coordinate vector confusion)

Question

Question is why's Q equal to an identity map from $\beta'$ to $\beta$ and equal to the matrix of $\beta$? I need some clarification for change of basis concept.

Answer 1

Let $v_1 = \begin{pmatrix} 1\\1 \end{pmatrix}$ and $v_2 = \begin{pmatrix} 1\\2 \end{pmatrix}$. Also, let $e_1 = \begin{pmatrix} 1\\0 \end{pmatrix}$, $e_2 = \begin{pmatrix} 0\\1 \end{pmatrix}$ and put $\gamma = \{ e_1,e_2 \}$.

It is easy to see that $$[L_A]_\gamma := [L_A]_\gamma^\gamma = A$$ since $Ae_1$ and $Ae_2$ are the columns of $A$, right?, Now, note that $$\begin{align} [L_A]_\beta = [L_A]_\beta^\beta &= [\textrm{id}_{\mathbb R^2} \circ L_A \circ \textrm{id}_{\mathbb R^2}]_\beta \\ &= [\textrm{id}_{\mathbb R^2}]_\gamma^\beta [L_A]_\gamma^\gamma [\textrm{id}_{\mathbb R^2}]_\beta^\gamma \\ &= ([\textrm{id}_{\mathbb R^2}]_\beta^\gamma)^{-1} A [\textrm{id}_{\mathbb R^2}]_\beta^\gamma \end{align}$$ so, the matrix $Q$ that you are trying to find is $[\textrm{id}_{\mathbb R^2}]_\beta^\gamma$, which is easy to compute. Can you see why this matrix can be obtained by taking $v_1$ and $v_2$ as its columns?