Exercise 2.5.13

Since $Q$ is invertible, we have that $L_Q$ is invertible. We try to check ${\beta }^{\prime }$ is an independent set and hence a basis since $V$ has dimension $n$. Suppose that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^n{a}_j{x}_j^{\prime }=0$. And it means that

Since $\beta$ is a basis, we have that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^n{a}_j{Q}_{\mathit{ij}}=0$ for all $i$. Actually this is a system of linear equations and can be written as

where $v=\left(\begin{array}{cccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill \text{…}\hfill & \hfill a_n\hfill \end{array}\right)$. But since $Q$ is invertible and so $Q^{-1}$ exist, we can deduce that $v=\mathit{vQ}{Q}^{-1}=0Q^{-1}=0$. So we know that $\beta$ is a basis. And it’s easy to see that $Q={[I]}_{\beta }^{\beta }$ is the change of coordinate matrix changing ${\beta }^{\prime }$-coordinates into $\beta$-coordinates.