As the title, let $(V,\langle,\rangle)$ be a complex inner product space and assume $S_1=(u_1,\ldots,u_n)$, $S_2=(v_1,\ldots,v_n)$ are orthonormal bases of $V$. Prove that the change of basis matrix $M_ IV(S_2,S_1)$ is a unitary matrix.
(There is a hint that let $S$ be the operator s.t. $S(u_i)=v_i$ and prove this is a unitary operator.)
Best Answer
We need to show that $\langle Mx,My\rangle=\langle x,y\rangle$ for all $x,y$. By definition, the change of basis matrix $M$ has the property that
$$ Mu_i=v_i $$
Let $x=\alpha_1u_1+\ldots+\alpha_nu_n$ and $y=\beta_1u_1+\ldots\beta_nu_n$. Then, since $\{u_i\}$ is an orthonormal basis, $\langle u_i,u_j\rangle=\delta_{ij}$, the Kronecker delta. Thus, by expanding out the inner product $\langle x,y\rangle$, we see that $$ \langle x,y\rangle=\sum_j\alpha_j\bar{\beta}_j $$
Now compute $\langle Mx,My\rangle$:
$$ \begin{align*} \langle Mx,My\rangle&=\langle M(\alpha_1u_1+\ldots+\alpha_nu_n),M(\beta_1u_1+\ldots+\beta_nu_n)\rangle\\ &=\langle \alpha_1Mu_1+\ldots+\alpha_nMu_n,\beta_1Mu_1+\ldots+\beta_nMu_n\rangle\\ &=\langle \alpha_1v_1+\ldots+\alpha_nv_n,\beta_1v_1+\ldots+\beta_nv_n\rangle\\ &=\sum_i\sum_j\alpha_i\bar{\beta}_j\langle v_i,v_j\rangle \end{align*} $$
Since $\{v_i\}$ is orthonormal, $\langle v_i,v_j\rangle=\delta_{ij}$, and so the above is exactly $\langle x,y\rangle$