Prove: if A is a transformation matrix of orthonormal bases then A is unitary

Question

I'm stuck proving this statement, perhaps because too many indexes are involved. $A \in M \tiny nxn$ (F) Prove that if there exist B', B orthonormal bases of $F^n$, such that A is a transformation matrix from B' to B, then A is unitary.

I was trying to calculate what is the i,j-th cell of A, then to calculate the i,j-th cell of A*, and then show AA* = I, but I can't keep up with this many indexes.

Thanks in advance, Barak

Answer 1

It is easy to verify that $||A(x)||=||x||$ for all $x$. It is also possible to express $\langle x, y \rangle$ in terms of norms: $\langle x, y \rangle =\frac {||x+y||^{2}+||x-y||^{2}} 2$. This gives $\langle Ax, Ay \rangle =\langle x, y \rangle$ for all $x,y$. Hence $\langle A^{*}Ax, y \rangle = \langle Ax, Ay \rangle =\langle x, y \rangle$ which gives $A^{*}A=I$. Hence $AA^{*}=I$ too and $A$ is unitary.