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If $B = (e_1,e_2,\ldots, e_n)$ is a basis for an inner product space $V$ and $B' = (f_1,f_2,\ldots,f_n)$ is an orthonormal basis of $V$.

Is the change of basis matrix $P$ necessarily orthogonal?

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The change basis matrix $P$ is $$P=\left(f_1 f_2\cdots f_n\right)$$ where $f_i$ is a column vector and since the basis $B'$ is orthonormal we have $$f_i \ . f_j=\delta_{ij}$$ where $\delta$ is the Kroneker symbol and hence $P$ is orthogonal

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  • $\begingroup$ oh my god i am stupid, thank you so much $\endgroup$
    – JC784
    May 30 '13 at 20:12
  • $\begingroup$ But the basis $e_i$ is not necessarily orthonormal, is it? $\endgroup$
    – J. J.
    May 30 '13 at 20:15
  • $\begingroup$ @J.J. Yes $e_i$ is not necessarily orthonormal. $\endgroup$
    – user63181
    May 30 '13 at 20:19
  • $\begingroup$ @Sami Ben Romdhane: So if I read this right, $P$ maps from basis $f$ to basis $e$, i.e. the column vectors $f_i$ are the coordinates of $f_i$ in the basis $e$. But why does the inner product in $V$ being $0$ imply that the dot product of the coordinates in the (non-orthonormal) basis $e$ is also $0$? $\endgroup$
    – J. J.
    May 30 '13 at 20:23
  • $\begingroup$ @J.J. To see that $P$ is orthogonal with different viewpoint think to this questions: did you know that an orthogonal matrix transform an orthonormal basis to an orthonormal basis? did you that the change matrix is a representation of the identity endomorphism? Do you understand? $\endgroup$
    – user63181
    May 30 '13 at 20:31
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Take any non-orthogonal invertible matrix $Q$. We wish to find a basis $e_1,\dots,e_n$ such that $Q$ is the change of basis from $(e_1,\dots,e_n)$ to $(f_1,\dots,f_n)$. But for this we may just set $e_i = Q_{1i} f_1 + \dots + Q_{ni} f_n$.

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