If $Av_1$ $Av_2$ ... $Av_k$ are orthonormal vectors in $R^n$ and $v_1$ $v_2$ ... $v_k$ are also orthonormal vectors in $R^n$. Show that the Matrix A must be orthogonal i.e. $A^TA=I$. I can prove it the other way round but can't this way. Thanks for your help in advance.

  • $\begingroup$ Hey @CareBear thanks for the explanation. I meant I can prove that if A is orthogonal then the vectors $Av_i$ are orthonormal but not if I have to prove it the other way round. $\endgroup$ – shaktiman Oct 12 '14 at 6:59
  • $\begingroup$ If $v$ has all zeroes except a one in the $k$th entry, then $Av$ is the $k$th column of $A$. Apply this to the standard basis. $\endgroup$ – Gerry Myerson Oct 12 '14 at 7:00
  • $\begingroup$ I don't understand what you mean by applying to standard basis. Does somehow multiplying with the standard basis give the result? The v's are general vectors. $\endgroup$ – shaktiman Oct 12 '14 at 7:06
  • $\begingroup$ The $v$ are whatever you want them to be. I want them to be the standard basis, because of what happens when you compute $Av$ when $v$ is part of that basis. $\endgroup$ – Gerry Myerson Oct 12 '14 at 7:46
  • $\begingroup$ Ok, but how does the example of a standard basis serve as a generalized proof for all $v$. With each $ v $ the weighting changes so we don't have the argument valid for other $v$'s. Can we somehow show $V^T (AA^T-I) V=0$ implies $A^TA=I$? Here $V$ is the matrix with column vectors $v$. $\endgroup$ – shaktiman Oct 12 '14 at 8:47

Let $e_1,\dots,e_n$ be the standard basis for ${\bf R}^n$. These vectors are orthonormal, so, by hypothesis, the vectors $Ae_1,Ae_2,\dots,Ae_n$ are also orthonormal. But these vectors are the columns of $A$. Thus, the columns of $A$ are orthonormal. Thus, $A$ is an orthogonal matrix.

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    $\begingroup$ Thanks for the answer. This approach is much simpler than the ab initio derivation I was thinking about and I understood that it is correct. Sorry for not understanding earlier. $\endgroup$ – shaktiman Oct 12 '14 at 20:51

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