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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.

shaktiman
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  • 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. – shaktiman Oct 12 '14 at 06:59
  • 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. – Gerry Myerson Oct 12 '14 at 07:00
  • 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. – shaktiman Oct 12 '14 at 07:06
  • 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. – Gerry Myerson Oct 12 '14 at 07:46
  • 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 $v$. – shaktiman Oct 12 '14 at 08:47
  • I give up. Solution posted. – Gerry Myerson Oct 12 '14 at 09:59

2 Answers2

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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.

Gerry Myerson
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    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. – shaktiman Oct 12 '14 at 20:51
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If we can show that $A^TAv=v, \space\forall v\in \mathbb{R}^n$, we will prove that $A^TA$ is the identity transformation and hence $A^TA=I$. Now since $v_1,\dots,v_n$ are orthonormal, they form a set of basis vectors. So proving $A^TAv_i=v_i \space \forall i={1,\dots, n}$ will suffice.

For an arbitrary $i$, let $A^TAv_i = \sum\limits_{j=1}^n a_jv_j \space [\because v_1,\dots v_n \text{ are basis vectors}] \tag{1}\label{1}$

Now, \begin{align} \langle Av_k, Av_i\rangle = v_k^T(A^TAv_i) &= v_k^T(\sum_{j=1}^n a_jv_j) & \text{using (1)}\\ &=\sum_{j=1}^n a_j(v_k^Tv_j)\\ &=a_k & \because v_1,\dots,v_n \text{ are orthonormal} \\ \end{align}

But since by hypothesis $Av_1,\dots,Av_n$ are orthonormal, we got $$a_k =\begin{cases} 0, &if \space k\neq i\\ 1, &if \space k = i \end{cases} $$

$\therefore A^TAv_i = \sum\limits_{j=1}^n a_jv_j=v_i$

Since we started with an arbitrary $v_i$, this proves that $A^TA=I$

Chayan Ghosh
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