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Let $A$ be an orthogonal $n\times n$ matrix. Show that $\|A\vec x\|=\|A^{-1}\vec x\|$ for any vector $\vec x$ in $\mathbb R^2$

I want to show that $\|A\vec x\|=\|A^{-1}\vec x\|=\|\vec x\|$

I tried to show that since $A^TA=I$, then using $A^T=A^{-1}$,

$\|A^{-1}\vec x\|=(A^{-1}\vec x)\cdot(A^{-1}\vec x)=(A^{-1}\vec x)^T(A^{-1}\vec x)=\vec x^T(A^{-1})^TA^{-1}\vec x=\vec x^T(A^{T})^TA^{-1}\vec x=\vec x^TAA^{-1}\vec x$.

I got stuck here since by definition $A^TA\neq AA^T$ (or is it)?

Any hint is appreciated!

Michael Hardy
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user95087
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    $AA^{-1}=A^{-1}A=I$ – vadim123 Nov 20 '13 at 22:04
  • @user95087 : I don't understand what the problem is. Everything you have is correct (except a typo), you just need to use $AA^{-1}=I$ as vadim123 wrote and finish up. Your typo is that your displayed equation should begin with the square of the norm of the vector. That gives you $|A \mathbf{v}| = |\mathbf{v}|$ if $A$ is orthogonal. Now prove $A^{-1}$ is also orthogonal, so $|A^{-1} \mathbf{v}| = |\mathbf{v}|$. – Stefan Smith Nov 20 '13 at 23:33
  • @user95087 : I don't know offhand if there is an easier way to prove $|A\mathbf{x}|=|A^{-1}\mathbf{x}|$ for all $\mathbf{x}$ (which is apparently all you had to do) without proving $|A\mathbf{x}|=|A^{-1}\mathbf{x}|=|\mathbf{x}|$ for all $\mathbf{x}$. – Stefan Smith Nov 20 '13 at 23:39
  • @user95087 : by the way, since $A$ is orthogonal, $A^T A = I$ (this is often used as a definition of orthogonal), and by properties of matrix inverses, $A A^T = I = A^T A$. – Stefan Smith Nov 20 '13 at 23:48

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Hint: You need the fact

a matrix $Q$ is orthogonal if its transpose is equal to its inverse: $$Q^\mathrm{T}=Q^{-1}, \,$$ which entails $$ Q^\mathrm{T} Q = Q Q^\mathrm{T} = I.$$

Mhenni Benghorbal
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