Mathematics Asked by a9302c on December 15, 2020

I am given that **A** is an orthogonal matrix of order $n$, and **$u, v$** are Vectors in the $R^n $ space.

I need to prove that $||u|| = ||Au||$. The first step of the solution hint I am given is that $$||Au||^2 = (Au)^T(Au)$$. Why is this so? I know that $A^{-1} = A^T$ in the definition of an orthogonal matrix, but how does this contribute to the above statement? Or is there some other property I’m missing out on?

Assume that $A$ is orthogonal, i.e. that $A^T A = I$ (observe that this is equivalent to $A^{-1} = A^T$). We consider

$$ || A x ||^2 = (Ax)^T (Ax) = x^T A^T A x = x^T x = || x || ^2, $$

which shows that $|| A x || = || x ||$.

Comment: On advise from another user, I posted a more thorough version of my earlier comment as this answer.

Correct answer by Fenris on December 15, 2020

If $u=(u_1,u_2,ldots,u_n)$, thenbegin{align}u^top u&=begin{pmatrix}u_1\u_2\vdots\u_nend{pmatrix}(u_1,u_2,ldots,u_n)\&=u_1^{,2}+u_2^{,2}+cdots+u_n^{,2}\&=|u|^2.end{align}

Answered by José Carlos Santos on December 15, 2020

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