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I have tried to solve this problem but I don't have an idea how to begin, any hints?

For any vector $x$ in $\mathbb{R}^n$ let $(x,x) =\sum\limits_{i=1}^n x_i^2 $ . Let $A$ be a matrix of size $n \times n$ such that $(Ax,Ax) = (x,x)$ for all $x \in \mathbb{R}^n$. Prove that either $\det A = 1$ or $\det A = −1$.

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just observe that transpose(A)*A = Identity and then (detA)^2 = 1 –  rohit May 10 '13 at 9:00
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Note that we have $$x^*A^*Ax = x^*x \implies x^*(A^*A-I)x = 0$$ Now show that if $C$ is a symmetric matrix satisfying $x^*Cx = 0$, then $C=0$. This means $$A^*A = I$$ Now recall the fact that $\det(AB) = \det(A) \cdot \det(B)$ and $\det(A^*) = \overline{\det(A)}$. This gives us $$\vert \det(A) \vert^2 = 1$$ If $A$ is a real matrix, then $\det(A)$ is real and hence $\det(A) = \pm 1$.

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