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Let $x$ is give numbers.and define the matrix $A$ such $$A^T\cdot\begin{bmatrix} 1&x\\ x&1 \end{bmatrix}\cdot A=1$$

this book say it is clear have $$A=\dfrac{1}{\sqrt{2(1+x)}}\begin{bmatrix} 1\\ 1 \end{bmatrix}$$

I think this not true,let $$A=\begin{bmatrix} a\\ b \end{bmatrix}$$ then we have $$A^T\cdot\begin{bmatrix} 1&x\\ x&1 \end{bmatrix}\cdot A=\binom{a+bx}{b+ax}\cdot\begin{bmatrix} a\\ b \end{bmatrix}=a^2+b^2+2abx=1$$

First which is wrong?

why is "clear" can you explain? Thank you,If not it is clear,can you solve this equation

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  • $\begingroup$ Is there some additional condition that means that $A = (1\ \ 0)^T$ doesn't work? $\endgroup$ – Ben Millwood Jan 4 '15 at 12:56
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    $\begingroup$ It seems like the given answer does work and is the only answer such that $a = b$. Is that necessary for some reason? Or is it just the case that the book is trying to give a solution rather than the solution? $\endgroup$ – Ben Millwood Jan 4 '15 at 12:59
  • $\begingroup$ oh,maybe this book is not true, if $a=b$ is true $\endgroup$ – china math Jan 4 '15 at 13:07
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This is true and clear

if you let $$A=\begin{bmatrix} a\\ b \end{bmatrix}$$ then we have $$A^T\cdot\begin{bmatrix} 1&x\\ x&1 \end{bmatrix}\cdot A=\binom{a+bx}{b+ax}\cdot\begin{bmatrix} a\\ b \end{bmatrix}=a^2+b^2+2abx=1$$

the book assumed a case where $a=b$.

This will then give $a^2+a^2+2a^2x=1$ and solving for a gives: $$a=\dfrac{1}{\sqrt{2(1+x)}}$$ Therefore we will have $$A=\dfrac{1}{\sqrt{2(1+x)}}\begin{bmatrix} 1\\ 1 \end{bmatrix}$$

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Is the above relation meant to be true for all real x? Because otherwise, there is some problem. For example, take x=0 and see that a = (1,0)' satisfies the above relation, although a is not of the above form.

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  • $\begingroup$ In fact, I just realized from Ben's comment that (1,0)' anyway works, whether or not the relation is required to be true for all x! $\endgroup$ – Somabha Mukherjee Jan 4 '15 at 13:04
  • $\begingroup$ The expression the book gave is true $\endgroup$ – shola Otitoju Jan 4 '15 at 13:20

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