Why does x-axis have eigenvalue = 1

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Why has the x-axis have an eigenvalue = 1 and the y-axis an eigenvalue =-1? (please stay simple in your answers)
 
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Could you be more specific? An eigenvalue as I understand it is generally associated with a function and a vector. I don't understand what the function is here.
 
LeonhardEuler said:
Could you be more specific? An eigenvalue as I understand it is generally associated with a function and a vector. I don't understand what the function is here.

Our lecturer, just drew a diagram the other day on the board with an arrow pointing to the x-axis with lambda = 1 and another arrow with lambda = -1 pointing to the y-axis? Does anyone get what he was trying to explain? Many thanks
 
Maybe he was talking about the eigenvalues of some particular function which had the unit vectors in the x- and y-directions as eigenvectors. For example, the function f([x,y])=[x,-y] has the unit vectors [1,0] and [0,1] as eigenvectors with these eigenvalues because f([1,0])=[1,0]=1*[1,0] and f([0,1])=[0,-1]=-1*[0,1]. Or maybe he was talking about something entirely different. It's hard to say since I don't know what subject he was discussing.
 
LeonhardEuler said:
Maybe he was talking about the eigenvalues of some particular function which had the unit vectors in the x- and y-directions as eigenvectors. For example, the function f([x,y])=[x,-y] has the unit vectors [1,0] and [0,1] as eigenvectors with these eigenvalues because f([1,0])=[1,0]=1*[1,0] and f([0,1])=[0,-1]=-1*[0,1]. Or maybe he was talking about something entirely different. It's hard to say since I don't know what subject he was discussing.

I think that was it. Thanks :-)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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