Solving Zero Eigenvector: A Homework Problem

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Homework Statement



I can calculate the proper eigenvalues, but when I plug them back into the matrix, I get x1=0 and x2=0. But this is not the answer Maple gives me! How do I solve for the eigenvector when it appears that a zero vector is the only solution?

Homework Equations



For example, for the matrix {1,1},{1,-1} (rows shown), Maple gives me: eigenvalue of sqrt(2) with eigenvector {1/((sqrt(2)-1),1} and eigenvalue of -sqrt(2) with eigenvector {1/(-(sqrt(2)-1),1}

The Attempt at a Solution



But I can't get these eigenvectors when I try to solve by hand! How do you solve in these situations?
What are these situations called?



 
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For sqrt(2), You end up with the eigenvector equation:

\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\binom{x}{y}=\sqrt{2}\binom{x}{y}

or:

(1-\sqrt{2}) x+y=0

x-y(1+\sqrt{2})=0

which are redundant so any vector (x,y) that satisfies either equation is a suitable eigenvector like:

\binom{1+\sqrt{2}}{1}

and even:

\binom{\frac{1}{\sqrt{2}-1}}{1}
 
Thanks! Oh duh! I didn't realize that the 2 equations were the same. That's why I was getting x=0 and y=0 as solutions. I had to multiply through by the appropriate constant to make the 2 equations look the same.
 
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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