Solving Zero Eigenvector: A Homework Problem

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The discussion centers on solving for eigenvectors when the initial calculations yield a zero vector. The user successfully calculates the eigenvalues for the matrix {1,1},{1,-1} but struggles to derive the corresponding eigenvectors by hand, leading to confusion. It is clarified that the equations derived from the eigenvector equation are redundant, which caused the misunderstanding. The user realizes that multiplying through by a constant can help align the equations, allowing for valid eigenvector solutions. This highlights the importance of recognizing equivalent equations in eigenvector calculations.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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