Finding eigenvectors of a simple 2x2 matrix

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The discussion revolves around finding eigenvectors for the 2x2 matrix |1 2| |3 4| after determining the eigenvalues as (5±√33)/2. The user struggles to solve the resulting equations derived from substituting the eigenvalue back into the matrix, leading to confusion over whether the only solution is x=y=0. Another participant clarifies that the two equations are essentially multiples of each other, indicating an infinite number of solutions. They suggest setting x to a constant value, such as 1, to simplify the process of finding y. The user expresses gratitude for the clarification and guidance provided.
The thinker
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Homework Statement



The matrix is:
|1 2|
|3 4|

The Attempt at a Solution



I've worked out the eigenvalues to be \stackrel{\underline{5\pm\sqrt{33}}}{2}

But when I plug the first eigenvalue back in I get:

|1 - \stackrel{\underline{5+\sqrt{33}}}{2}......2 |
|3......4 - \stackrel{\underline{5+\sqrt{33}}}{2} |

Multiplied by (x,y) = (0,0)

[Sorry I couldn't fit that on the same line as the matrix.]Which in turn gives two equations:

x*(1-\stackrel{\underline{5+\sqrt{33}}}{2}) + 2y = 0

and3x + y*(4-\stackrel{\underline{5+\sqrt{33}}}{2}) = 0Which I can't work out how to solve for the eigenvector.

Is the only solution x=y=0?
 
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You solve them like you solve any two linear equations. Remember if (x,y) is an eigenvector then a*(x,y) is also an eigenvector. So you are going to have an undetermined parameter which you may as well may make x. Put say, x=1 and solve for y. You should get the same value of y from both equations, since one equation is really a multiple of the other.
 
Thanks for the reply!

Thats exactly what I've been trying to do but X and/or Y always end up canceling and I just get some useless number.
 
The thinker said:
Thanks for the reply!

Thats exactly what I've been trying to do but X and/or Y always end up canceling and I just get some useless number.

Are you trying to eliminate a variable from the two equations and solve for the other? That won't work. If you eliminate x then the resulting equation is 0*y=0. Is that your problem? Like I said before, that's happening because the two equations are really the same. They have an infinite number of solutions. Put x equal to any constant (like x=1) and then solve for y.
 
AH!

Thank you!
 
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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