Eigenvectors of Matrix: Solving Basics

daveyman
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Homework Statement



Find the eigenvectors of the following matrix:

<br /> \[ \left( \begin{array}{ccc}<br /> 1 &amp; 1 \\<br /> 4 &amp; -2 \end{array} \right)\] <br />

Homework Equations



N/A

The Attempt at a Solution



I already know how to find the solutions. They are {1 1} and {-1 4}. My question is this: could a solution also be {1 -4} (switching the minus sign)? I don't think it is a solution but I don't see why not...

Any help would be great. Thanks!
 
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Have you tried \begin{pmatrix}1 \\ -4\end{pmatrix} ? All it takes to be a solution is solving the eigenvalue problem for some eigenvalue \lambda.

Tom
 
your sol'n times any c in R is fine
 
davyjones said:
your sol'n times any c in R is fine

Oh I guess that's obvious huh. So if c=-1, I'm good.

Okay thanks!
 
Exactly.
 
One important thing you should have learned (or maybe this exercise is in preparation) the set of all eigenvectors (corresponding to a given eigenvalue) forms a subspace: it is closed under both vector addition and scalar multiplication.
 
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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