Linear Algebra practice final - Eigenvectors

Char. Limit
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Homework Statement



Let \[<br /> A =<br /> \begin{array}{ccc}<br /> 3 &amp; 5 &amp; 6 \\<br /> 0 &amp; 2 &amp; h \\<br /> 0 &amp; 0 &amp; 2 \\<br /> \end{array}<br /> \]. What should be the value of h so that there are two linearly independent eigenvectors of A corresponding to eigenvalue 2?

Homework Equations



(A-\lambda I) \textbf{x} = \textbf{0}

The Attempt at a Solution



So I tried A-2I, and got this as a matrix:

\[<br /> A-2I =<br /> \begin{array}{ccc}<br /> 1 &amp; 5 &amp; 6 \\<br /> 0 &amp; 0 &amp; h \\<br /> 0 &amp; 0 &amp; 0 \\<br /> \end{array}

Common sense would seem to suggest that h=0, but that just seems too easy for this...
\]
 
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looks right to me.. back yourself

if you want a further check, check the characteristic equation has algebraic multpliciy of 2 with & without h = 0 (necessary but not sufficient though)...

however h=0 gives you the equation of a plane which has a basis of 2 independent vectors. Any other value would give you a single vector upto multiplicative constant
 
All right then. Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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