Linear Algebra practice final - Eigenvectors

[Char. Limit]

Homework Statement

Let $$\[<br /> A =<br /> \begin{array}{ccc}<br /> 3 & 5 & 6 \\<br /> 0 & 2 & h \\<br /> 0 & 0 & 2 \\<br /> \end{array}<br /> \]$$. 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 & 5 & 6 \\<br /> 0 & 0 & h \\<br /> 0 & 0 & 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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[lanedance]

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

 

[Char. Limit]

All right then. Thanks!