What is the geometric multiplicity of \lambda=0 as an eigenvalue of A?

chuy52506
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Homework Statement


\lambda=0 is an eigenvalue of
A=
|1 1 1 1 1|
|1 1 1 1 1|
|1 1 1 1 1|
|1 1 1 1 1|
|1 1 1 1 1|

Homework Equations


Find the geometric multiplicity of \lambda=0 as an eigenvalue of A


The Attempt at a Solution


I row reduced it then got the last four rows of all 0s but don't know where to go from there??
 
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now i have x1=-x2-x3-x4-x5
does that mean it has geom mult of 5?
 
Now you have 4 free variables, namely x_2, x_3, x_4, x_5. If that doesn't tell you right away, set x_2 = t_1, x_3 = t_2, etc. This gives you
<br /> \begin{pmatrix}<br /> x_1 \\<br /> x_2 \\<br /> x_3 \\<br /> x_4 \\<br /> x_5<br /> \end{pmatrix}=<br /> \begin{pmatrix}<br /> -t_1 - t_2 - t_3 - t_4 \\<br /> t_1 \\<br /> t_2 \\<br /> t_3 \\<br /> t_4<br /> \end{pmatrix}<br />

Now separate out the different variables and see how many basis vectors this gives you.
 
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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