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Defunkt · #2 November 6th, 2009, 06:41 AM

Hi!

There really is no need to calculate the characteristic polynomial that way.

First, you can easily observe that 0 is an eigenvalue of geometric multiplicty 1 (since $rank(A) = 1$ . Now, note that $trace(A) = 1+1 = 2$ and A only has two eigenvalues, so the second one must be 2. Another way to know that 2 is an eigenvalue is to note that the sum of all rows is 2, therefore, according to a theorem which you should have learned, 2 is an eigenvalue with corresponding eigenvector $(1 \ 1)^T$ .

So we know that 2 is an eigenvalue of geometric multiplicity $\geq 1$ and 0 is an eigenvector with geometric multiplicity 1. Since algebraic multiplicity $\geq$ geometric multiplicity for each eigenvalue, we get that the algebraic and geometric multiplicities for each eigenvector are 1 and 1.

Therefore: $\Delta_A(x) = x(x-2)$
With 0,2 as eigenvalues
and $(1\ -1)^T, (1 \ 1)^T$ the corresponding eigenvectors.

However, if you still want to know what you did wrong --

$det(\lambda I-A) = (\lambda-1)(\lambda-1) - (-1)(-1) = (\lambda-1)^2 -1 \neq (\lambda-1)^2+1$

$=\Rightarrow det(\lambda I-A) = \lambda^2 -2\lambda +1 -1 = \lambda^2 -2\lambda = \lambda(\lambda-2)$