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Suppose I have the following matrix:

$$\begin{bmatrix} 1 & 0 & 0 \\ 1 & 3 & 1 \\ -2 & -4 & -1 \\ \end{bmatrix}$$

The only eigenvalue of which is $\lambda = 1$ with a multiplicity of 3. Solving the expression $(A - \lambda I)$ yields eigenvector solutions of the form $a = -2b - c$. Where

$$ x = \begin{bmatrix} a\\ b\\ c\\ \end{bmatrix}$$

We can easily find two linearly independent vectors of this form, such as

$$ x_1 = \begin{bmatrix} -2\\ 1\\ 0\\ \end{bmatrix}$$

and

$$ x_2 = \begin{bmatrix} -1\\ 0\\ 1\\ \end{bmatrix}$$

My question: Is there any way we can pre-determine the defect of an eigenvalue given some kind of relation among the components of its associated eigenvectors (e.g. $a = -2b - c$)? Is the defect of the eigenvalue in this particular example 1? And how do I determine a third linearly independent eigenvector (I am looking for an intuitive answer here, as my book has a weird way of explaining things), and what conditions must that eigenvector satisfy?

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    $\begingroup$ The eigenspace is only two-dimensional, so you’re not going to find a third linearly independent eigenvector. $\endgroup$ – amd May 30 '16 at 22:50
  • $\begingroup$ Thank you! That answers my first and second questions! $\endgroup$ – user312437 May 30 '16 at 23:06

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