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I have a matrix $A$ that is expressed as $$A=\begin{bmatrix} 6& 0& 3& 0& 3&0 \\ 2& -6& 1& -3& 1&-3 \\ 0& 0& 6& 0& 0& 0\\ 0& 0& 2& -6& 0&0 \\ 0& 0& 0& 0& 3& 0\\ 0& 0& 0& 0& 1& -3 \end{bmatrix}$$ I now want to find the eigenvectors that correspond to a repeated eigenvalue. For example, I consider the eigenvalues $\lambda =6$ which is repeated twice.

Let's denote $x=[x_1,..,x_6]^T$ the eigenvectors of this eigenvalue.

First, I calculate $$\ (A - 6I_6)^2 x=0 $$ After that, I get the solution

$$ \left\{\begin{matrix} -x_1 + 6x_2 -x_3 + 6x_4=0\\ -x_3 + 6x_4 =0\\ x_5=0\\ x_6 =0 \end{matrix}\right.$$ However, when I check the results by using Matlab. It gives $x=[0.9864 \; 0.1644 \;0 \;0 \;0 \;0]^T$ and $x=[-0.9864\; -0.1644 \;0 \;0 \;0 \;0]^T$.

Could you please point out what is wrong in my thinking?

Thank you very much!

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  • $\begingroup$ Your 'solution' is not yet finished, you first have to calculate the eigenvectors. Then note that if $x$ is an eigenvector, then $ax$ is also an eigenvector for every $a\neq 0$. $\endgroup$ – flawr Nov 29 '15 at 20:28
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Searching for proper eigenvectors of the eigenvalue $\lambda=6$ you have to solve the equation: $$ (A-6I)x=0 $$that gives the system $$ \begin{cases} 3x_3+3x_5=0\\ 2x_1-12x_2+x_3-3x_4+x_5-3x_6=0\\ 2x_3-12x_4=0\\ -3x_5=0\\ x_5-9x_6=0 \end{cases} $$ that has solution: $x_1=6x_2 \quad \forall \quad x_3,x_4,x_5,x_6 \in \mathbb{R}$ So, the eigenspace has dimension $1$ ( the eigenvector $v_{\lambda=6}=(6,1,0,0,0,0)^T$).

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  • $\begingroup$ If there is only one eigenvector, how can I define the Jordan form? $\endgroup$ – user272011 Nov 29 '15 at 20:56
  • $\begingroup$ Your matrix have the other eigenvalues: $\lambda=-6$ with algebraic multiplicity $2$ and geometric multiplicity $1$ and $\lambda=\pm 3$ each with algebraic and geometric multiplicity $1$ . So the the Jordan form has these six values on the diagonal, and an $1$ upper the diagonal in the two places right and upper the doubles eigenvalues. ( see wolframalpha.com/input/…. $\endgroup$ – Emilio Novati Nov 29 '15 at 21:06
  • $\begingroup$ The calculus of the generalized eigenvectors is a bit more tricky. See: en.wikipedia.org/wiki/Generalized_eigenvector. $\endgroup$ – Emilio Novati Nov 29 '15 at 21:10
  • $\begingroup$ Yeah. In my solution, I am trying to calculate generalized eigenvectors for $\lambda =6$ . But It seems that I am confusing $\endgroup$ – user272011 Nov 29 '15 at 21:30
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The eigenvalue $6$ has algebraic multiplicity $2$, but geometric multiplicity $1$: the null space of $A - 6 I$ is one-dimensional, spanned by $(6,1,0,0,0,0)^T$. $(0,0,6,1,0,0)^T$ is in the null space of $(A-6I)^2$ but is not an eigenvector of $A$.

Numerically, this is an unstable situation: a very small perturbation of the matrix is likely to have two distinct (but very close) eigenvalues, each with their own eigenvectors, rather than a double eigenvalue. Thus Matlab, which uses numerical methods, is likely to be unreliable in this situation.

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  • $\begingroup$ Could you tell me how to select two eigenvectors corresponding to $\lambda =6$? I mean that how can I select the second one? $\endgroup$ – user272011 Nov 29 '15 at 20:35
  • $\begingroup$ There don't exist two linearly independent eigenvectors corresponding to $\lambda = 6$. $\endgroup$ – Robert Israel Nov 29 '15 at 20:38
  • $\begingroup$ Haizzz. Now, it makes me confuse. In this case, how can I define the Jordan form? $\endgroup$ – user272011 Nov 29 '15 at 20:54

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