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I have some problems to determine the eigenvectors of a given matrix:

The matrix is:

$$ A = \left( \begin{array}{ccc} 1 & 0 &0 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array} \right) $$

I calculated the eigenvalues first and got $$ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 1$$ There was no problem for me so far. But I do not know how to determine the eigenvectors. The formula I have to use is $$ (A-\lambda_i E)u=0, \lambda_i = \{1,2,3\}, u\ is\ eigenvector$$ When I determined the eigenvector with $ \lambda_2=2$ there was not a problem. I got the result that $x_3 = variable$ and $x_2 = x_3$, so: $$ EV_2= \left( \begin{array}{ccc} 0 \\ \beta \\ \beta \end{array} \right) \ \beta\ is\ variable,\ so\ EV = span\{\left( \begin{array}{ccc} 0 \\ 1 \\ 1 \end{array} \right)\} $$

But when I used $ \lambda_1 = \lambda_3 = 1 $, I had to calculate: $$ \left( \begin{array}{ccc} 0 & 0 &0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right) * \left( \begin{array}{ccc} x_1 \\ x_2 \\ x_3 \end{array} \right) =0 $$

what in my opinion means that $x_3 = 0 $ and $x_1$ and $x_2$ are variable, but not necessarily the same as in the case above, so $ EV_{1,3} = \left( \begin{array}{ccc} \alpha \\ \beta \\ 0 \end{array} \right) $

What does that mean for my solution? is it $$ EV_{1,3} = span\{\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 1 \\ 1 \\ 0 \end{array} \right)\} $$

What exactly is now my solution in this case for the eigenvectors $ \lambda_1, \lambda_3 $? In university we just had one variable value in the matrix so I don't know how to handle two of them being different.

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  • $\begingroup$ I deleted my post because I am in error. Sorry, and disregard my comments. However. The Wolf gives 3 eigenvectors wolframalpha.com/input/… $\endgroup$ – neofoxmulder Mar 6 '14 at 14:34
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Every linear combination of $EV_{1}=\pmatrix{1\\0\\0}$ and $EV_3=\pmatrix{0\\1\\0}$ is a eigenvector with eigenvalue $1$.

$EV_{1,3} = span\{\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 1 \\ 1 \\ 0 \end{array} \right)\}$ is the same as $EV_{1,3} = span\{\left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right), \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right)\}$.

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  • $\begingroup$ Thanks. I think my main problem is that I don't see why $(1,1,0)^T$ is not really necessary. Could you explain it to my why it is the same with or without $(1,1,0)^T$ ? $\endgroup$ – Drudge Mar 6 '14 at 13:07
  • $\begingroup$ @Drudge because $\pmatrix{1\\1\\0}$ lies in the vector space that is spanned by the other two... $\endgroup$ – draks ... Mar 6 '14 at 13:44
  • $\begingroup$ Yes, you are right. I did not know about this context here. But I think I got it now. Thanks! $\endgroup$ – Drudge Mar 6 '14 at 14:04
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Update: I have undeleted my answer because I think it is fixed now.

You got $$ V_{\lambda_2} = \left(\begin{array}{ccc} 0 \\ 1 \\ 1 \end{array} \right) $$

correct but then copied it down wrongly.(I think..)

Then you correctly wrote down the case $\lambda_1$. From

$$ \left(\begin{array}{ccc } 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right) $$

you should easily conclude (I think you did) that $ z = 0 $ , x and y can be anything leading to

$$ V_{\lambda_{1 \ or \ 3}} = \left(\begin{array}{ccc } 1 \\ 1 \\ 0 \end{array} \right) $$

Now , since the dimension of the nullspace is 2 we can decompose this into 2 seperate eigenvectors corresponding to the repeated eigenvalue of 1

$$ V_{ \lambda_1} = \left(\begin{array}{ccc } 1 \\ 0 \\ 0 \end{array} \right) \ \ , \ \ V_{ \lambda_2 } = \left(\begin{array}{ccc } 0 \\ 1 \\ 1 \end{array} \right) \ \ , \ \ V_{ \lambda_3} = \left(\begin{array}{ccc } 0 \\ 1 \\ 0 \end{array} \right) $$

You can check all three are independent and satisfy

$$AV_i = \lambda_iV_i$$

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  • $\begingroup$ Thank you for your answer. Actually, I didn't used $ V_{\lambda_2} $ anymore after I determined it. Where exactly did I copy it down wrongly? I also thought that $ (1,1,0)^T $ was a correct solution but my excecise instructor striked my solution and wrote $$ \{[(1,0,0)^T, (0,1,0)^T]\} $$ beneath it. That was my main intention to ask the question. But when I read the first answer, it seems that he has made a mistake. $\endgroup$ – Drudge Mar 6 '14 at 14:26
  • $\begingroup$ I didn't see you present $(0,1,1)^T$ as an eigenvector so I thought you copied it down as a typo $(1,1,0)$ $\endgroup$ – neofoxmulder Mar 6 '14 at 17:05
  • $\begingroup$ I kind of thought that, no problem. But thank you anyway! Edit: Oh sorry, I see your edit only now. Thank you for explaining the case with $ (1,1,0)^T $! $\endgroup$ – Drudge Mar 6 '14 at 17:56
  • $\begingroup$ You are welcome. Sometimes when we get a repeated eigenvalue we can't get enough vectors , it all depends on the dimension of the nullspace. In other words , if you have $n$ eigenvalues ALL DISTINCT then you get $n$ independent eigenvectors , but let's say we have an eigenvalue of $k$ that repeats 3 times , then that eigenvalue is gauranteed to produce 1 eigenvector , you may get 2 or 3 but not more than 3 ( corresponding to that k) and depending on the dimension of the nullspace. $\endgroup$ – neofoxmulder Mar 6 '14 at 18:19

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