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Find the eigenvalues and eigenvectors of $A= \left( \begin{matrix} 3 & 1 \\ -3 & 7 \\ \end{matrix} \right) $

So:

$$\det (A-\lambda I_2)= \left| \begin{matrix} 3-\lambda & 1 \\ -3 & 7-\lambda \\ \end{matrix} \right| =\lambda^2-10\lambda+24 $$ $$\lambda^2-10\lambda+24=(\lambda-6)\cdot(\lambda-4) \implies \text{eigenvalues =}\{6,4\}$$

Replacing in the eigenbasis formula: $$E(\lambda)=\{x\in\Bbb R^n: \text{Ax} = \lambda\text{x}\}=\text{N}(\text{A} -\lambda I_n)$$

Then

$E(4)=N\left( \begin{matrix} -1 & 1 \\ -3 & 3 \\ \end{matrix} \right) $

$ E(6)= N\left( \begin{matrix} -3 & 1 \\ -3 & 1 \\ \end{matrix} \right) $

This is easy, but my professor wrote this:

We see that $v_1= \left[ \begin{matrix} 1 \\ 1 \\ \end{matrix} \right] $ give a basis for $E(4)$ and $v_2= \left[ \begin{matrix} 1 \\ 3 \\ \end{matrix} \right] $ gives a basis for $E(6)$

The truth is that I can't see nothing, because I don't know about "basis". And I can't understand how to get $v_1$ and $v_2$

Someone can explain with some details about $v_1$ and $v_2$?

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    $\begingroup$ $E(\lambda)$ is the null space of $A-\lambda I_n$; it is not the matrix itself. $\endgroup$ – Casteels Nov 10 '13 at 20:29
  • $\begingroup$ Sorry, edited!!! $\endgroup$ – Tomi Nov 10 '13 at 20:30
  • $\begingroup$ That's ok! Now are you saying you don't know what a "basis" is, or that you aren't sure how to calculate the null space? $\endgroup$ – Casteels Nov 10 '13 at 20:31
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If we add the component of each row of the matrix we find $4$ so we see that $v_1= \left[ \begin{matrix} 1 \\ 1 \\ \end{matrix} \right] $ give a basis for $E(4)$ but this observation isn't a general method to find the eigenvector.

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  • $\begingroup$ Have a great day my dear brother Sami. :+) $\endgroup$ – mrs Nov 11 '13 at 5:37
  • $\begingroup$ Nice work, Sami! $\endgroup$ – Namaste Feb 20 '14 at 13:37

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