# Eigenvalues and eigenvectors doubt

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 $v_2$?

• $E(\lambda)$ is the null space of $A-\lambda I_n$; it is not the matrix itself. – Casteels Nov 10 '13 at 20:29
• Sorry, edited!!! – Tomi Nov 10 '13 at 20:30
• 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? – Casteels Nov 10 '13 at 20:31

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.