Quick way to find Eigenvectors

Method

Once we have the eigenvalues of a given matrix, find the eigenvectors by solving:

$$(\mathbf{A}-\lambda \mathbf{I})\mathbf{v}=0$$

The eigen vector can be found out by solving the matrix system.

Shortcut

If $(A-\lambda I)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, then $v=\left(\begin{array}{c}-b\\ a\end{array}\right)$

Example

• For ${\lambda }_1=2+\sqrt{5}$ Solve $(A-{\lambda }_{1}I)v=0$:

  $$\left(\begin{array}{cc}3-(2+\sqrt{5})& 1\\ 4& 1-(2+\sqrt{5})\end{array}\right)=\left(\begin{array}{cc}1-\sqrt{5}& 1\\ 4& -1-\sqrt{5}\end{array}\right)$$

  Using the shortcut: If $(A-\lambda I)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, then $v=\left(\begin{array}{c}-b\\ a\end{array}\right)$

$$\boxed{v_1=\left(\begin{array}{c}-1\\ 1-\sqrt{5}\end{array}\right)}$$

• For ${\lambda }_2=2-\sqrt{5}$

  $$(A-{\lambda }_{2}I)=\left(\begin{array}{cc}1+\sqrt{5}& 1\\ 4& -1+\sqrt{5}\end{array}\right)$$

$$\boxed{v_2=\left(\begin{array}{c}-1\\ 1+\sqrt{5}\end{array}\right)}$$