CDS - Eigenvalues and Eigenvectors

Remember:

• To calculate the eigenvalues: solve the characteristic polynomial $\det (A-\lambda I)=0$
• For each eigenvalue ${\lambda }_i$, to find the corresponding eigenvector, solve:$(A-{\lambda }_{i}I)\mathbf{v}=\vec{0}$

Here’s an example:$A=\left(\begin{array}{lc}4& 1\\ 2& 3\end{array}\right)$The characteristic polynomial $-$ is found as:$\det \left(\begin{array}{lc}4-\lambda & 1\\ 2& 3-\lambda \end{array}\right)=(4-\lambda )(3-\lambda )-(2\cdot 1)$If we solve it we obtain:${\lambda }^2-7\lambda +10$And finally:$\begin{array}{l}{\lambda }_1=5\\ {\lambda }_2=2\end{array}$

not-sure-about-this (ChatGPT 4o) Shouldn’t I find 2 vectors? I did not review this answer, I do not care. After finding: ${\lambda }_1=5$ and ${\lambda }_2=2$, we can then calculate the eigenvectors from:$\left(\begin{array}{lc}4-5& 1\\ 2& 3-5\end{array}\right)\left(\begin{array}{l}{v}_1\\ v_2\end{array}\right)=\left(\begin{array}{l}0\\ 0\end{array}\right)$And from:$\left(\begin{array}{lc}4-2& 1\\ 2& 3-2\end{array}\right)\left(\begin{array}{l}{v}_1\\ v_2\end{array}\right)=\left(\begin{array}{l}0\\ 0\end{array}\right)$And after some calcualtions we obtain:$v=\left(\begin{array}{c}1\\ -2\end{array}\right)$