# Eigenvalues and Eigenfunctions of an Operator

Suppose I am given two orthonormal vectors $v_1$ and $v_2$ which constitute a base of a two-dimensional space. Suppose also that an operator $T$ defined on this space is satisfying some equations of the form: $Tv_1=av_2+bv_2$ and $Tv_2=cv_1+dv_2$. How do I calculate the eigenvalues and the orthonormal eigenfunctions of $T$?

• @ Andrew I'm not sure what you mean under orthonormal eigenfunctions. Could you, please, explain that? – user539887 Mar 25 '18 at 18:53

## 2 Answers

In that basis the matrix corresponding to $T$ is $\begin{pmatrix}a&c\\b&d\end{pmatrix}$. To see this, I encourage you to see the definition of a matrix of an operator in a given basis. The eigenvalues and eigenfunctions in this basis can be found by the usual means for matrices. In the end you just need to interpret $\begin{pmatrix}x\\y\end{pmatrix}$ as $xv_1+yv_2$.

• So, calculating the eigenfunctions is all about expressing the eigenvectors in a more 'analytic' form like the one you described? – Jevaut Mar 25 '18 at 18:29
• @AndrewTzevas In this case, yes. When the dimension is low and everything is explicit, this is often the best approach. (Did you notice that you now have the privilege to vote?) – Joonas Ilmavirta Mar 25 '18 at 18:33

This is a finite dimensional question, so you may write the operator in matrix form (using the basis $v_1,v_2$) as $$\begin{bmatrix}a&c\\b&d \end{bmatrix}$$ then solve for the roots of characteristic polynomial to find eigenvalues $\lambda$ (there may be none, one, or two depending on what field you are over!) solving $$\lambda^2-(a+d)\lambda+\det(T)=0$$ If you get an eigenvalue $\lambda$, you find eigenvectors by examining $$\ker\left(\begin{bmatrix}a-\lambda&c\\b&d-\lambda \end{bmatrix}\right)$$