Some lecture notes I’m reading present the following matrix:
\begin{equation*} L = \begin{pmatrix} 0 & a \\ b & c\\ \end{pmatrix} \end{equation*}
It then says that the dominant eigenvalue $ \lambda$ is the solution to the characteristic equation $\lambda^2-c\lambda-ab=0$, which I think I follow.
My problem is that it then says the dominant left eigenvector is proportional to $[b \: \lambda]$ and the dominant right eigenvector is proportional to $[a \: \lambda]$. I am fairly new to matrix algebra, and I do not follow these steps. I would appreciate a simple explanation.
I’ve tried solving for these eigenvectors in Mathematica using the Eigensystem[] function (as ultimately I want to use a similar step on any 2x2 matrix as an automatic subsection in a larger script), but the output doesn’t seem to match the notes.
As previously noted, the characteristic equation $\lambda^2 - c\lambda - ab = 0$ gives both eigenvalues. The roots are $$ \lambda = \frac{c}{2} \pm \sqrt{\left(\frac{c}{2}\right)^2 + ab}, $$ and I assume if these parameters are positive, both eigenvalues are real, and the dominant one is the positive case.
The eigenvalue equations for left and right eigenvectors: $$ \mathbf{r}L = \lambda \mathbf{r}, \qquad L\mathbf c = \lambda \mathbf{c} $$ are equivalent to $$ \mathbf{r}(L-\lambda I) = \mathbf 0, \qquad (L-\lambda I)\mathbf c = \mathbf 0, $$ where $I$ is the identity matrix. Since $$ L - \lambda I = \begin{bmatrix} -\lambda & a \\ b & c-\lambda \end{bmatrix}, $$ we have $$ \begin{bmatrix} b & \lambda \end{bmatrix} \begin{bmatrix} -\lambda & a \\ b & c-\lambda \end{bmatrix} = \begin{bmatrix}0 & ab - \lambda(c-\lambda) \end{bmatrix} = \boldsymbol 0, $$ and $$ \begin{bmatrix} -\lambda & a \\ b & c-\lambda \end{bmatrix} \begin{bmatrix} a \\ \lambda \end{bmatrix} = \begin{bmatrix} 0 \\ ab - \lambda(c-\lambda) \end{bmatrix} = \boldsymbol 0. $$ These formulas for the eigenvectors are true for both eigenvalues, not just the dominant (substitute in the given eigenvalue to get the given eigenvector).