# Considering the linear system $Y'=AY$

What would be an equation that I can use when I compute the eigenpairs for the coefficient matrix $A.$

-
$AX=\lambda X$? – 1015 May 3 '13 at 3:52
Would I need to find the det of A – user1988 May 3 '13 at 3:55
oh is that where I would subtract the lambda from a and d? – user1988 May 3 '13 at 3:59
If you're talking about $2\times 2$ matrices with $a,d$ on the diagonal, yes. – 1015 May 3 '13 at 4:01
Yes, I should have been more specific. Thank you! – user1988 May 3 '13 at 4:02

In the two dimensional case, the eigenvalues of the matrix $A=\begin{pmatrix}a& b\\ c& d\end{pmatrix}$ are the roots of the characteristic polynomial $$\chi_A(\lambda)=\lambda^2-\mathrm{tr}(A)\lambda+\det(A),$$ where $$\mathrm{tr}(A)=a+d,\qquad\det(A)=ad-bc.$$ Let me assume you know how to compute these roots. To compute an eigenvector $u_\lambda=\begin{pmatrix}x_\lambda\\ y_\lambda\end{pmatrix}$ for the eigenvalue $\lambda$, note that $Au_\lambda=\lambda u_\lambda$ if and only if $$by_\lambda=(\lambda-a)x_\lambda,\qquad cx_\lambda=(\lambda-d)y_\lambda,$$ and that these two equations are actually equivalent (this is the sign that $\lambda$ is indeed an eigenvalue), hence a first choice for $u_\lambda$ is $$u_\lambda=\begin{pmatrix}b\\ \lambda-a\end{pmatrix}.$$ This choice fails if the RHS is the zero vector, that is, if $b=0$ and $\lambda=a$, then choose $$u_\lambda=\begin{pmatrix}\lambda-d\\ c\end{pmatrix}.$$ The only case when both choices fail is when $A=aI$, then both eigenvalues are $a$ and every nonzero vector is an eigenvector.

-

Here is the fundamental idea for solving a system of linear ode's with constant coefficients.

To solve the system $Y'(t)=AY(t)$, assuming for simplicity $A_{2\times 2}$ matrix, we assume the solution to have the form

$$Y(t) = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\rm e^{\lambda t} \implies Y'(t) = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\lambda \rm e^{\lambda t}.$$

Now, we substitute back in the Diff. eq. which gives

$$Y'(t) = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\lambda \rm e^{\lambda t} = A \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\rm e^{\lambda t}\implies \begin{bmatrix} \lambda k_1 \\ \lambda k_2 \end{bmatrix}= \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} k_1 \\ k_2 \end{bmatrix}.$$

Solve the last system for $k_1$ and $k_2$. In order to get a non trivial solution for the system we force the determinant of the matrix of the coefficients to be zero which results in getting $\lambda's$ the eigenvalues.

-