As we saw in last class, the system $$ \frac{d}{dt}\vec{y} = A \vec{y}$$ has solutions in the form $\vec{y}(t) = e^{At}\vec{C}.$ If $A$ is a diagonalizable 2x2 matrix, then $$\vec{y}(t) = C_1 e^{\lambda_1 t}\vec{u}+C_2 e^{\lambda_2 t}\vec{v},$$ where $\vec{u}$ and $\vec{v}$ are eigenvectors of $A$ with eigenvaues $\lambda_1$ and $\lambda_2$ respectively.

If $A$ is not diagonalizable, then we will only be able to find one linearly independent eigenvector. In this case, there will be one repeated eigenvalue. We still want to follow a similar procedure to calculate $e^{At},$ but instead of diagonalizing, we will just get as close as we can.

Definition: Suppose $u$ is an eigenvector for $A$ with eigenvalue $\lambda$. A generalize eigenvector chained to $u$ is a non-zero vector $v$ that satisfies $$ (A-\lambda I)v = u.$$

We can see that $u$ and $v$ are linearly independent. If not, then $v$ would be a scalar multiple of $u$ and would satisfy $(A-\lambda I)v=0.$ This means the matrix $Q = [u|v]$ is invertible, and we can use the matrix $Q$ to find a matrix similar to $A$. \begin{align} B &= Q^{-1}AQ\\ &= Q^{-1}A[u|v]\\ &= Q^{-1}[Au | Av]\\ &= Q^{-1}[\lambda u | \lambda v + u]\\ &= [\lambda Q^{-1} u | \lambda Q^{-1}v +Q^{-1} u ]\\ &= \begin{bmatrix}\lambda & 1 \\ 0 & \lambda \end{bmatrix} \end{align} This matrix is almost diagonal, in the sense that the eigenvalue is still along the diagonal, but we have an extra $1$ along the top row. We'll notice that we can write $$B = \begin{bmatrix}\lambda & 0 \\ 0 & \lambda \end{bmatrix}+\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = D + N.$$ This first matrix is a diagonal matrix, while the second matrix is called a nilpotent matrix and satisfies $$ N^2 = 0.$$ By splitting up the matrix $B$, we have \begin{align} e^{At} &= e^{QBQ^{-1}t}\\ &= Qe^{Bt}Q^{-1}\\ &= Qe^{(D+N)t}Q^{-1}\\ &= Qe^{Dt}e^{Nt}Q^{-1}. \end{align} Note: The relationship $e^{D+N}=e^{D}e^{N}$ doesn't hold for every pair of matrices $D$ and $N$, only if the matrices commute, i.e. $ND = DN$. Luckily for us, these matrices happen to commute.

Using the definition of $e^{Nt}$ we have $$e^{Nt} = I + Nt+ \frac{1}{2}(Nt)^2 + ...$$ since $N$ is nilpotent, we have $N^2 = 0$ and therefore $$e^{Nt} = I + Nt = \begin{bmatrix} 1 & t \\ 0 & 1\end{bmatrix}. $$ Therefore the solution to $$\frac{d\vec{y}}{dt} = A \vec{y}$$ when $A$ is not diagonalizable is \begin{align} \vec{y}(t) &= e^{At}\vec{C}\\ & = Q e^{Dt}e^{Nt}Q^{-1}\vec{C}\\ &= [u|v]\begin{bmatrix}e^{\lambda t} & 0\\ 0 & e^{\lambda t}\end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}\begin{bmatrix}C_1 \\ C_2 \end{bmatrix} \\ &= [u|v]\begin{bmatrix}e^{\lambda t} & 0 \\ 0 & e^{\lambda t}\end{bmatrix}\begin{bmatrix} C_1+tC_2\\ C_2 \end{bmatrix}\\ &= [u|v]\begin{bmatrix}(C_1+C_2t)e^{\lambda t}\\ C_2 e^{\lambda t}\end{bmatrix}\\ &= (C_1+C_2 t)e^{\lambda t}u + C_2 e^{\lambda t}v. \end{align}

Example: Find the solution to \begin{align}\frac{d}{dt}\vec{y} = \begin{bmatrix} 2 & 2 \\ 0 & 2\end{bmatrix}\vec{y} && \vec{y}(0) = \begin{bmatrix}0 \\ 1\end{bmatrix}. \end{align} The eigenvalues for the matrix $A$ are 2 and 2. Finding eigenvector(s) for this eigenvector yields \begin{align} (A-2I)u &= 0 \\ \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{align} Letting $u_1 = 1$, it follows that $u_2 = 0$, and therefore $u = [1,0]^T$.

Since we found only one eigenvector, we need to find a generalized eigenvector $v$. \begin{align} (A-2I)v &= u \\ \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} &= \begin{bmatrix} 1 \\ 0 \end{bmatrix}\\ v_2 &= \frac{1}{2} \end{align} Therefore the generalized eigenvector is $v = [0,1/2]^T$, and the general solution is $$ \vec{y}(t) = \left(C_1+ C_2 t \right)\begin{bmatrix}1 \\ 0 \end{bmatrix} e^{2t} + C_2 \begin{bmatrix}0 \\ \frac{1}{2}\end{bmatrix}e^2t.$$ Substituting in the initial condition gives us $$ \vec{y}(0) = \begin{bmatrix}0\\1\end{bmatrix} = C_1\begin{bmatrix}1\\ 0 \end{bmatrix} + C_2 \begin{bmatrix} 0 \\ \frac{1}{2}\end{bmatrix},$$ which implies $C_2 = 0$. Therefore the solution to the initial value problem is $$ \vec{y}(t) = 2t\begin{bmatrix} 1 \\ 0 \end{bmatrix}e^{2t}+\begin{bmatrix} 0 \\ 1\end{bmatrix} e^2t$$

Practice Problems

Find the general solution to each system of ODEs

1. $$ \frac{d\vec{y}}{dt} = \begin{bmatrix} 1 & 0 \\ 3 & 1\end{bmatrix}\vec{y}$$
2. $$ \frac{d\vec{y}}{dt} = \begin{bmatrix} -1 & 1 \\ -1 & 3\end{bmatrix}\vec{y}$$
3. $$ \frac{d\vec{y}}{dt} = \begin{bmatrix} -1 & 0 \\ 0 & -1\end{bmatrix}\vec{y}$$