# Matrix form of differential equation, not diagonalizable

So I came across this problem while studying and I'm a little confused.

$$y'(t) = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}y(t)$$

Usually I solve for the eigenvalues and the corresponding eigenvectors and then basically plug into:

$$y = c_{1} e^{\lambda_{1} t} \begin{bmatrix} e1 \\ e2 \end{bmatrix}+ c_{2} e^{\lambda_{2} t} \begin{bmatrix} e3 \\ e4 \end{bmatrix}$$

But in this case it isn't diagonalizable and I only get one eigenvector. How do you solve this?

• The general theory relates to Jordan Canonical Forms. – Jack Dec 14 '16 at 20:17

For a $2\times 2$ matrix with repeated eigenvalue, and only one linearly independent eigenvector, one can obtain a second solution as $\vec y_2(t)=te^{\lambda t}\vec v_1+e^{\lambda t}\vec v_2$, where $\vec v_1$ is the eigenvector for the matrix, and $\vec v_2$ is a "generalized eigenvector", which satisfies $(A-\lambda I)\vec v_2=\vec v_1$, or equivalently satisfies $(A-\lambda I)^2\vec v_2=0$. So the general solution becomes $$\vec y(t)=c_1e^{\lambda t}\vec v_1+c_2(te^{\lambda t}\vec v_1+e^{\lambda t}\vec v_2)$$ If you are familiar with the matrix exponential, then you could also solve this problem using $e^{At}$ as a fundamental matrix. Notice here that $A$ can be written as the sum of the identity matrix, and a nilpotent matrix (i.e. $N^k=0$ for some $k\in\Bbb N$), so $A=I+N$, and $e^{At}=e^{It}e^{Nt}$.
• And in this case $e^{Nt} = I + Nt$. – copper.hat Dec 14 '16 at 20:43
The common way to deal with the case of repeated eigenvalues is to assume a first solution of \begin{align} v_1e^{\lambda t} \end{align} and a second solution of the form \begin{align} v_1te^{\lambda t} + v_2e^{\lambda t} \end{align} where $v_2$ is another vector that must be solved for. This guess of a second solution results in two linearly independent solutions (what we are after, so that we can form a general solution). See this example for more details.
Solve for eigenvalues. $$det\begin{bmatrix} 1-r & 1 \\ 0 & 1-r \end{bmatrix}=0$$ $$1-2r+r^2=0\rightarrow r=1$$ with multiplicity of 2.
Find eigenvector.Solve $$\begin{bmatrix} 1-1 & 1 \\ 0 & 1-1 \end{bmatrix} \begin{bmatrix} a\\ b \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$$ From this we know $b=0$ and $a$ is any real number. Choose $a=1$ and then one solution of this system is $$y_1(t)=e^t\begin{bmatrix} 1\\ 0 \end{bmatrix}$$. The second solution will be of form $$y_2(t)=e^t\begin{bmatrix} p_1\\ p_2 \end{bmatrix}+te^t\begin{bmatrix} 1\\ 0 \end{bmatrix}$$ where $$\begin{bmatrix} p_1\\ p_2 \end{bmatrix}$$ is a vector that satisfies the condition $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} p_1\\ p_2 \end{bmatrix}=\begin{bmatrix} 1\\ 0 \end{bmatrix}$$.