Examples

Eigenvector approach - real eigenvalues, enough eigenvectors

Eigenvector approach: real eigenvalues, enough eigenvectors

Problem: Solve the system given by ${\mathbf{y}}^{\prime }=\left(\begin{array}{rrr}4& 0& 1\\ -2& 1& 0\\ -2& 0& 1\end{array}\right)\mathbf{y}$.
Solution: First find the eigenvalues:

$$\begin{array}{rl}\det A_{\lambda }& =\left|\begin{array}{ccc}4-\lambda & 0& 1\\ -2& 1-\lambda & 0\\ -2& 0& 1-\lambda \end{array}\right|\\ & \\ & =(4-\lambda )\left|\begin{array}{cc}1-\lambda & 0\\ 0& 1-\lambda \end{array}\right|-0\left|\begin{array}{cc}-2& 0\\ -2& 1-\lambda \end{array}\right|+1\left|\begin{array}{cc}-2& 1-\lambda \\ -2& 0\end{array}\right|\\ & \\ & =(4-\lambda )(1-\lambda {)}^2-0+2(1-\lambda )\\ & =-{\lambda }^3+6{\lambda }^2-11\lambda +6.\end{array}$$

The roots are $\lambda =1,2,3$. Now we seek eigenvectors by solving $A_{\lambda }\mathbf{x}=0$ for each $\lambda$ root. When $\lambda =1$:

$$\left(\begin{array}{rrr}3& 0& 1\\ -2& 0& 0\\ -2& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$$

This is equivalent to the system $3x_1+x_3=0$, $-2x_1=0$, and $-2x_1=0$. Solving this system, we have $x_1=0=x_3$ and $x_2$ can be anything. Plugging these in we have $\mathbf{x}=\left(\begin{array}{c}0\\ x_2\\ 0\end{array}\right)$, and we choose $x_2=1$ obtaining $\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$.

Next for $\lambda =2$ we have:

$$\left(\begin{array}{rrr}2& 0& 1\\ -2& -1& 0\\ -2& 0& -1\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$

and the equivalent system is $2x_1+x_3=0$, $-2x_1-x_2=0$, and $-2x_1-x_3=0$. Solving this system, we have $x_2=x_3=-2x_1$. Plugging these in we have $\mathbf{x}=\left(\begin{array}{c}{x}_1\\ -2x_1\\ -2x_1\end{array}\right)$ and we choose $x_1=1$ obtaining $\left(\begin{array}{c}1\\ -2\\ -2\end{array}\right)$.

Finally for $\lambda =3$ we have:

$$\left(\begin{array}{rrr}1& 0& 1\\ -2& -2& 0\\ -2& 0& -2\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$

and the equivalent system is $x_1+x_3=0$, $-2x_1-2x_2=0$ and $-2x_1-2x_3=0$. Solving this system, we have $x_2=x_3=-x_1$. Plugging in we have $\mathbf{x}=\left(\begin{array}{c}{x}_1\\ -x_1\\ -x_1\end{array}\right)$, and we choose $x_1=1$ obtaining $\left(\begin{array}{c}1\\ -1\\ -1\end{array}\right)$.

(Note: the only reason we have been choosing $x_1=1$ is for convenience. Anything except zero will work.)

Notice that we have found three independent eigenvectors for a $3$D system, so we are done. Here is the complete set of solutions:

$$C_1\left(\begin{array}{r}0\\ 1\\ 0\end{array}\right)e^t+C_2\left(\begin{array}{r}1\\ -2\\ -2\end{array}\right)e^{2t}+C_3\left(\begin{array}{r}1\\ -1\\ -1\end{array}\right)e^{3t}.$$

System with complex eigenvalues and eigenvectors

System with complex eigenvalues and eigenvectors

Problem: Solve the system ${\mathbf{y}}^{\prime }=\left(\begin{array}{rr}a& -b\\ b& a\end{array}\right)\mathbf{y}$. (Assume $a\ne 0,b\ne 0$.) What are the solutions for $a=b=1$ that take the initial value ${\mathbf{y}}_0=\left(\begin{array}{c}A\\ B\end{array}\right)$ at $t=0$?

Solution: The matrix $\left(\begin{array}{rr}a& -b\\ b& a\end{array}\right)$ has $\det A_{\lambda }=(\lambda -a{)}^2=b^2$ so $\lambda =a\pm bi$. To find the eigenvectors:

$$A_{a+bi}\cdot \mathbf{x}=\left(\begin{array}{cc}-bi& -b\\ b& -bi\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

Therefore $-bi{x}_1-b{x}_2=0$ and $b{x}_1-bi{x}_2=0$. These equations are redundant since the second is $i$ times the first. Dividing by $b$ and then plugging in $x_1=i{x}_2$ we have $\left(\begin{array}{c}i{x}_2\\ x_2\end{array}\right)$, and we choose $x_2=1$ obtaining:

$${\mathbf{v}}_{+}=\left(\begin{array}{c}i\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}1\\ 0\end{array}\right)i={\mathbf{v}}_a+{\mathbf{v}}_{b}i,{\mathbf{v}}_a=\left(\begin{array}{c}0\\ 1\end{array}\right),{\mathbf{v}}_b=\left(\begin{array}{c}1\\ 0\end{array}\right).$$

We treat $\lambda =a-bi$ similarly. To find the eigenvectors:

$$A_{a-bi}\cdot \mathbf{x}=\left(\begin{array}{cc}bi& -b\\ b& bi\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

Therefore $bi{x}_1-b{x}_2=0$ and $b{x}_1+bi{x}_2=0$. These equations are redundant since the second is $-i$ times the first. Dividing by $b$ and then plugging in $x_1=-i{x}_2$ we have $\left(\begin{array}{c}-i{x}_2\\ x_2\end{array}\right)$, and we choose $x_2=1$ obtaining:

$${\mathbf{v}}_{-}=\left(\begin{array}{c}-i\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)-\left(\begin{array}{c}1\\ 0\end{array}\right)i={\mathbf{v}}_a-{\mathbf{v}}_{b}i,{\mathbf{v}}_a,{\mathbf{v}}_b\text{as above}.$$

The solutions are therefore

$$\begin{array}{rl}{\mathbf{y}}_1(t)& =\left(\begin{array}{c}0\\ 1\end{array}\right)e^{at}\cos (bt)-\left(\begin{array}{c}1\\ 0\end{array}\right)e^{at}\sin (bt)\\ & \\ {\mathbf{y}}_2(t)& =\left(\begin{array}{c}0\\ 1\end{array}\right)e^{at}\sin (bt)+\left(\begin{array}{c}1\\ 0\end{array}\right)e^{at}\cos (bt).\end{array}$$

When $a=b=1$ and $t=0$ these solutions reduce to:

$${\mathbf{y}}_1(0)=\left(\begin{array}{c}0\\ 1\end{array}\right),{\mathbf{y}}_2(0)=\left(\begin{array}{c}1\\ 0\end{array}\right).$$

We seek $\mathbf{y}(t)=C_1{\mathbf{y}}_1+C_2{\mathbf{y}}_2$ satisfying $\mathbf{y}(0)=\left(\begin{array}{c}A\\ B\end{array}\right)$. Clearly we need $C_1=B$ and $C_2=A$. So our final answer (with $a=b=1$) is:

$$\begin{array}{rl}\mathbf{y}(t)& =B\left(\left(\begin{array}{c}0\\ 1\end{array}\right)e^t\cos (t)-\left(\begin{array}{c}1\\ 0\end{array}\right)e^t\sin (t)\right)+A\left(\left(\begin{array}{c}0\\ 1\end{array}\right)e^t\sin (t)+\left(\begin{array}{c}1\\ 0\end{array}\right)e^t\cos (t)\right)\\ & \\ & =\left(\begin{array}{c}A\\ B\end{array}\right)e^t\cos (t)+\left(\begin{array}{c}-B\\ A\end{array}\right)e^t\sin (t).\end{array}$$

System with too few eigenvectors

System with too few eigenvectors

Problem: Solve the system ${\mathbf{y}}^{\prime }=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\mathbf{y}$.

Solution: The matrix $A=\left(\begin{array}{rc}1& 1\\ 0& 1\end{array}\right)$ has $\det A_{\lambda }=(1-\lambda {)}^2$. Therefore it has a “double” eigenvalue $\lambda =+1$. To find the eigenvectors, solve $A_{\lambda }\mathbf{x}=0$:

$$A_{+1}\cdot \mathbf{x}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

This tells us $x_2=0$ and $x_1$ can be anything. Therefore we have the eigenvectors $\left(\begin{array}{c}{x}_1\\ 0\end{array}\right)$ and we may choose $x_1=1$, obtaining $\left(\begin{array}{c}1\\ 0\end{array}\right)$.

However, no more independent eigenvectors are available (other than scalar multiples of this one, which would not be independent). Therefore it is impossible to find a complete set of solutions using eigenvectors.

Solving a system with insufficient eigenvectors

Solving a system with insufficient eigenvectors

Problem: Solve the system ${\mathbf{y}}^{\prime }=\left(\begin{array}{ccc}1& 0& 0\\ 1& 3& 0\\ 0& 1& 1\end{array}\right)\mathbf{y}$.

Solution: The matrix $A=\left(\begin{array}{ccc}1& 0& 0\\ 1& 3& 0\\ 0& 1& 1\end{array}\right)$ has $\det A_{\lambda }=-(\lambda -1{)}^2(\lambda -3)$.

For $\lambda =3$ we solve:

$$\left(\begin{array}{rrr}-2& 0& 0\\ 1& 0& 0\\ 0& 1& -2\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$

and obtain $x_1=0$, $x_2=2x_3$. So we get an eigenvector $\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)$ and a solution ${\mathbf{y}}_{+3}(t)=\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^{3t}$.

For $\lambda =1$ we first compute $A_{\lambda }^2$:

$$A_{+1}^2={\left(\begin{array}{ccc}0& 0& 0\\ 1& 2& 0\\ 0& 1& 0\end{array}\right)}^2=\left(\begin{array}{ccc}0& 0& 0\\ 2& 4& 0\\ 1& 2& 0\end{array}\right).$$

Next seek solutions $\mathbf{x}$ to:

$$A_{+1}^2\mathbf{x}=\left(\begin{array}{ccc}0& 0& 0\\ 2& 4& 0\\ 1& 2& 0\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$$

We want ${\mathbf{v}}_0$ and ${\mathbf{v}}_1$ independent vector solutions such that $A_{+1}{\mathbf{v}}_0={\mathbf{v}}_1$. These vectors must satisfy $x_1+2x_2=0$, and that is the only condition. Consider the vector ${\mathbf{v}}_0=\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)$. Notice that $A_{+1}{\mathbf{v}}_0=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$, so we define ${\mathbf{v}}_1=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$. This also satisfies $x_1+2x_2=0$, and it is independent of ${\mathbf{v}}_0$.

Now that we have ${\mathbf{v}}_0$ and ${\mathbf{v}}_1$, we can write the pair of solutions we need:

$$\begin{array}{rl}{y}_{+1}(t)& ={\mathbf{v}}_1{e}^t\\ y_{+1^{\prime }}(t)& ={\mathbf{v}}_0{e}^t+{\mathbf{v}}_{1}t{e}^t.\end{array}$$

Finally, our complete family of solutions is the set of linear combinations of those we have found:

$$A\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^{3t}+B\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)e^t+C\left(\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)e^t+\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)t{e}^t\right)=\left(\begin{array}{c}-2C{e}^t\\ 2A{e}^{3t}+C{e}^t\\ A{e}^{3t}+B{e}^t+Ct{e}^t\end{array}\right).$$

Explicit matrix exponential using series termination - revisited system

Explicit matrix exponential using series termination

Problem: Solve the system ${\mathbf{y}}^{\prime }=\left(\begin{array}{ccc}1& 0& 0\\ 1& 3& 0\\ 0& 1& 1\end{array}\right)\mathbf{y}$. (Same system as a previous example.)

Solution: This time we solve the system using $e^{At}$ and the generalized eigenvectors. Recall that the matrix has $\det A_{\lambda }=-(\lambda -1{)}^2(\lambda -3)$. In the previous solution to this example we already found the eigenvector $\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)$ with eigenvalue $\lambda =3$, and we found the generalized eigenvectors ${\mathbf{v}}_0=\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)$ and ${\mathbf{v}}_1=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$ with eigenvalue $\lambda =1$. (We had also arranged that $A_{+}{\mathbf{v}}_0={\mathbf{v}}_1$, but we do not need this fact here.)

Now we have:

$$\begin{array}{rl}{e}^{At}\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)& =e^{3t}{e}^{A_{3}t}\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)\\ & =e^{3t}\left(I+t{A}_3+\frac{t^2}{2!}{A}_3^2+\frac{t^3}{3!}{A}_3^3+\cdots \right)\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)\\ & =e^{3t}\left(\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)+\text{cancelto}0t{A}_3\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)+\text{cancelto}0\frac{t^2}{2}{A}_3^2\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)+\cdots \right)=e^{3t}\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right).\end{array}$$

Similarly, since ${\mathbf{v}}_1$ is a genuine eigenvector, we have the same calculation but for $A_1$ and ${\mathbf{v}}_1$:

$$\begin{array}{rl}{e}^{At}\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)& =e^t\left(I+t{A}_1+\frac{t^2}{2!}{A}_1^2+\frac{t^3}{3!}{A}_1^3+\cdots \right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\\ & =e^t\left(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+\text{cancelto}0t{A}_1\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+\text{cancelto}0\frac{t^2}{2}{A}_1^2\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+\cdots \right)=e^t\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).\end{array}$$

Now for the generalized eigenvector, the summation has one additional nonzero term:

$$\begin{array}{rl}{e}^{At}\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)& =e^t\left(I+t{A}_1+\frac{t^2}{2!}{A}_1^2+\frac{t^3}{3!}{A}_1^3+\cdots \right)\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)\\ & =e^t\left(\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)+t{A}_1\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)+\text{cancelto}0\frac{t^2}{2}{A}_1^2\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)+\cdots \right)=e^t\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)+t{e}^t\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).\end{array}$$

Now remember that $A_1{\mathbf{v}}_0={\mathbf{v}}_1$, so the extra term which isn’t cancelled is actually $t{\mathbf{v}}_1$. So we arrive at the solution:

$$e^{At}\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)=e^t\left(\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)+t\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right)=\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right)e^t+\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)t{e}^t=\left(\begin{array}{c}-2e^t\\ e^t\\ t{e}^t\end{array}\right).$$

The middle formulation may be recognized as the same result we got by the previous method. So our approach here justifies the solutions of the form $\mathbf{y}(t)={\mathbf{v}}_0{e}^{at}+{\mathbf{v}}_{1}t{e}^{at}$ studied previously.

Explicit matrix exponential using series termination - triple eigenvalue

Explicit matrix exponential using series termination

Problem: Solve the system ${\mathbf{y}}^{\prime }=\left(\begin{array}{rrr}2& 1& -1\\ -3& -1& 1\\ 9& 3& -4\end{array}\right)\mathbf{y}$.

Solution: The matrix has $\det A_{\lambda }=-{\lambda }^3-3{\lambda }^2-3\lambda -1=-(\lambda +1{)}^3$. So $(\lambda --1{)}^3$ is a factor, and we expect three independent generalized eigenvectors with eigenvalue $-1$.

Next we compute that $A_{-1}^2=\left(\begin{array}{rrr}-3& 0& 1\\ 0& 0& 0\\ -9& 0& 3\end{array}\right)$ and $A_{-1}^3=0$. Therefore ${\mathbf{e}}_1$, ${\mathbf{e}}_2$, and ${\mathbf{e}}_3$ are independent eigenvectors that satisfy $A_{-1}^3\mathbf{x}=0$, so they are a basis of generalized eigenvectors. Now we compute $e^{A_{-1}t}$ acting on these vectors:

$$\begin{array}{rl}{e}^{At}{\mathbf{e}}_1& =e^{-t}\left(I+t{A}_{-1}+\frac{t^2}{2!}{A}_{-1}^2+\frac{t^3}{3!}{A}_{-1}^3+\cdots \right){\mathbf{e}}_1\\ & =e^{-t}\left({\mathbf{e}}_1+t{A}_{-1}{\mathbf{e}}_1+\frac{t^2}{2!}{A}_{-1}^2{\mathbf{e}}_1\right)\\ & =e^{-t}\left(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+t\left(\begin{array}{c}3\\ -3\\ 9\end{array}\right)+\frac{t^2}{2!}\left(\begin{array}{c}-3\\ 0\\ -9\end{array}\right)\right).\end{array}$$

Similarly we have:

$$\begin{array}{rl}{e}^{At}{\mathbf{e}}_2& =e^{-t}\left(I+t{A}_{-1}+\frac{t^2}{2!}{A}_{-1}^2+\frac{t^3}{3!}{A}_{-1}^3+\cdots \right){\mathbf{e}}_2\\ & =e^{-t}\left({\mathbf{e}}_2+t{A}_{-1}{\mathbf{e}}_2+\frac{t^2}{2!}{A}_{-1}^2{\mathbf{e}}_2\right)\\ & =e^{-t}\left(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+t\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)+\frac{t^2}{2!}\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\right),\end{array}$$

and:

$$\begin{array}{rl}{e}^{At}{\mathbf{e}}_3& =e^{-t}\left(I+t{A}_{-1}+\frac{t^2}{2!}{A}_{-1}^2+\frac{t^3}{3!}{A}_{-1}^3+\cdots \right){\mathbf{e}}_3\\ & =e^{-t}\left({\mathbf{e}}_3+t{A}_{-1}{\mathbf{e}}_3+\frac{t^2}{2!}{A}_{-1}^2{\mathbf{e}}_3\right)\\ & =e^{-t}\left(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)+t\left(\begin{array}{c}-1\\ 1\\ -3\end{array}\right)+\frac{t^2}{2!}\left(\begin{array}{c}1\\ 0\\ 3\end{array}\right)\right).\end{array}$$

These three functions give us independent solutions. In fact, since $\left(\begin{array}{ccc}{\mathbf{e}}_1& {\mathbf{e}}_2& {\mathbf{e}}_3\end{array}\right)=I_3$, and $e^{At}\cdot I_3=e^{At}$, we can simply write a fundamental matrix by putting these solutions as the columns:

$$X(t)=e^{At}=\left(\begin{array}{ccc}(1+3t-\frac{3t^2}{2!})e^{-t}& t{e}^{-1}& (-t+\frac{t^2}{2!})e^{-t}\\ -3t{e}^{-t}& e^{-t}& t{e}^{-t}\\ (9t-\frac{9t^2}{2!})e^{-t}& 3t{e}^{-t}& (1-3t+\frac{3t^2}{2!})e^{-t}\end{array}\right).$$