Chapter 06 - Linear Systems

Linear systems I: matrix descriptions

Systems from higher order equations

Systems from higher order equations

A system of ODEs looks like this:

$$\begin{array}{rl}{y}_1^{\prime }& =f_1(t,y_1,\dots ,y_n)\\ y_2^{\prime }& =f_2(t,y_1,\dots ,y_n)\\ ⋮& \\ y_n^{\prime }& =f_n(t,y_1,\dots ,y_n).\end{array}$$

The derivative of each variable $y_i$ is determined by its own function of the other variables, $f_i(t,y_1,\dots ,y_n)$.

One reason systems are important is that they offer an alternative to the language of order of a differential equation. Consider a general ODE of order $n$ given by the equation

$$F(t,y,y^{\prime },\dots ,y^{(n)})=0.$$

This ODE (and its complete families of solutions as well as IVPs) is equivalent to a first-order system in which higher derivatives are replaced with new variables:

$$\begin{array}{rl}F(t,y_0,y_1,\dots ,y_n)& =0\\ y_1& =y_0^{\prime }\\ y_2& =y_1^{\prime }\\ y_3& =y_2^{\prime }\\ ⋮& \\ y_n& =y_{n-1}^{\prime }.\end{array}$$

In particular, consider the $n^{\text{th}}$ order linear ODE:

$$y^{(n)}+a_{n-1}(t)y^{(n-1)}+\cdots +a_1(t)y^{\prime }+a_0(t)y=f(t).$$

By defining $y_0,\dots ,y_{n-1}$ with the formulas $y_0=y$ and $y_i=y^{(i)}$, we find an equivalent linear system:

$$\begin{array}{rl}{y}_0^{\prime }& =y_1\\ y_1^{\prime }& =y_2\\ ⋮& \\ y_{n-1}^{\prime }& =-a_{n-1}(t)y_{n-1}-\cdots -a_0(t)y_0+f(t).\end{array}$$

Notice that this system can be represented using matrix multiplication:

$${\left(\begin{array}{c}{y}_0\\ y_1\\ ⋮\\ y_{n-1}\end{array}\right)}^{\prime }=\left(\begin{array}{cccccc}0& 1& 0& 0& \dots & 0\\ 0& 0& 1& 0& \dots & 0\\ ⋮& & & ⋮& \dots & ⋮\\ -a_0(t)& -a_1(t)& -a_2(t)& -a_3(t)& \dots & -a_{n-1}(t)\end{array}\right)\left(\begin{array}{c}{y}_0\\ y_1\\ ⋮\\ y_{n-1}\end{array}\right)+\left(\begin{array}{c}0\\ 0\\ ⋮\\ f(t)\end{array}\right).$$

The last equation suggests vector-matrix formalism for systems of linear equations:

$${\mathbf{y}}^{\prime }(t)=\mathbf{A}(t)\mathbf{y}(t)+\mathbf{f}(t),\mathbf{A}(t)=\left(\begin{array}{ccc}{a}_{11}(t)& \dots & a_{1n}(t)\\ ⋮& & ⋮\\ a_{n1}(t)& \dots & a_{nn}(t)\end{array}\right),\mathbf{y}(t)=\left(\begin{array}{c}{y}_1(t)\\ ⋮\\ y_n(t)\end{array}\right),\mathbf{f}(t)=\left(\begin{array}{c}{f}_1(t)\\ ⋮\\ f_n(t)\end{array}\right).$$

This format is called the normal form for the linear system of ODEs.

Later sections, starting with homogeneous systems with constant coefficients, address the problem of extending our previous methods for second-order linear equations to the formalism for linear systems.

The methods we will extend:

• Homogeneous – complete family of solutions
• Inhomogeneous – variation of parameters for a particular solution
• Constant coefficients – real, repeated, and complex: exponential, linear factor, and sinusoidal solutions

Some new methods will be introduced to address the complexities of linear interaction between variables:

• Independence – to explain ‘complete’ family of solutions
• Eigenvectors – to describe proportionality between variables having the same overall exponential / sinusoidal behavior
• Matrix calculus – to write a general solution formula for constant coefficient systems

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Homogeneous system

Homogeneous system

When the normal form of the system has nso $\mathbf{f}(t)$ term, it is called homogeneous:

$${\mathbf{y}}^{\prime }(t)=\mathbf{A}(t)\mathbf{y}(t),$$

or simply ${\mathbf{y}}^{\prime }=\mathbf{Ay}$.

The important thing about homogeneous systems is that the solutions form a vector space, meaning that solutions can be added and scaled while remaining valid solutions. For example, if ${\mathbf{y}}_1(t)$ and ${\mathbf{y}}_2(t)$ are vector solutions, then

$$C_1{\mathbf{y}}_1(t)+C_2{\mathbf{y}}_2(t)$$

is also a valid vector solution for any constants $C_1$ and $C_2$. Here’s a proof:

$$\begin{array}{rl}(C_1{\mathbf{y}}_1(t)+C_2{\mathbf{y}}_2(t){)}^{\prime }& =C_1{\mathbf{y}}_1^{\prime }(t)+C_2{\mathbf{y}}_2^{\prime }(t)\\ & =C_1\mathbf{A}(t){\mathbf{y}}_1(t)+C_2\mathbf{A}(t){\mathbf{y}}_2(t)\text{(separate solutions)}\\ & =\mathbf{A}(t)(C_1{\mathbf{y}}_1(t)+C_2{\mathbf{y}}_2(t))\text{(linearity of A).}\end{array}$$

The fact that solutions can be added to form valid solutions is called the superposition principle and it applies to linear homogeneous systems in the same way as it applies to $2^{\text{nd}}$ order linear equations.

Since the complete family of solutions of a system like ${\mathbf{y}}^{\prime }=\mathbf{Ay}$ forms a vector space, we would like to describe this family as the span of a specific set of independent solutions. Just as for any other vectors, a set of solutions ${\mathbf{y}}_1,\dots ,{\mathbf{y}}_n$ is an independent set when any equation

$$C_1{\mathbf{y}}_1+\cdots +C_n{\mathbf{y}}_n=0$$

can only be valid when $C_1=\cdots =C_n=0$.

A set of independent solutions is a complete set when every other solution vector can be written as a linear combination of them, i.e. it is in their span:

$$\mathbf{y}(t)={\alpha }_1{\mathbf{y}}_1+\cdots +{\alpha }_n{\mathbf{y}}_n,\text{some }{\alpha }_1,\dots ,{\alpha }_n.$$

In linear algebra, a complete set of independent vectors for a space is called a basis for the space. In ODE theory, a complete set of independent solution vectors is called a fundamental set of solutions. To reiterate: a fundamental set of solutions ${\mathbf{y}}_1,\dots ,{\mathbf{y}}_n$ is both independent and complete.

Solving ODE system

Solving an ODE system (finding the complete family of solutions) means: finding a fundamental set of solutions.

Given a fundamental set of solution vectors ${\mathbf{y}}_1,\dots ,{\mathbf{y}}_n$, we can arrange these vectors as the columns of a matrix called the fundamental matrix:

$$\mathbf{Y}(t)=\left(\begin{array}{ccc}{\mathbf{y}}_1& \cdots & {\mathbf{y}}_n\end{array}\right).$$

By linear algebra theory, these columns form a basis if and only if this matrix has nonzero determinant. The determinant of any fundamental matrix is called the Wronskian matrix $W(t)$.

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Example

Verifying a fundamental matrix

Verifying a fundamental matrix

Problem: Consider the ODE system:

$${\mathbf{y}}^{\prime }(t)=\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\mathbf{y}(t).$$

Verify that the following is a fundamental matrix for this ODE:

$$\mathbf{Y}=\left(\begin{array}{ccc}{e}^{2t}& -e^{-t}& 0\\ e^{2t}& 0& e^{-t}\\ e^{2t}& e^{-t}& -e^{-t}\end{array}\right).$$

Solution: First we must check that each column of $\mathbf{Y}$ is a solution vector for the ODE:

$$\begin{array}{rl}{\mathbf{y}}_1^{\prime }={\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right)}^{\prime }& =\left(\begin{array}{c}2e^{2t}\\ 2e^{2t}\\ 2e^{2t}\end{array}\right)=2\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right),\\ & \\ \mathbf{A}{\mathbf{y}}_1=\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right)& =\left(\begin{array}{c}0+e^{2t}+e^{2t}\\ e^{2t}+0+e^{2t}\\ e^{2t}+e^{2t}+0\end{array}\right)=2\left(\begin{array}{c}{e}^{2t}\\ e^{2t}\\ e^{2t}\end{array}\right).\end{array}$$

Similarly for the second column vector:

$$\begin{array}{rl}{\left(\begin{array}{c}-e^{-t}\\ 0\\ e^{-t}\end{array}\right)}^{\prime }& =\left(\begin{array}{c}{e}^{-t}\\ 0\\ -e^{-t}\end{array}\right)\\ & \\ \left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)\left(\begin{array}{c}-e^{-t}\\ 0\\ e^{-t}\end{array}\right)& =\left(\begin{array}{c}0+0+e^{-t}\\ -e^{-t}+0+e^{-t}\\ -e^{-t}+0+0\end{array}\right)=\left(\begin{array}{c}{e}^{-t}\\ 0\\ -e^{-t}\end{array}\right).\end{array}$$

And similarly for the third column vector.

Secondly, we must verify that these solutions are independent. For that, we compute the determinant of the Wronskian matrix using the formula for $3\times 3$ matrices:

$$\begin{array}{rl}W(t)=\det \mathbf{Y}& =\left|\begin{array}{ccc}{e}^{2t}& -e^{-t}& 0\\ e^{2t}& 0& e^{-t}\\ e^{2t}& e^{-t}& -e^{-t}\end{array}\right|\\ & \\ & =e^{2t}\left|\begin{array}{cc}0& e^{-t}\\ e^{-t}& -e^{-t}\end{array}\right|--e^{-t}\left|\begin{array}{cc}{e}^{2t}& e^{-t}\\ e^{2t}& -e^{-t}\end{array}\right|+0\left|\begin{array}{cc}{e}^{2t}& 0\\ e^{2t}& e^{-t}\end{array}\right|\\ & \\ & =e^{2t}(-e^{-2t})+e^{-t}(-2e^t)+0=-1-2=-3.\end{array}$$

Therefore $W(t)=-3$ and is never $0$, so the columns of $\mathbf{Y}$ are independent. Combining this fact with the first fact, we know that $\mathbf{Y}$ is a fundamental matrix.

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Linear systems: homogeneous, constant coefficients

Introduction

We are interested in systems of equations that can be written in the form ${\mathbf{y}}^{\prime }=A\mathbf{y}$, where $\mathbf{y}(t)$ is a vector function of $t$, and $A$ is a matrix with constant entries (independent of $t$).

Our goal is to find a complete set of solutions to ${\mathbf{y}}^{\prime }=A\mathbf{y}$. It is not enough to find just one. How many solutions constitute a complete set?

Picard-Lindelöf for first-order systems

The system: ${\mathbf{y}}^{\prime }(t)=A\mathbf{y}(t),\mathbf{y}(0)={\mathbf{y}}_0$
always has existence and uniqueness of solutions, for any initial vector ${\mathbf{y}}_0$ and constant coefficient matrix $A$.

This implies: a complete set of solutions is a vector space (by linearity) of the same dimension as ${\mathbf{y}}_0$ (by the theorem).

Eigenvectors
The eigenvector approach to solving the system starts from the observation that if $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, meaning that $A\mathbf{v}=\lambda \mathbf{v}$, then the vector function $\mathbf{y}(t)=\mathbf{v}{e}^{\lambda t}$ is a solution to the system.

The downside of this approach is that the solution cannot be written down until all the eigenvalues have been computed; the eigenvalues frequently involve imaginary numbers; and it is possible there are simply not enough eigenvectors to give a complete family.

Matrix exponentials
The matrix exponential approach to solving the same system starts by constructing the function $Y(t)=e^{At}$ using power series. This function is a matrix-valued function of $t$ that satisfies $Y^{\prime }=AY$, and because $e^{At}$ is always an invertible matrix, the columns of $Y(t)$ automatically give a complete set of solutions.

The downside of this approach is that it is hard to compute $e^{At}$, and the typical method requires finding the eigenvalues of $A$ anyway. On the other hand, it is not necessary to use imaginary numbers because the matrix $e^{At}$ can be put into what is called “Real Jordan Form” (with only real numbers). This form uses an extension of the idea of eigenvector that is designed specifically to handle the situation of missing eigenvectors. “Putting into” this form simply means changing variables $\mathbf{z}=P\mathbf{y}$ according to some invertible transformation $P$. The matrix $PA{P}^{-1}$ of the new equation ${\mathbf{z}}^{\prime }=(PA{P}^{-1})\mathbf{z}$ will have the Real Jordan Form.

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Eigenvector approach

Eigenvector approach

Suppose we have any system of the form ${\mathbf{y}}^{\prime }=A\mathbf{y}$ for a matrix with constant entries $a_{ij}$. Suppose $\mathbf{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, meaning that $A\mathbf{v}=\lambda \mathbf{v}$. (Note that $\mathbf{v}$ also has constant entries, like $A$.) Then the function $\mathbf{y}(t)=\mathbf{v}{e}^{\lambda t}$ solves the system:

$$\begin{array}{rl}{\mathbf{y}}^{\prime }(t)& =(\mathbf{v}{e}^{\lambda t}{)}^{\prime }={\mathbf{v}}^{\prime }{e}^{\lambda t}+\mathbf{v}(e^{\lambda t}{)}^{\prime }=0+\mathbf{v}\lambda e^{\lambda t}=\lambda (\mathbf{v}{e}^{\lambda t})=\lambda \mathbf{y}(t)\\ & \\ A\mathbf{y}(t)& =A(\mathbf{v}{e}^{\lambda t})=(A\mathbf{v})e^{\lambda t}=(\lambda \mathbf{v})e^{\lambda t}=\lambda (\mathbf{v}{e}^{\lambda t})=\lambda \mathbf{y}(t).\end{array}$$

Eigenvector-eigenvalue solutions

Any eigenvector-eigenvalue pair $\mathbf{v},\lambda$ determines a solution $\mathbf{y}(t)=\mathbf{v}{e}^{\lambda t}$.

Remember that if $\mathbf{v}$ is an eigenvector of $A$, then any multiple $a\mathbf{v}$ is also an eigenvector. In terms of solutions, this simply means that if $\mathbf{y}(t)$ is a solution of the form $\mathbf{v}{e}^{\lambda t}$, then $a\mathbf{y}(t)$ is also a solution. We already knew this: it is part of linearity – the fact that solutions are vectors in a vector space.

The Picard-Lindelöf theorem implies that the space of solutions has dimension $n$ equal to the dimension of the vectors $\mathbf{y}$ and ${\mathbf{y}}_0$. (The dimension of $\mathbf{y}$ is just the number of components it has.) Therefore a set of solutions ${\mathbf{y}}_1(t),\dots ,{\mathbf{y}}_k(t)$ is a complete set when (a) we have $k=n$, and (b) they are independent when considered as vectors. The Wronskian (using the determinant) can tell us when $n$ vectors having $n$ components are independent.

When there are $n$ independent eigenvectors, and all of their eigenvalues are real numbers, the eigenvector strategy works perfectly without further complication. Just find all eigenvector-eigenvalue pairs and write down all the solutions they determine, and we are done.

The main challenge of the eigenvector strategy becomes the challenge of handling situations of two types. It can happen that (a) some eigenvalues involve imaginary numbers, or (b) some systems do not have enough eigenvectors. In applications, the former tends to arise for oscillatory systems, and the latter arises for analogues of critical damping.

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Example

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:

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

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Complex eigenvalues

Complex eigenvalues

Suppose the solution to the eigenvalue equation $\det A_{\lambda }=0$ is a complex eigenvalue $\lambda =a+bi$.

Because the matrix $A_{\lambda }$ has real-valued entries, such eigenvalue roots must come in conjugate pairs, meaning we would always find both eigenvalues $\lambda =a\pm bi$. Furthermore, the corresponding eigenvectors also come in a complex conjugate pair ${\mathbf{v}}_{+}={\mathbf{v}}_a+{\mathbf{v}}_{b}i$ and ${\mathbf{v}}_{-}={\mathbf{v}}_a-{\mathbf{v}}_{b}i$:

$$\begin{array}{rl}A{\mathbf{v}}_{+}& =(a+bi){\mathbf{v}}_{+}\\ A{\mathbf{v}}_{-}& =(a-bi){\mathbf{v}}_{-}.\end{array}$$

(This follows by taking the complex conjugate of both sides of the equation, using the fact that conjugation and multiplication can be performed in either order.)

We proceed as in the $2^{\text{nd}}$ order linear theory, and combine the solutions ${\mathbf{y}}_{+}(t)={\mathbf{v}}_{+}{e}^{(a+bi)t}$ and ${\mathbf{y}}_{-}(t)={\mathbf{v}}_{-}{e}^{(a-bi)t}$ in such a way as to extract vector functions having exclusively real-valued entries. We have:

$$\begin{array}{rlrl}{\mathbf{y}}_1(t)& :=\frac{1}{2}({\mathbf{y}}_{+}+{\mathbf{y}}_{-})& & ={\mathbf{v}}_a{e}^{at}\cos (bt)-{\mathbf{v}}_b{e}^{at}\sin (bt)\\ & & & \\ {\mathbf{y}}_2(t)& :=\frac{1}{2i}({\mathbf{y}}_{+}-{\mathbf{y}}_{-})& & ={\mathbf{v}}_a{e}^{at}\sin (bt)+{\mathbf{v}}_b{e}^{at}\cos (bt).\end{array}$$

In summary: whenever $\lambda =a+bi$ is an eigenvalue for the eigenvector $\mathbf{v}={\mathbf{v}}_a+{\mathbf{v}}_{b}i$, the formulas above for ${\mathbf{y}}_1$ and ${\mathbf{y}}_2$ give two independent solutions to the system ${\mathbf{y}}^{\prime }=A\mathbf{y}$.

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Question

01

Extracting real solutions

Verify the real-valued formulas for ${\mathbf{y}}_1$ and ${\mathbf{y}}_2$ by plugging in ${\mathbf{y}}_{+}$ and ${\mathbf{y}}_{-}$ and then expanding and simplifying.

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Example

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:

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

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Exercise

01

Complex eigenvalues

Solve the system ${\mathbf{y}}^{\prime }=\left(\begin{array}{cc}3& -13\\ 5& 1\end{array}\right)\mathbf{y}$. First find the complex solutions, then find the real solutions.

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Insufficient eigenvectors

Insufficient eigenvectors

Suppose the solution to the eigenvalue equation $\det A_{\lambda }=0$ is a repeated eigenvalue, so for example the equation has a factor like $(\lambda -a{)}^k=0$. In this case, sometimes it happens that the eigenvalue $\lambda =a$ does not have enough eigenvectors associated to it.

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Example

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.

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Insufficient eigenvectors (part 2)

Let us consider the way around this difficulty.
The way around is not very simple! Still, we try to take the simplest path.

Suppose our situation is that $(\lambda -a{)}^2=0$ as a factor in $\det A_{\lambda }$, and that we have a single eigenvector $\mathbf{v}$ satisfying $A_a\mathbf{v}=0$. So we start with the solution $\mathbf{v}{e}^{at}$. We seek a second solution.

There is a concept of generalized eigenvector, which is any $\mathbf{x}$ that satisfies $A_{\lambda }^k\mathbf{x}=0$ for higher powers $k\ge 1$. According to linear algebra theory, whenever $(\lambda -a{)}^k=0$ from a factor in $\det A_{\lambda }$, there will be enough independent generalized eigenvectors satisfying $A_{\lambda }^k\mathbf{x}=0$. Notice that if $A_{\lambda }\mathbf{x}=0$ then $A_{\lambda }^2\mathbf{x}=0$, so regular eigenvectors always count as generalized eigenvectors.

Now let us consider the system ${\mathbf{y}}^{\prime }=A\mathbf{y}$ again to see how to extract new solutions from generalized eigenvectors with $k\ge 2$. We start by computing two linearly independent solutions to $A_{\lambda }^k\mathbf{x}=0$, called ${\mathbf{v}}_0$ and ${\mathbf{v}}_1$, which are connected according to multiplication by $A_a=(A-aI)$:

$${\mathbf{v}}_0\stackrel{(A-aI)}{⟶}{\mathbf{v}}_1\stackrel{(A-aI)}{⟶}0.$$

So ${\mathbf{v}}_1$ is a regular eigenvector, while ${\mathbf{v}}_0$ is only a generalized one with $k=2$. (In a sense, ${\mathbf{v}}_0$ is “deeper down” than ${\mathbf{v}}_1$. So regular eigenvectors do not always reach deep enough down into preimages of $A_a$.)

We already know that $({\mathbf{v}}_1{e}^{at}{)}^{\prime }=A({\mathbf{v}}_1{e}^{at})$ since ${\mathbf{v}}_1$ is a regular eigenvector with eigenvalue $a$.

Now look what happens when we plug ${\mathbf{v}}_{1}t{e}^{at}$ into the equation:

$$\begin{array}{rl}({\mathbf{v}}_{1}t{e}^{at}{)}^{\prime }& ={\mathbf{v}}_1{e}^{at}+a{\mathbf{v}}_{1}t{e}^{at},\\ & \\ A\left({\mathbf{v}}_{1}t{e}^{at}\right)& =(A-aI+aI){\mathbf{v}}_{1}t{e}^{at}\\ & =(A-aI){\mathbf{v}}_{1}t{e}^{at}+a{\mathbf{v}}_{1}t{e}^{at}\\ & =a{\mathbf{v}}_{1}t{e}^{at}.\end{array}$$

And look what happens when we plug ${\mathbf{v}}_0{e}^{at}$ into the equation:

$$\begin{array}{rl}({\mathbf{v}}_0{e}^{at}{)}^{\prime }& =a{\mathbf{v}}_0{e}^{at},\\ & \\ A\left({\mathbf{v}}_0{e}^{at}\right)& =(A-aI+aI){\mathbf{v}}_0{e}^{at}\\ & =(A-aI){\mathbf{v}}_0{e}^{at}+a{\mathbf{v}}_0{e}^{at}\\ & ={\mathbf{v}}_1{e}^{at}+a{\mathbf{v}}_0{e}^{at}.\end{array}$$

Insufficient eigenvectors

If we add the solutions to form $\mathbf{y}(t)={\mathbf{v}}_0{e}^{at}+{\mathbf{v}}_{1}t{e}^{at}$, the ODE equation holds.

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Example

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:

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

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Exercise

02

Insufficient eigenvectors

Solve the IVP: $\begin{array}{rl}{x}^{\prime }(t)& =7x+y\\ y^{\prime }(t)& =-4x+3y,\end{array}$ $\left(\begin{array}{c}x(0)\\ y(0)\end{array}\right)=\left(\begin{array}{c}2\\ -5\end{array}\right)$.

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Matrix exponential approach

Matrix exponential approach

Recall that the matrix exponential approach gives us the formula $Y(t)=e^{At}$ for the fundamental matrix solution to ${\mathbf{y}}^{\prime }=A\mathbf{y}$. Each column of $Y=\left(\begin{array}{ccc}{\mathbf{y}}_1& \cdots & {\mathbf{y}}_n\end{array}\right)$ is a solution vector function, and these columns are independent of each other. The complete family of solutions to ${\mathbf{y}}^{\prime }=A\mathbf{y}$ is:

$$Y\mathbf{c}=Y\left(\begin{array}{c}{C}_1\\ C_2\\ ⋮\\ C_n\end{array}\right)=C_1{\mathbf{y}}_1+\cdots +C_n{\mathbf{y}}_n.$$

This approach has no difficulty in finding a formula for the solution. The challenge for this approach is: How do we describe the formula $e^{At}$ in more practical terms?

A complete description of the answer to this challenge gets too complicated to cover here and requires more linear algebra than this course assumes. Instead of trying to give a complete description, our goal will be to see a glimpse of the main ideas and then let your imagination fill in the rest.

Key observation

The central observation we need is this. First, we can rewrite the series in an ‘expansion around $\lambda$’:

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

Now suppose we have a basis of generalized eigenvectors ${\mathbf{u}}_1,\dots ,{\mathbf{u}}_k$. This means $A_{\lambda }^k{\mathbf{u}}_i=0$ for $i=1,\dots ,k$ and ${\mathbf{u}}_1,\dots ,{\mathbf{u}}_k$ are independent of each other.

Then when $e^{At}$ is applied to any of the ${\mathbf{u}}_i$, the terms of the second factor series are eventually just zeros:

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

where the zeros occur because $A_{\lambda }^j{\mathbf{u}}_i=0$ whenever $j\ge k$.

This finite series can then be calculated. No limits of matrix sums need be taken.

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Example

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.

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Matrix exponential approach (part 2)

In order to generalize the previous method to all cases you could encounter, it suffices to be able to find a set of $k$ independent generalized eigenvectors whenever $\det A_{\lambda }=(\lambda -a{)}^k$ (or has the latter as a factor). The best general method for this is to use “row reduction” from linear algebra to find a basis of the null space of $A_a$, i.e. to solve for $\mathbf{x}$ in the vector equation $(A-\lambda I)\mathbf{x}=0$ when $\lambda =a$.

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Example

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:

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

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