Distinct real eigenvalues case

For a constant-coefficient linear system ${\mathbf{x}}^{\prime }=A\mathbf{x}$ where $A$ is $n\times n$ with distinct real eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$, the general solution is a linear combination of exponentials:

$\mathbf{x}(t)=c_1{\mathbf{u}}_1{e}^{{\lambda }_{1}t}+c_2{\mathbf{u}}_2{e}^{{\lambda }_{2}t}+\cdots +c_n{\mathbf{u}}_n{e}^{{\lambda }_{n}t}$

where ${\mathbf{u}}_i$ is an eigenvector corresponding to ${\lambda }_i$.

This is the cleanest case — distinct eigenvalues automatically give $n$ linearly independent eigenvectors, hence $n$ linearly independent solutions, hence a complete general solution.

Why eigenvectors give solutions

Plug $\mathbf{x}(t)=\mathbf{u}{e}^{\lambda t}$ into ${\mathbf{x}}^{\prime }=A\mathbf{x}$:

$\lambda \mathbf{u}{e}^{\lambda t}=A\mathbf{u}{e}^{\lambda t}$

Cancel $e^{\lambda t}$:

$A\mathbf{u}=\lambda \mathbf{u}$

So $\lambda$ must be an eigenvalue of $A$ and $\mathbf{u}$ a corresponding eigenvector. Lemma: $\mathbf{x}=\mathbf{u}{e}^{\lambda t}$ is a solution of ${\mathbf{x}}^{\prime }=A\mathbf{x}$ if and only if $(\lambda ,\mathbf{u})$ is an eigenvalue/eigenvector pair of $A$.

The procedure

1. Find eigenvalues: solve $\det (A-\lambda I)=0$.
2. For each eigenvalue ${\lambda }_i$, find an eigenvector: solve $(A-{\lambda }_{i}I){\mathbf{u}}_i=0$.
3. Write general solution: $\mathbf{x}(t)=\sum c_i{\mathbf{u}}_i{e}^{{\lambda }_{i}t}$.
4. Apply initial conditions to find the $c_i$.

Worked example

Solve ${\mathbf{x}}^{\prime }=\left[\begin{array}{cc}1& 1\\ 4& 1\end{array}\right]\mathbf{x}$.

Eigenvalues: $\det \left[\begin{array}{cc}1-\lambda & 1\\ 4& 1-\lambda \end{array}\right]=(1-\lambda {)}^2-4={\lambda }^2-2\lambda -3=0$.

Roots: ${\lambda }_1=-1$, ${\lambda }_2=3$. Distinct real.

Eigenvector for ${\lambda }_1=-1$: solve $(A+I){\mathbf{u}}_1=0$, i.e., $\left[\begin{array}{cc}2& 1\\ 4& 2\end{array}\right]{\mathbf{u}}_1=0$.

The two rows are dependent (second is twice the first), giving $2u_{1,1}+u_{1,2}=0$. Pick $u_{1,1}=1$, $u_{1,2}=-2$:

${\mathbf{u}}_1=\left[\begin{array}{c}1\\ -2\end{array}\right]$

Eigenvector for ${\lambda }_2=3$: solve $(A-3I){\mathbf{u}}_2=0$, i.e., $\left[\begin{array}{cc}-2& 1\\ 4& -2\end{array}\right]{\mathbf{u}}_2=0$.

Gives $-2u_{2,1}+u_{2,2}=0$. Pick $u_{2,1}=1$, $u_{2,2}=2$:

${\mathbf{u}}_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$

General solution:

$\mathbf{x}(t)=c_1\left[\begin{array}{c}1\\ -2\end{array}\right]e^{-t}+c_2\left[\begin{array}{c}1\\ 2\end{array}\right]e^{3t}$

In components:

$\{\begin{array}{l}{x}_1(t)=c_1{e}^{-t}+c_2{e}^{3t}\\ x_2(t)=-2c_1{e}^{-t}+2c_2{e}^{3t}\end{array}$

Diagonal special case

If $A$ is diagonal, the eigenvalues are the diagonal entries and the eigenvectors are the standard basis vectors. The system decouples completely:

$A=\left[\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& {\lambda }_2& 0\\ 0& 0& {\lambda }_3\end{array}\right]\Rightarrow \{\begin{array}{l}{x}_1^{\prime }={\lambda }_1{x}_1\\ x_2^{\prime }={\lambda }_2{x}_2\\ x_3^{\prime }={\lambda }_3{x}_3\end{array}$

Each component is a 1D ODE: $x_i=c_i{e}^{{\lambda }_{i}t}$. The general solution is

$\mathbf{x}(t)=c_1{e}^{{\lambda }_{1}t}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]+c_2{e}^{{\lambda }_{2}t}\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+c_3{e}^{{\lambda }_{3}t}\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

When $A$ isn’t diagonal but has distinct eigenvalues, the eigenvector basis effectively diagonalizes the system in a rotated frame — same idea, different coordinates.

Stability and qualitative behavior

Long-term behavior depends on the signs of the eigenvalues:

• All ${\lambda }_i<0$: solutions decay to zero. Origin is asymptotically stable (a stable node in 2D).
• All ${\lambda }_i>0$: solutions grow exponentially. Origin is unstable.
• Mixed signs: solutions in some directions decay, others grow. Origin is a saddle point (unstable).

For 2D phase portraits, see Phase plane behaviour cases 1, 2, 3.

For other eigenvalue scenarios, see Complex conjugate eigenvalues case and Repeated eigenvalues case.