Repeated eigenvalues case

When the matrix $A$ has a repeated eigenvalue $r$ (so the characteristic polynomial has $(\lambda -r{)}^k$ as a factor for $k>1$) but doesn’t have enough linearly independent eigenvectors to span the corresponding eigenspace, the system ${\mathbf{x}}^{\prime }=A\mathbf{x}$ requires generalized eigenvectors to construct a full set of solutions.

The key concept: even though $r$ is an eigenvalue with multiplicity $k$, you might only have one linearly independent eigenvector. In that case, $\mathbf{x}(t)=\mathbf{u}{e}^{rt}$ alone doesn’t give $k$ solutions — you need additional solutions involving $t{e}^{rt}$, $t^2{e}^{rt}$, etc.

Algebraic vs geometric multiplicity

For an eigenvalue $r$:

• Algebraic multiplicity $m_a(r)$: number of times $r$ appears as a root of $\det (A-\lambda I)=0$.
• Geometric multiplicity $m_g(r)$: dimension of the eigenspace $\ker (A-rI)$, i.e., the number of linearly independent eigenvectors.

In general, $m_a(r)\ge m_g(r)$. When equal, the matrix is “well-behaved” for $r$ — you have enough eigenvectors. When $m_a>m_g$, the matrix is defective at $r$ and you need generalized eigenvectors.

Example: $A=\left[\begin{array}{cc}1& -1\\ 1& 3\end{array}\right]$ has $\det (A-\lambda I)=(\lambda -2{)}^2$, so $r=2$ with $m_a=2$. The eigenspace is one-dimensional ($m_g=1$) — only one eigenvector $\mathbf{u}=(1,-1{)}^T$ exists.

Generalized eigenvectors

To get a second linearly independent solution when $m_a>m_g$, look for a solution of the form

${\mathbf{x}}_2(t)=\xi t{e}^{rt}+\eta e^{rt}$

where $\xi$ and $\eta$ are vectors to be determined.

Compute the derivative:

${\mathbf{x}}_2^{\prime }(t)=\xi (e^{rt}+rt{e}^{rt})+r\eta e^{rt}$

Set equal to $A{\mathbf{x}}_2=t{e}^{rt}A\xi +e^{rt}A\eta$. Match coefficients of $t{e}^{rt}$ and $e^{rt}$:

$\{\begin{array}{l}(A-rI)\xi =0\\ (A-rI)\eta =\xi \end{array}$

So $\xi$ is an eigenvector (set it to $\mathbf{u}$, the one we already found), and $\eta$ is a generalized eigenvector — a vector that maps to $\mathbf{u}$ under $A-rI$.

Worked example

For $A=\left[\begin{array}{cc}1& -1\\ 1& 3\end{array}\right]$:

Step 1: $r=2$ is the repeated eigenvalue.

Step 2 (eigenvector): $(A-2I)\mathbf{u}=0$, i.e., $\left[\begin{array}{cc}-1& -1\\ 1& 1\end{array}\right]\mathbf{u}=0$. Pick $\mathbf{u}=(1,-1{)}^T$.

Step 3 (generalized eigenvector): $(A-2I)\eta =\mathbf{u}$, i.e., $\left[\begin{array}{cc}-1& -1\\ 1& 1\end{array}\right]\eta =\left[\begin{array}{c}1\\ -1\end{array}\right]$.

Both equations give $-{\eta }_1-{\eta }_2=1$. Pick ${\eta }_1=-1$, ${\eta }_2=0$:

$\eta =\left[\begin{array}{c}-1\\ 0\end{array}\right]$

Step 4 (general solution):

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

The two solutions are ${\mathbf{x}}_1=e^{2t}\mathbf{u}$ and ${\mathbf{x}}_2=t{e}^{2t}\mathbf{u}+e^{2t}\eta$. Linearly independent (Wronskian nonzero).

Higher multiplicities

For $m_a=3$ with $m_g=1$, you’d need three solutions. Following the pattern:

${\mathbf{x}}_3(t)=\frac{t^2}{2}{e}^{rt}\xi +t{e}^{rt}\eta +e^{rt}\zeta$

Plugging in and matching coefficients gives:

$(A-rI)\xi =0,(A-rI)\eta =\xi ,(A-rI)\zeta =\eta$

So you build a “Jordan chain” of generalized eigenvectors, each mapping to the previous one under $A-rI$.

In general, for $m_a=k$ with $m_g<k$, you find a Jordan chain of length $k$ (or several chains summing to total length $k$).

Why this happens

For a “diagonalizable” matrix (where $m_a=m_g$ for every eigenvalue), the matrix has a basis of eigenvectors and the system decouples. For a “defective” matrix, the generalized eigenvectors fill in the missing dimensions of the solution space.

The general theory: any matrix has a Jordan canonical form $J=P^{-1}AP$ where $J$ is a block-diagonal matrix of Jordan blocks. Each Jordan block of size $k$ corresponds to one generalized eigenvector chain of length $k$.

Phase portrait

For 2D systems with a repeated eigenvalue:

• If $m_g=2$ (two eigenvectors): trajectories are straight lines through origin scaled by $e^{rt}$. Equilibrium is a proper node.
• If $m_g=1$ (one eigenvector): trajectories curve, tangent to the eigenvector direction. Equilibrium is an improper node.

See Phase plane behaviour case 4 for diagrams.

For other eigenvalue scenarios, see Distinct real eigenvalues case and Complex conjugate eigenvalues case. For algebraic linear-algebra context (eigenspaces, Jordan form), see your linear algebra notes.