Defective Matrices and Generalized Eigenvectors

When $𝐀$ has fewer than $n$ linearly independent eigenvectors, the eigenvector method comes up short and extra solutions must be built. The natural guess $t e^{λ_{1} t} 𝐯$ , copied from the repeated-root rule for single equations, does not work here. What does work is $e^{λ_{1} t} (𝐯 t + 𝐮)$ , where $𝐮$ is a generalized eigenvector satisfying $(𝐀 - λ_{1} 𝐈) 𝐮 = 𝐯$ .

Quick Reference

Item	Statement
Defective matrix	Fewer than $n$ linearly independent eigenvectors
Only possible when	Some eigenvalue is repeated; distinct eigenvalues always give independent eigenvectors
Double eigenvalue, one eigenvector	$𝐲^{(1)} = e^{λ_{1} t} 𝐯^{(1)}$ , $𝐲^{(2)} = e^{λ_{1} t} (𝐯^{(1)} t + 𝐮)$
Equation for $𝐮$	$(𝐀 - λ_{1} 𝐈) 𝐮 = 𝐯^{(1)}$
Triple eigenvalue, one eigenvector	Add $𝐲^{(3)} = e^{λ_{1} t} (\frac{1}{2} 𝐯^{(1)} t^{2} + 𝐮 t + 𝐰)$
Equation for $𝐰$	$(𝐀 - λ_{1} 𝐈) 𝐰 = 𝐮$
Generalized eigenvector property	${(𝐀 - λ_{1} 𝐈)}^{2} 𝐮 = 0$
Common warning	$t e^{λ_{1} t} 𝐯$ alone is not a solution

When Does a Matrix Become Defective?

This case can only happen if some eigenvalues of $𝐀$ are repeated, because eigenvectors corresponding to distinct eigenvalues are always linearly independent. So if $𝐀$ has $n$ distinct eigenvalues, it certainly has $n$ linearly independent eigenvectors.

Even if the $n \times n$ matrix $𝐀$ has repeated eigenvalues, it might still have $n$ linearly independent eigenvectors. For example, if $𝐀$ is symmetric or skew-symmetric, it has $n$ independent eigenvectors regardless of repetition. Repetition is necessary for defectiveness, not sufficient.

The relevant vocabulary: the algebraic multiplicity of an eigenvalue is how many times it appears as a root of $det (𝐀 - λ 𝐈) = 0$ , and its geometric multiplicity is the number of independent eigenvectors it has. A matrix is defective exactly when some eigenvalue has geometric multiplicity strictly less than its algebraic multiplicity.

The Solutions

1. If $λ_{1} = λ_{2}$ and only one linearly independent eigenvector $𝐯^{(1)}$ is associated with this eigenvalue, then two linearly independent solutions corresponding to it are

𝐲^{(1)} = e^{λ_{1} t} 𝐯^{(1)}

and

𝐲^{(2)} = e^{λ_{1} t} (𝐯^{(1)} t + 𝐮)

where $𝐮$ is a constant column vector determined by

(𝐀 - λ_{1} 𝐈) 𝐮 = 𝐯^{(1)} .

2. If

λ_{1} = λ_{2} = λ_{3}

with only one independent eigenvector, then three linearly independent solutions are

𝐲^{(1)} = e^{λ_{1} t} 𝐯^{(1)}, 𝐲^{(2)} = e^{λ_{1} t} (𝐯^{(1)} t + 𝐮),

𝐲^{(3)} = e^{λ_{1} t} (\frac{1}{2} 𝐯^{(1)} t^{2} + 𝐮 t + 𝐰),

where

𝐮

and

𝐰

satisfy

(𝐀 - λ_{1} 𝐈) 𝐮 = 𝐯^{(1)} and (𝐀 - λ_{1} 𝐈) 𝐰 = 𝐮 .

Derivation for the double eigenvalue

If $λ_{1} = λ_{2}$ and $𝐯^{(1)}$ is the only linearly independent eigenvector corresponding to $λ_{1}$ , one solution is obviously $𝐲^{(1)} = e^{λ_{1} t} 𝐯^{(1)}$ .

Experience with double roots in single equations may lead us to expect a second solution of the form $t e^{λ_{1} t} 𝐯^{(1)}$ . Unfortunately that is not enough. Substituting it gives

\frac{d}{d t} (t e^{λ_{1} t} 𝐯^{(1)}) = e^{λ_{1} t} 𝐯^{(1)} + λ_{1} t e^{λ_{1} t} 𝐯^{(1)}

while the right-hand side gives $𝐀 t e^{λ_{1} t} 𝐯^{(1)} = λ_{1} t e^{λ_{1} t} 𝐯^{(1)}$ . The leftover term $e^{λ_{1} t} 𝐯^{(1)}$ has nothing to cancel it. A constant vector must be added to absorb it, which is exactly what $𝐮$ does.

So look for a solution of the form

𝐲^{(2)} = e^{λ_{1} t} (𝐮 + 𝐯^{(1)} t),

where the constant vector $𝐮$ is found by substituting into \mathbf{y}' = \mathbf{A}\mathbf{y}. Differentiating by the product rule,

\frac{d}{d t} 𝐲^{(2)} = λ_{1} e^{λ_{1} t} (𝐮 + 𝐯^{(1)} t) + e^{λ_{1} t} 𝐯^{(1)} = 𝐀 e^{λ_{1} t} (𝐮 + 𝐯^{(1)} t) .

Since $𝐀 𝐯^{(1)} = λ_{1} 𝐯^{(1)}$ , the terms containing $t$ on both sides are both equal to $λ_{1} e^{λ_{1} t} 𝐯^{(1)} t$ and cancel. What remains is

λ_{1} e^{λ_{1} t} 𝐮 + e^{λ_{1} t} 𝐯^{(1)} = 𝐀 e^{λ_{1} t} 𝐮 .

Dividing both sides by $e^{λ_{1} t}$ and rearranging,

(𝐀 - λ_{1} 𝐈) 𝐮 = 𝐯^{(1)} .

Because $(𝐀 - λ_{1} 𝐈) 𝐯^{(1)} = 0$ , applying $(𝐀 - λ_{1} 𝐈)$ to both sides of this equation gives

{(𝐀 - λ_{1} 𝐈)}^{2} 𝐮 = 0 .

The vector $𝐮$ is called a generalized eigenvector of $𝐀$ : it is killed by the second power of $𝐀 - λ_{1} 𝐈$ though not by the first.

The general pattern. If there is only one linearly independent eigenvector $𝐯_{1}^{(1)}$ corresponding to an eigenvalue $λ_{1}$ of algebraic multiplicity $m$ , then $m$ linearly independent solutions are

\begin{aligned} \mathbf{y}^{(1)} &= e^{\lambda_1t}\mathbf{v}_1^{(1)} \\ \mathbf{y}^{(2)} &= e^{\lambda_1t}\left(\mathbf{v}_1^{(1)}t + \mathbf{v}_2^{(1)}\right) \\ \mathbf{y}^{(3)} &= e^{\lambda_1t}\left(\mathbf{v}_1^{(1)}\frac{t^{2}}{2} + \mathbf{v}_2^{(1)}t + \mathbf{v}_3^{(1)}\right) \\ &\ \ \vdots \\ \mathbf{y}^{(m)} &= e^{\lambda_1t}\left(\mathbf{v}_1^{(1)}\frac{t^{m-1}}{(m-1)!} + \mathbf{v}_2^{(1)}\frac{t^{m-2}}{(m-2)!} + \cdots + \mathbf{v}_{m-1}^{(1)}t + \mathbf{v}_m^{(1)}\right) \end{aligned}

where the vectors are obtained from the chain

\begin{aligned} \left(\mathbf{A} - \lambda\mathbf{I}\right)\mathbf{v}_1^{(1)} &= \mathbf{0} \\ \left(\mathbf{A} - \lambda\mathbf{I}\right)\mathbf{v}_2^{(1)} &= \mathbf{v}_1^{(1)} \\ \left(\mathbf{A} - \lambda\mathbf{I}\right)\mathbf{v}_3^{(1)} &= \mathbf{v}_2^{(1)} \\ &\ \ \vdots \\ \left(\mathbf{A} - \lambda\mathbf{I}\right)\mathbf{v}_m^{(1)} &= \mathbf{v}_{m-1}^{(1)} \end{aligned}

Each vector in the chain is the input for finding the next, and the factorials $\frac{t^{k}}{k!}$ appear because differentiating that term reproduces the previous one exactly.

Worked Example

Find the solution of the system

\left\{ \begin{aligned} y_1' &= 3y_1 \\ y_2' &= 2y_1 + 3y_2 \\ y_3' &= -2y_1 + 3y_3 \end{aligned} \right.

subject to $y_{1} (0) = - 1$ , $y_{2} (0) = 2$ , and $y_{3} (0) = 1$ .

Solution

In matrix form,

\mathbf{y}' = \mathbf{A}\mathbf{y}, \qquad \mathbf{A} = \begin{bmatrix} 3 & 0 & 0 \\ 2 & 3 & 0 \\ -2 & 0 & 3 \end{bmatrix} .

Step 1: eigenvalues. The matrix is lower triangular except for the zero in position $(2, 3)$ , and expanding the determinant gives

det (𝐀 - λ 𝐈) = | \begin{matrix} 3 - λ & 0 & 0 \\ 2 & 3 - λ & 0 \\ - 2 & 0 & 3 - λ \end{matrix} | = (3 - λ)^{3} = 0 .

Therefore $λ_{1} = λ_{2} = λ_{3} = 3$ , an eigenvalue of algebraic multiplicity $3$ .

Step 2: eigenvectors. Solve $(𝐀 - 3 𝐈) 𝐯 = 0$ with $𝐯 = (a, b, c)^{T}$ :

\underset{[\begin{matrix} 0 \\ 2 a \\ - 2 a \end{matrix}]}{\underset{⏟}{[\begin{matrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ - 2 & 0 & 0 \end{matrix}] [\begin{matrix} a \\ b \\ c \end{matrix}]}} = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]

This forces $a = 0$ , while $b$ and $c$ are arbitrary. Choosing $b = 1$ , $c = 0$ and then $b = 0$ , $c = 1$ gives two linearly independent eigenvectors

𝐯^{(1)} = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}], 𝐯^{(2)} = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] .

So the geometric multiplicity is $2$ while the algebraic multiplicity is $3$ : the matrix is defective, short by exactly one solution.

Two solutions are

𝐲^{(1)} = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] e^{3 t}, 𝐲^{(2)} = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] e^{3 t} .

Step 3: the third solution. It has the form

𝐲^{(3)} = (t 𝐯 + 𝐮) e^{3 t}

where $(𝐀 - 3 𝐈) 𝐮 = 𝐯$ and $𝐯$ is some linear combination of the two eigenvectors:

𝐯 = A 𝐯^{(1)} + B 𝐯^{(2)} = [\begin{matrix} 0 \\ A \\ B \end{matrix}] .

Not every combination will do. We must choose $A$ and $B$ so that the equation for $𝐮$ is solvable:

\underset{𝐀 - 3 𝐈}{\underset{⏟}{[\begin{matrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ - 2 & 0 & 0 \end{matrix}]}} [\begin{matrix} u_{1} \\ u_{2} \\ u_{3} \end{matrix}] = [\begin{matrix} 0 \\ A \\ B \end{matrix}] ⟹ [\begin{matrix} 0 \\ 2 u_{1} \\ - 2 u_{1} \end{matrix}] = [\begin{matrix} 0 \\ A \\ B \end{matrix}]

The left side always has its third component equal to minus its second, so we need $B = - A$ . Setting $A = 2$ and $B = - 2$ makes $u_{1} = 1$ , while $u_{2}$ and $u_{3}$ remain arbitrary. Choosing $u_{2} = u_{3} = 0$ ,

𝐲^{(3)} = (t [\begin{matrix} 0 \\ 2 \\ - 2 \end{matrix}] + [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]) e^{3 t} = [\begin{matrix} 1 \\ 2 t \\ - 2 t \end{matrix}] e^{3 t} .

Step 4: general solution.

\begin{aligned} \mathbf{y}(t) &= c_1\mathbf{y}^{(1)} + c_2\mathbf{y}^{(2)} + c_3\mathbf{y}^{(3)} \\ &= c_1\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}e^{3t} + c_2\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}e^{3t} + c_3\begin{bmatrix} 1 \\ 2t \\ -2t \end{bmatrix}e^{3t} \\ &= \begin{bmatrix} c_3e^{3t} \\ c_1e^{3t} + 2c_3te^{3t} \\ c_2e^{3t} - 2c_3te^{3t} \end{bmatrix} \end{aligned}

Step 5: initial conditions. At $t = 0$ the exponentials are $1$ and the terms with $t$ vanish, so

[\begin{matrix} - 1 \\ 2 \\ 1 \end{matrix}] = [\begin{matrix} c_{3} \\ c_{1} \\ c_{2} \end{matrix}] ⟹ c_{1} = 2, c_{2} = 1, c_{3} = - 1 .

Therefore

\begin{aligned} y_1(t) &= -e^{3t} \\ y_2(t) &= 2e^{3t} - 2te^{3t} \\ y_3(t) &= e^{3t} + 2te^{3t} \end{aligned}

Check. y_1' = -3e^{3t} = 3y_1. ✓ And y_2' = 6e^{3t} - 2e^{3t} - 6te^{3t} = 4e^{3t} - 6te^{3t}, while $2 y_{1} + 3 y_{2} = - 2 e^{3 t} + 6 e^{3 t} - 6 t e^{3 t} = 4 e^{3 t} - 6 t e^{3 t}$ . ✓

Exercises

Solve \mathbf{y}' = \begin{bmatrix} 3 & -1 \\ 1 & 1 \end{bmatrix}\mathbf{y}.

Solution

det [\begin{matrix} 3 - λ & - 1 \\ 1 & 1 - λ \end{matrix}] = (3 - λ) (1 - λ) + 1 = λ^{2} - 4 λ + 4 = (λ - 2)^{2},

so $λ = 2$ is a double eigenvalue.

For the eigenvector, $[\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}] 𝐯 = 0$ gives $v_{1} = v_{2}$ , so $𝐯 = (1, 1)^{T}$ and only one independent eigenvector exists. The matrix is defective.

For $𝐮$ , solve $[\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}] 𝐮 = [\begin{matrix} 1 \\ 1 \end{matrix}]$ , that is, $u_{1} - u_{2} = 1$ . Take $𝐮 = (1, 0)^{T}$ .

𝐲 = c_{1} e^{2 t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{2 t} (t [\begin{matrix} 1 \\ 1 \end{matrix}] + [\begin{matrix} 1 \\ 0 \end{matrix}]) = c_{1} e^{2 t} [\begin{matrix} 1 \\ 1 \end{matrix}] + c_{2} e^{2 t} [\begin{matrix} t + 1 \\ t \end{matrix}] .

Show that $𝐀 = [\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}]$ has a repeated eigenvalue but is not defective, and solve \mathbf{y}' = \mathbf{A}\mathbf{y}.

Solution

The characteristic polynomial is $(5 - λ)^{2}$ , so $λ = 5$ has algebraic multiplicity $2$ . But $𝐀 - 5 𝐈$ is the zero matrix, so every vector is an eigenvector and the geometric multiplicity is also $2$ . The matrix is non-defective.

𝐲 = c_{1} e^{5 t} [\begin{matrix} 1 \\ 0 \end{matrix}] + c_{2} e^{5 t} [\begin{matrix} 0 \\ 1 \end{matrix}],

that is, $y_{1} = c_{1} e^{5 t}$ and $y_{2} = c_{2} e^{5 t}$ . The system was uncoupled all along.

This confirms that repeated eigenvalues do not by themselves imply defectiveness.

Verify directly that $t e^{λ_{1} t} 𝐯$ alone is not a solution of \mathbf{y}' = \mathbf{A}\mathbf{y} when $𝐀 𝐯 = λ_{1} 𝐯$ .

Solution

Differentiating,

\frac{d}{d t} (t e^{λ_{1} t} 𝐯) = e^{λ_{1} t} 𝐯 + λ_{1} t e^{λ_{1} t} 𝐯 .

Applying $𝐀$ instead,

𝐀 (t e^{λ_{1} t} 𝐯) = t e^{λ_{1} t} 𝐀 𝐯 = λ_{1} t e^{λ_{1} t} 𝐯 .

The two differ by $e^{λ_{1} t} 𝐯$ , which is nonzero. So the trial function fails, and the discrepancy is exactly what the added term $e^{λ_{1} t} 𝐮$ is designed to cancel, giving the condition $(𝐀 - λ_{1} 𝐈) 𝐮 = 𝐯$ .

Frequently Asked Questions

What makes a matrix defective?

Having fewer independent eigenvectors than its size. Concretely, some eigenvalue's geometric multiplicity, the number of independent eigenvectors it has, is smaller than its algebraic multiplicity, the number of times it appears as a root of the characteristic polynomial.

Why does

t e^{λ t} 𝐯

fail here when

x e^{r x}

worked for single equations?

Differentiating $t e^{λ t} 𝐯$ produces an extra $e^{λ t} 𝐯$ that the right-hand side cannot match. In the scalar case the analogous leftover was absorbed by the fact that the characteristic polynomial and its derivative both vanished. Here it must be cancelled explicitly, which is what the constant vector $𝐮$ does.

What is a generalized eigenvector?

A vector $𝐮$ satisfying $(𝐀 - λ 𝐈) 𝐮 = 𝐯$ for an eigenvector $𝐯$ . Equivalently, ${(𝐀 - λ 𝐈)}^{2} 𝐮 = 0$ while $(𝐀 - λ 𝐈) 𝐮 \neq 0$ . It sits one rung above $𝐯$ on a chain that continues for higher multiplicities.

The equation for

𝐮

has no solution. What now?

Then the eigenvector you started with is the wrong one. When an eigenvalue has two or more independent eigenvectors but still falls short, only certain combinations $𝐯 = A 𝐯^{(1)} + B 𝐯^{(2)}$ put the right-hand side in the range of $𝐀 - λ 𝐈$ . Leave $A$ and $B$ symbolic and impose the solvability condition, as the worked example does.

𝐮

unique?

No. Any multiple of the eigenvector $𝐯$ may be added to $𝐮$ , since $(𝐀 - λ 𝐈) 𝐯 = 0$ . Different choices change $𝐲^{(2)}$ by a multiple of $𝐲^{(1)}$ , which is absorbed into the arbitrary constants. Pick the simplest, usually by setting free components to zero.

How is this related to Jordan form?

Directly. The chain $𝐯_{1}, 𝐯_{2}, \dots, 𝐯_{m}$ is a Jordan chain, and using it as a basis puts $𝐀$ into a Jordan block. The polynomial-times-exponential solutions listed here are exactly what the matrix exponential of a Jordan block produces. Nothing in this section requires that theory, but it explains why the pattern is what it is.

Defective Coefficient Matrices

Quick Reference

When Does a Matrix Become Defective?

The Solutions

Worked Example

Exercises

Frequently Asked Questions

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