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Eigenvalues

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1. Core idea

An eigenvector of a square matrix $A$ is a nonzero vector $v$ whose direction is unchanged by the linear transformation defined by $A$.

It satisfies

$$ Av = \lambda v $$

where $\lambda$ is the corresponding eigenvalue.

Interpretation:

  • $A$ acts on most vectors by changing both magnitude and direction.

  • On an eigenvector, $A$ only scales the vector by $\lambda$.

  • If $\lambda < 0$, the vector is reversed as well as scaled.

  • If $\lambda = 0$, the eigenvector is sent to the zero vector.

Eigenvalues only make sense for square matrices.

Geometric meaning

For a $2 \times 2$ or $3 \times 3$ matrix, eigenvectors identify invariant directions or invariant lines. These are the directions that survive the transformation without rotating away.

Examples:

  • A stretch in the $x$-direction has eigenvectors on the coordinate axes.

  • A projection matrix has eigenvalue $1$ on the subspace preserved by the projection and eigenvalue $0$ on the directions collapsed.

  • A pure rotation in $\mathbb{R}^2$ usually has no real eigenvectors unless the angle is $0$ or $\pi$.

Interactive visual

Eigenvector directions

Use a symmetric matrix so the real eigenvectors stay visible as the transformation changes.

Eigenvalue 1 1.87
Eigenvalue 2 -0.12

2. Characteristic equation

Starting from

$$ Av = \lambda v $$

move everything to one side:

$$ (A - \lambda I)v = 0 $$

For a nonzero solution $v$ to exist, the matrix $A - \lambda I$ must be singular. Therefore

$$ \det(A - \lambda I) = 0 $$

This is the characteristic equation. Its left-hand side is the characteristic polynomial.

Why the determinant condition works

The homogeneous system

$$ (A - \lambda I)v = 0 $$

has a nontrivial solution exactly when the coefficient matrix is not invertible. That is equivalent to determinant zero.

Example

Let

$$ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} $$

Then

$$ A - \lambda I = \begin{bmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{bmatrix} $$

and

$$ \det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 $$

So

$$ \lambda^2 - 7\lambda + 10 = 0 $$

which factors as

$$ (\lambda - 5)(\lambda - 2) = 0 $$

Hence the eigenvalues are

$$ \lambda = 5,\quad \lambda = 2 $$

3. Eigenvectors and eigenspaces

Once an eigenvalue $\lambda$ is known, find its eigenvectors by solving

$$ (A - \lambda I)v = 0 $$

The set of all eigenvectors for $\lambda$, together with the zero vector, forms the eigenspace

$$ E_\lambda = \operatorname{Null}(A - \lambda I) $$

This is a subspace of the vector space.

Example continued

For

$$ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} $$

and $\lambda = 5$,

$$ A - 5I = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} $$

Solve

$$ \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

This gives

$$ y = x $$

so one eigenvector is

$$ v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

For $\lambda = 2$,

$$ A - 2I = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} $$

which gives

$$ 2x + y = 0 $$

so one eigenvector is

$$ v_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix} $$

Any nonzero scalar multiple of an eigenvector is still an eigenvector for the same eigenvalue.

Important restriction

The zero vector is never called an eigenvector. If it were allowed, every scalar would become an eigenvalue candidate, which would destroy the definition.

4. Multiplicity

Eigenvalues can repeat.

Algebraic multiplicity

The algebraic multiplicity of $\lambda$ is its multiplicity as a root of the characteristic polynomial.

Geometric multiplicity

The geometric multiplicity of $\lambda$ is

$$ \dim(E_\lambda) = \dim(\operatorname{Null}(A - \lambda I)) $$

Fundamental inequality

For each eigenvalue,

$$ 1 \le \text{geometric multiplicity} \le \text{algebraic multiplicity} $$

Repeated eigenvalue example

Consider

$$ A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix} $$

Then

$$ \det(A - \lambda I) = (3-\lambda)^2 $$

So $\lambda = 3$ has algebraic multiplicity $2$.

But

$$ A - 3I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$

and solving

$$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

gives

$$ y = 0 $$

So the eigenspace is

$$ \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\} $$

which has geometric multiplicity $1$.

This matrix has too few linearly independent eigenvectors to diagonalize.

5. Determinant, trace, and invertibility

Eigenvalues encode several global properties of a matrix.

Determinant

For an $n \times n$ matrix,

$$ \det(A) = \lambda_1 \lambda_2 \cdots \lambda_n $$

counting algebraic multiplicity and allowing complex eigenvalues when needed.

Trace

The trace is the sum of diagonal entries:

$$ \operatorname{tr}(A) = a_{11} + a_{22} + \cdots + a_{nn} $$

It also equals the sum of the eigenvalues:

$$ \operatorname{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n $$

Invertibility test

A matrix $A$ is invertible if and only if $0$ is not an eigenvalue.

Reason:

  • $0$ is an eigenvalue exactly when $Av = 0$ for some nonzero $v$.

  • That means $\operatorname{Null}(A)$ is nontrivial.

  • So $A$ is singular.

Triangular matrices

If $A$ is upper or lower triangular, its eigenvalues are the diagonal entries.

This is one of the fastest ways to read off eigenvalues by inspection.

6. Diagonalization

A matrix $A$ is diagonalizable if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that

$$ A = PDP^{-1} $$

The columns of $P$ are eigenvectors of $A$, and the corresponding diagonal entries of $D$ are the eigenvalues.

Equivalently,

$$ P^{-1}AP = D $$

Why diagonalization matters

Diagonal matrices are easy to work with:

$$ D^k = \begin{bmatrix} \lambda_1^k & 0 & \cdots & 0 \\ 0 & \lambda_2^k & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n^k \end{bmatrix} $$

So if

$$ A = PDP^{-1} $$

then

$$ A^k = PD^kP^{-1} $$

This is useful in differential equations, dynamical systems, and repeated matrix multiplication.

Diagonalizability criterion

An $n \times n$ matrix is diagonalizable if it has $n$ linearly independent eigenvectors.

Sufficient conditions:

  • $A$ has $n$ distinct eigenvalues.

  • $A$ is real symmetric.

  • For each eigenvalue, geometric multiplicity equals algebraic multiplicity, and the total number of independent eigenvectors is $n$.

Example

Let

$$ A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} $$

with eigenpairs

$$ \lambda_1 = 5,\quad v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

and

$$ \lambda_2 = 2,\quad v_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix} $$

Then

$$ P = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}, \qquad D = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix} $$

and

$$ A = PDP^{-1} $$

7. Special matrix classes

Symmetric matrices

If $A$ is real symmetric, then:

  • All eigenvalues are real.

  • Eigenvectors for distinct eigenvalues are orthogonal.

  • $A$ is diagonalizable by an orthogonal matrix.

This gives the spectral theorem:

$$ A = QDQ^T $$

where $Q$ is orthogonal and $D$ is diagonal.

This is stronger and more stable than ordinary diagonalization.

Projection matrices

If $P$ satisfies

$$ P^2 = P $$

then its eigenvalues can only be

$$ 0 \quad \text{or} \quad 1 $$

Reason:

$$ Pv = \lambda v \quad \Rightarrow \quad P^2v = \lambda^2 v $$

but also

$$ P^2v = Pv = \lambda v $$

so

$$ \lambda^2 = \lambda $$

and therefore

$$ \lambda(\lambda - 1) = 0 $$

Nilpotent matrices

If

$$ A^k = 0 $$

for some positive integer $k$, then every eigenvalue of $A$ is $0$.

Reason:

$$ Av = \lambda v \quad \Rightarrow \quad A^k v = \lambda^k v $$

Since $A^k v = 0$ and $v \ne 0$, we must have

$$ \lambda^k = 0 $$

so $\lambda = 0$.

Orthogonal matrices

If $Q^TQ = I$, then over $\mathbb{C}$ every eigenvalue satisfies

$$ |\lambda| = 1 $$

For real orthogonal matrices, eigenvalues may be real or complex. Real eigenvalues can only be

$$ 1 \quad \text{or} \quad -1 $$

8. Complex eigenvalues and real matrices

Not every real matrix has real eigenvalues.

Example:

$$ A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $$

This is rotation by $90^\circ$ in $\mathbb{R}^2$.

Its characteristic equation is

$$ \det(A - \lambda I) = \begin{vmatrix} -\lambda & -1 \\ 1 & -\lambda \end{vmatrix} = \lambda^2 + 1 = 0 $$

So the eigenvalues are

$$ \lambda = i,\quad \lambda = -i $$

There are no real eigenvectors because no nonzero real vector stays on its own line under a $90^\circ$ rotation.

Conjugate pairs

If a real matrix has a complex eigenvalue

$$ \lambda = a + bi $$

then its complex conjugate

$$ \bar{\lambda} = a - bi $$

is also an eigenvalue.

This happens because the characteristic polynomial has real coefficients.

9. How to compute eigenvalues by hand

For a general matrix

  1. Form $A - \lambda I$.

  2. Compute $\det(A - \lambda I)$.

  3. Solve the characteristic equation.

  4. For each eigenvalue, solve $(A - \lambda I)v = 0$.

  5. Check how many independent eigenvectors you obtained.

For a $2 \times 2$ matrix

If

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

then

$$ \det(A - \lambda I) = \lambda^2 - (a+d)\lambda + (ad-bc) $$

So the eigenvalues satisfy

$$ \lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = 0 $$

This is worth memorizing.

Sanity checks

After solving, verify:

  • The sum of eigenvalues equals $\operatorname{tr}(A)$.

  • The product of eigenvalues equals $\det(A)$.

  • Repeated roots require an eigenspace check before claiming diagonalizability.

Common computational shortcut

For triangular matrices, do not expand a full determinant. Read the eigenvalues directly from the diagonal.

10. Applications

Dynamical systems

For

$$ x_{k+1} = Ax_k $$

the eigenvalues of $A$ control long-term behavior.

  • If $|\lambda| < 1$, the associated mode decays.

  • If $|\lambda| > 1$, the associated mode grows.

  • If $|\lambda| = 1$, the mode is neutral in magnitude and may oscillate.

Differential equations

For systems such as

$$ x'(t) = Ax(t) $$

eigenvalues describe growth, decay, and oscillation. Diagonalization can reduce the system to decoupled scalar equations.

Principal directions

In quadratic forms and symmetric matrices, eigenvectors give principal directions and eigenvalues give the stretch or curvature in those directions.

This underlies:

  • Principal component analysis

  • Normal modes in mechanics

  • Stress and strain analysis

  • Conic sections and quadric surfaces

Markov chains

For a stochastic matrix, $\lambda = 1$ is especially important. The corresponding eigenvectors help describe steady states and long-run distributions.

11. Problem-solving workflow

Use this checklist for most eigenvalue problems.

Step 1: Confirm the matrix is square

If $A$ is not square, standard eigenvalues are not defined.

Step 2: Form the characteristic polynomial

Compute

$$ \det(A - \lambda I) $$

and solve for $\lambda$.

Step 3: Use structure before brute force

Check whether the matrix is:

  • Triangular

  • Diagonal

  • Symmetric

  • A projection

  • Nilpotent

  • Block structured

These often reveal eigenvalues immediately.

Step 4: Find eigenspaces

For each eigenvalue, solve

$$ (A - \lambda I)v = 0 $$

using row reduction.

Step 5: Compare multiplicities

If an eigenvalue repeats, check whether its eigenspace has enough dimension.

Step 6: Decide whether diagonalization is possible

Count the number of linearly independent eigenvectors.

Step 7: Verify with trace and determinant

Use

$$ \sum \lambda_i = \operatorname{tr}(A), \qquad \prod \lambda_i = \det(A) $$

to catch arithmetic mistakes.

12. Formula sheet

Definitions

$$ Av = \lambda v,\qquad v \ne 0 $$
$$ (A - \lambda I)v = 0 $$
$$ \det(A - \lambda I) = 0 $$
$$ E_\lambda = \operatorname{Null}(A - \lambda I) $$

Multiplicity

$$ 1 \le \dim(E_\lambda) \le \text{algebraic multiplicity of } \lambda $$

Determinant and trace

$$ \det(A) = \lambda_1 \lambda_2 \cdots \lambda_n $$
$$ \operatorname{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n $$

Invertibility

$$ A \text{ invertible } \iff 0 \text{ is not an eigenvalue} $$

Diagonalization

$$ A = PDP^{-1} $$
$$ A \text{ diagonalizable } \iff A \text{ has } n \text{ linearly independent eigenvectors} $$
$$ A^k = PD^kP^{-1} $$

Real symmetric matrices

$$ A = QDQ^T $$

with $Q$ orthogonal and $D$ diagonal.

Special cases

For triangular $A$, eigenvalues are the diagonal entries.

For

$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

the characteristic polynomial is

$$ \lambda^2 - (a+d)\lambda + (ad-bc) $$

Common mistakes to avoid

  • Forgetting that eigenvalues are defined for square matrices.

  • Solving $\det(A)=0$ instead of $\det(A-\lambda I)=0$.

  • Calling the zero vector an eigenvector.

  • Finding eigenvalues but not computing eigenvectors.

  • Assuming a repeated eigenvalue guarantees enough eigenvectors to diagonalize.

  • Forgetting that complex eigenvalues can occur for real matrices.

  • Mixing up algebraic multiplicity and geometric multiplicity.

  • Ignoring trace and determinant checks after solving the characteristic equation.

  • Claiming $A = PDP^{-1}$ without putting eigenvectors in the columns of $P$ in the same order as the eigenvalues in $D$.

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