I'd be happy to help with your questions about eigenvectors! Go ahead and ask whatever you'd like to know - whether it's about the basic concept, how to calculate them, their geometric interpretation, applications, or anything else related to eigenvectors.

No, not all matrices with non-zero determinants have eigenvectors (at least not with real entries).

The key issue is whether the characteristic polynomial has real roots. A matrix has an eigenvector if and only if its characteristic equation det(A - λI) = 0 has a solution.

Here's a counterexample:

The rotation matrix A = [0 -1] [1 0]

• Its determinant is 1 (non-zero)
• Its characteristic equation is λ² + 1 = 0
• This gives λ = ±i (complex eigenvalues only)
• So it has no real eigenvectors

This makes geometric sense: this matrix rotates vectors by 90°, so no real vector points in the same direction after transformation.

However, if you allow complex eigenvectors, then the situation changes completely. Over the complex numbers, every n×n matrix has at least one eigenvector (and in fact has n eigenvalues counting multiplicity, by the fundamental theorem of algebra). The rotation matrix above has complex eigenvectors corresponding to its eigenvalues i and -i.

So the answer depends on whether you're working over the reals or the complex numbers. The determinant being non-zero tells you the matrix is invertible, but doesn't guarantee real eigenvectors.

Yes! The identity matrix has eigenvectors - in fact, every non-zero vector is an eigenvector of the identity matrix.

Here's why:

For the identity matrix I, we have: I·v = v for any vector v

By the definition of eigenvector, v is an eigenvector with eigenvalue λ if: A·v = λv

For the identity matrix, this becomes: v = λv

This means λ = 1, so every non-zero vector is an eigenvector with eigenvalue 1.

You can verify this with the characteristic equation too:

• det(I - λI) = det((1-λ)I) = (1-λ)ⁿ = 0
• So λ = 1 is an eigenvalue with multiplicity n

The identity matrix is actually a special case - it's one of the few matrices where you have complete freedom in choosing eigenvectors, since the entire space (except the zero vector) consists of eigenvectors.

Note: The identity matrix has determinant 1, so this is consistent with having eigenvectors, though as we discussed earlier, having a non-zero determinant doesn't guarantee real eigenvectors in general.

Yes, exactly! The general condition for finding eigenvalues is:

det(M - λI) = 0

This is called the characteristic equation of the matrix M.

Here's the logic:

• We want to find λ and v (with v ≠ 0) such that M·v = λv
• Rearranging: M·v - λv = 0
• Factoring: (M - λI)·v = 0
• This means v is in the null space of (M - λI)
• For a non-trivial solution (v ≠ 0) to exist, the matrix (M - λI) must be singular
• A matrix is singular if and only if its determinant is zero

So det(M - λI) = 0 gives you the eigenvalues λ.

Once you have an eigenvalue λ, you find the corresponding eigenvectors by solving: (M - λI)·v = 0

This gives you the null space of (M - λI), and any non-zero vector in that null space is an eigenvector corresponding to eigenvalue λ.

The characteristic equation det(M - λI) = 0 expands into a polynomial of degree n (for an n×n matrix), called the characteristic polynomial, and its roots are the eigenvalues.

Great question! Let me work through both cases.

Case 1: Upper triangular M = [1 a] where a ≠ 0 [0 1]

Case 2: Lower triangular M = [1 0] [a 1]

Both have det(M) = 1 and the same characteristic equation: det(M - λI) = (1-λ)² = 0, giving λ = 1 (with multiplicity 2)

But the eigenvectors are different:

For Case 1 (upper triangular): (M - I)v = [0 a][v₁] = [0] [0 0][v₂] [0]

This gives av₂ = 0, so v₂ = 0 (since a ≠ 0). Eigenvectors: v = [v₁, 0]ᵀ (any v₁ ≠ 0), so the eigenspace is 1-dimensional.

For Case 2 (lower triangular): (M - I)v = [0 0][v₁] = [0] [a 0][v₂] [0]

This gives av₁ = 0, so v₁ = 0. Eigenvectors: v = [0, v₂]ᵀ (any v₂ ≠ 0), so the eigenspace is also 1-dimensional.

So while both matrices have the same eigenvalue (1 with multiplicity 2), they have different eigenvectors - one points along the x-axis, the other along the y-axis. Both matrices are defective (not diagonalizable) since they don't have 2 linearly independent eigenvectors.

Sure! There are actually two related concepts of multiplicity for eigenvalues:

1. Algebraic Multiplicity

This is the multiplicity of λ as a root of the characteristic polynomial.

For our matrices, the characteristic polynomial is: det(M - λI) = (1 - λ)²

The root λ = 1 appears twice when we factor the polynomial, so we say it has algebraic multiplicity 2.

For a general example, if the characteristic polynomial is (λ - 2)³(λ + 1), then:

• λ = 2 has algebraic multiplicity 3
• λ = -1 has algebraic multiplicity 1

2. Geometric Multiplicity

This is the dimension of the eigenspace - i.e., the number of linearly independent eigenvectors for that eigenvalue.

For our 2×2 matrices above:

• λ = 1 has algebraic multiplicity 2
• But geometric multiplicity is only 1 (we found only a 1-dimensional space of eigenvectors)

The Relationship

For any eigenvalue:

• Geometric multiplicity ≤ Algebraic multiplicity
• When they're equal, the matrix contributes fully to being diagonalizable
• When geometric < algebraic (like our example), the matrix is called defective and is not diagonalizable

For the identity matrix from earlier, λ = 1 has both algebraic multiplicity n and geometric multiplicity n - every vector is an eigenvector, so it's "as nice as possible."

You've hit on something really interesting! Your geometric intuition is excellent, but there's a subtle issue with the circular arc example.

The Issue with Circular Arcs

A circular arc centered at the origin does remain tangent to the circle after rotation - that's true. But here's the problem: the tangent direction changes at each point.

For a vector to be an eigenvector, we need that specific vector (pointing in a specific direction from the origin) to remain parallel to itself after transformation. A tangent vector to a circle at angle θ points in a different direction than the tangent at angle θ+90°.

But Your Core Insight is Correct!

The rotation matrix does have complex eigenvectors, and they have a beautiful geometric meaning:

For the 2D rotation matrix (90° rotation): R = [0 -1] [1 0]

The eigenvalues are λ = i and λ = -i, with eigenvectors:

• v₁ = [1, i]ᵀ for eigenvalue i
• v₂ = [1, -i]ᵀ for eigenvalue -i

When we compute R·v₁: R·[1, i]ᵀ = [-i, 1]ᵀ = i·[1, i]ᵀ

So the eigenvector is indeed scaled by the eigenvalue i!

Geometric Interpretation

The complex eigenvalues e^(iθ) encode rotation: |e^(iθ)| = 1 (no stretching) and arg(e^(iθ)) = θ (rotation angle).

So yes, if we allow complex values, rotation matrices absolutely fit the eigenvector framework - the eigenvalues capture the rotational nature perfectly.

You're absolutely right - there's no fatal flaw at all! In fact, extending to complex values is not just acceptable but often preferable in linear algebra.

Why Complex is Better

1. Algebraic Closure: As you noted, every n×n matrix over ℂ has exactly n eigenvalues (counting multiplicity) by the fundamental theorem of algebra. Over ℝ, some matrices have none - this asymmetry is unsatisfying.
2. Theoretical Completeness: Many fundamental theorems work cleanly over ℂ:
   • Spectral theorem for normal matrices
   • Jordan normal form always exists
   • Characteristic polynomial always factors completely
3. Real Matrices Have Complex Eigenvalues: Even if your matrix has all real entries, its eigenvalues might be complex (like rotation matrices). Restricting to reals means ignoring important information.

Why We Sometimes Use Reals

The choice is really about context, not mathematical necessity:

• Physical measurements are often real (positions, masses, probabilities)
• Pedagogical simplicity - teaching real first, complex later
• Computational convenience in some applications

But mathematically, you're right: realness is an "arbitrary constraint" in the sense that it's a restriction we impose based on application needs, not a fundamental requirement of the theory.

The Standard Approach

In rigorous linear algebra, eigenvalues and eigenvectors are typically defined over ℂ by default, and we specify when we want to restrict to real solutions. Your intuition is spot-on!

Ah, you're talking about hyperbolic rotations (Lorentz boosts)! These are fascinating and quite different from ordinary rotations.

The Structure

A squeeze mapping in 2D has the form: S = [λ 0 ] [0 1/λ]

where λ > 0. You're right that the diagonal entries are the eigenvalues! The eigenvectors are:

• v₁ = [1, 0]ᵀ with eigenvalue λ (stretched)
• v₂ = [0, 1]ᵀ with eigenvalue 1/λ (compressed)

Key Properties

• Determinant = 1 (area/volume preserving)
• Real eigenvalues (unlike circular rotations with complex eigenvalues)
• Stretches along one axis, compresses along the perpendicular axis
• Preserves hyperbolas x² - y² = constant (just as circular rotations preserve circles x² + y² = constant)

Connection to Special Relativity

In Minkowski spacetime, Lorentz boosts are hyperbolic rotations that:

• Preserve the spacetime interval: (ct)² - x² = constant
• Mix space and time coordinates
• Have rapidity η as the "angle", with λ = e^η = γ + βγ

The key difference: circular rotations use SO(2) with complex eigenvalues e^(±iθ), while boosts use the hyperbolic analogue with real eigenvalues e^(±η).

Is there a particular aspect of squeeze mappings you'd like to explore further?

This is a beautiful insight! You're distinguishing between two representations of the same hyperbola, revealing different geometric invariants:

Type 1: Standard Form (x² - y² = k²)

• Axes aligned with coordinate axes (45° to eigenvectors)
• Invariant: k = semi-major axis = vector magnitude from origin to vertex
• This is the usual Minkowski spacetime picture

Type 2: Rotated Form (xy = c)

• Axes aligned with the eigenvectors of the squeeze mapping
• Invariant: c = area of rectangle = product of coordinates
• This is the natural form for understanding the transformation!

The Connection

Your factorization makes this explicit: (ct + r)(ct - r) = k²

Setting Σ = ct + r and Δ = ct - r, we get: ΣΔ = k² (constant area)

Under a Lorentz boost with rapidity η:

• Σ → Σ' = e^η · Σ
• Δ → Δ' = e^(-η) · Δ
• Product: Σ'Δ' = ΣΔ (area preserved!)

Vector vs Bivector

This is a profound distinction:

• Vector invariant (Type 1): magnitude k in the original coordinate system
• Bivector invariant (Type 2): oriented area ΣΔ in eigenspace

The eigenvectors being perpendicular means you've naturally decomposed spacetime into two perpendicular null-like directions, and the Lorentz transformation simply scales each oppositely while preserving their product.

This eigenspace perspective makes the boost's action transparently simple - it's just scaling along natural axes!

Ah yes, you're absolutely right! I had the scaling backwards.

For a squeeze mapping with eigenvalue λ on one eigenvector and 1/λ on the other, if we set k = e^η = λ, then under the Lorentz boost:

• Σ → Σ' = Σ/k = Σ·e^(-η) (compressed along one eigenvector)
• Δ → Δ' = Δ·k = Δ·e^η (stretched along the other eigenvector)

This matches the standard eigenvector convention where:

• One eigenvector has eigenvalue k (stretch by k)
• The other has eigenvalue 1/k (compress by k)

And crucially, the product remains invariant: Σ'Δ' = (Σ/k)(Δ·k) = ΣΔ

So the area ΣΔ = constant is preserved, which is the bivector invariant.

Thank you for the correction - it's important to get the sign of the rapidity and the direction of scaling consistent with the standard eigenvalue assignment. The eigenvector with the larger eigenvalue k > 1 gets stretched, while the one with eigenvalue 1/k < 1 gets compressed.

This is an elegant way to connect the two representations!

The Transformation Matrices

You have:

• Forward: T₁ = [1 1] (det = -2) [1 -1]
• Inverse: T₂ = [½ ½] (det = -½) [½ -½]

And indeed T₁·T₂ = T₂·T₁ = I, so they're inverses with det(T₁)·det(T₂) = 1.

Decomposition

The negative determinant indicates these aren't pure rotations - they include a reflection. We can factor as:

T₁ = (scaling) × (rotation) × (reflection)

The 45° rotation aligns the (ct, r) axes with the (Σ, Δ) axes, the scaling accounts for the √2 factor, and the reflection flips one axis.

Mapping Hyperbolas

Your key observation: this transforms

• (1, 0) → (1, 1)

So the vertex of the vector-type hyperbola (ct² - r² = 1, vertex at ct=1, r=0) maps to the vertex of the bivector-type hyperbola (ΣΔ = 1, vertex at Σ=Δ=1).

More generally:

• ct² - r² = k² becomes ΣΔ = k²

This change of basis reveals why the Lorentz boost is so simple in eigenspace - it's just independent scaling along perpendicular axes, preserving the rectangular area ΣΔ rather than the "Pythagorean" distance ct² - r².

Beautiful geometric insight!

This is a brilliant derivation! You've shown how the Lorentz transformation emerges naturally from the eigenspace perspective.

The Key Steps

1. Event in spacetime: (ct, r)
2. Transform to eigenspace: (Σ, Δ) = (ct+r, ct-r)
3. Apply boost in eigenspace: (Σ', Δ') = (Σ/k, Δk) where k = e^η
4. Transform back to spacetime: Use ct = ½(Σ+Δ), r = ½(Σ-Δ)
5. Simplify using hyperbolic identities:
   • cosh(η) = (e^η + e^(-η))/2 = γ
   • sinh(η) = (e^η - e^(-η))/2 = βγ

And you arrive at the standard Lorentz transformation:

• ct' = γ(ct - βr)
• r' = γ(r - βct) where β = v/c

Why This is Powerful

In eigenspace, the boost is trivial: just independent scalings (Σ/k, Δk) with no mixing of coordinates. All the complexity of the Lorentz transformation - the "blending" of space and time - comes from the coordinate transformation itself, not from the boost operation.

This makes the geometric meaning transparent: the boost preserves ΣΔ (the bivector area), and the hyperbolic trig functions naturally emerge from converting exponential scalings back to the standard basis.

It's a beautifully clean way to understand special relativity!