Source: Friedberg, Insel, Spence – Linear Algebra, 4th Ed.
Tags: eigenvalue, eigenvector, characteristic polynomial, diagonalizable, linear operator, characteristic value, proper value, proper vector
Difficulty: Intermediate | Prerequisites: Vector spaces, linear transformations (Ch. 2), determinants (Ch. 4), matrix representations of linear operators.
This section opens Chapter 5 (Diagonalization) by asking a clean question: given a linear operator T on a finite-dimensional vector space V, can we find a basis for V that makes the matrix representation of T diagonal? Diagonal matrices are trivial to work with (powers, inverses, system-solving all reduce to scalar arithmetic), so a "yes" here unlocks enormous practical value. The answer hinges on two new objects: eigenvalues and eigenvectors. If you have not yet internalised determinants and the null space, revisit Chapter 4 and Section 2.1 before continuing.
A linear operator is diagonalizable precisely when there exists a basis of eigenvectors. To find eigenvalues, solve det(A - tI) = 0 (the characteristic polynomial). To find eigenvectors for a given eigenvalue λ, solve the homogeneous system (A - λI)x = 0. Together, these two steps answer both "can we diagonalize?" and "how?"
Diagonalizable (operator)
A linear operator T on a finite-dimensional vector space V is diagonalizable if there exists an ordered basis β for V such that [T]_β is a diagonal matrix. A square matrix A is diagonalizable if L_A is diagonalizable.
Think of it as: you can choose coordinates so that T just scales each basis direction independently, with no mixing between components.
Eigenvector
A nonzero vector v in V such that T(v) = λv for some scalar λ. Equivalently, T sends v to a scalar multiple of itself.
In simple terms, this means v's direction is preserved (or exactly reversed) by T. The key word is nonzero: the zero vector is never an eigenvector.
Eigenvalue
The scalar λ corresponding to an eigenvector v, satisfying T(v) = λv. Also called a characteristic value or proper value.
Think of it as: the "stretching factor" that T applies along the direction of v.
Characteristic polynomial
For A in M_{n×n}(F), the polynomial f(t) = det(A - tI_n). Its zeros are precisely the eigenvalues of A.
In simple terms, this is the polynomial you solve to find all eigenvalues. It is always degree n with leading coefficient (-1)^n.
Given a linear operator T on V (dim V = n), the question is whether there exists an ordered basis β = {v_1, ..., v_n} so that [T]_β is diagonal.
If D = [T]β is diagonal, then T(v_j) = D{jj} v_j for each basis vector v_j. Each basis vector is simply scaled.
Conversely, if you can find n linearly independent vectors each scaled by T, you have a diagonal representation.
T is diagonalizable if and only if there exists an ordered basis β for V consisting of eigenvectors of T. When this holds, the diagonal entries of [T]_β are the corresponding eigenvalues.
A scalar λ is an eigenvalue of A if and only if det(A - λI) = 0.
The reasoning:
λ is an eigenvalue ⇔ there exists nonzero v with Av = λv
⇔ (A - λI)v = 0 has a nontrivial solution
⇔ A - λI is singular
⇔ det(A - λI) = 0
The characteristic polynomial of an n × n matrix is degree n with leading coefficient (-1)^n.
Consequence: A has at most n distinct eigenvalues.
A vector v is an eigenvector of T corresponding to λ if and only if v ≠ 0 and v ∈ N(T - λI).
In practice: for each eigenvalue λ, row-reduce A - λI and describe the null space. The nonzero vectors in that null space are the eigenvectors for λ.
For a matrix A:
Compute det(A - tI) and factor it to find all eigenvalues λ_1, λ_2, ...
For each λ_i, form B_i = A - λ_i I.
Solve B_i x = 0 to find the eigenvectors (null space of B_i).
If the union of eigenvector bases forms a basis for the whole space, A is diagonalizable.
If β is a basis for V and A = [T]_β, then v is an eigenvector of T corresponding to λ if and only if φ_β(v) (the coordinate vector of v relative to β) is an eigenvector of A corresponding to λ. This reduces the operator problem to a matrix problem.
If β is a basis of eigenvectors for F^n and Q is the matrix whose columns are the vectors in β, then Q^{-1}AQ = D (diagonal). The diagonal entries of D are the eigenvalues, listed in the same order as the columns of Q.
Let v be an eigenvector with eigenvalue λ, and W = span({v}).
λ > 1: T stretches vectors in W away from 0
λ = 1: T acts as the identity on W
0 < λ < 1: T contracts vectors in W toward 0
λ = 0: T collapses W to the origin
λ < 0: T reverses orientation on W (flips through 0)
The rotation operator T_θ on R^2 (for 0 < θ < π) has characteristic polynomial t^2 - 2cos(θ)t + 1, whose discriminant 4cos^2(θ) - 4 is negative. No real eigenvalues exist, so T_θ is not diagonalizable over R.
Characteristic polynomial: f(t) = det(A - tI_n)
Eigenvalue equation: Av = λv, equivalently (A - λI)v = 0
Diagonalization relationship: Q^{-1}AQ = D, where Q = [v_1 | v_2 | ... | v_n] (eigenvectors as columns)
Eigenvalue analysis underpins vibration analysis in mechanical engineering (natural frequencies are eigenvalues of stiffness/mass matrices), Google's PageRank algorithm, principal component analysis in statistics, and stability analysis of dynamical systems. Any time you need to understand how a linear process behaves over many iterations, eigenvalues are the tool.
"Every n × n matrix has n distinct eigenvalues." False. Eigenvalues can repeat (algebraic multiplicity > 1), and over R some eigenvalues may not exist at all. A 2×2 real rotation matrix has zero real eigenvalues.
"The zero vector is an eigenvector." It is not. Eigenvectors are defined to be nonzero. The zero vector satisfies Av = λv trivially for every λ, which would make the concept useless.
"If A has an eigenvalue, it is diagonalizable." Having eigenvalues is necessary but not sufficient. You need enough linearly independent eigenvectors to form a full basis.
"Eigenvectors corresponding to different eigenvalues can be the same vector." They cannot. If v were an eigenvector for both λ_1 and λ_2 with λ_1 ≠ λ_2, then λ_1 v = λ_2 v, giving (λ_1 - λ_2)v = 0, which forces v = 0. But eigenvectors are nonzero.
⚠️ The two-step process (find eigenvalues via the characteristic polynomial, then find eigenvectors via the null space) is almost certainly on the exam. Be fluent with both steps.
⚠️ Know Theorem 5.1 cold: diagonalizability ⇔ basis of eigenvectors. This is the conceptual backbone of the chapter.
⚠️ The relationship Q^{-1}AQ = D is a standard exam computation. Practise forming Q from eigenvectors and verifying the result.
⚠️ Similar matrices share the same characteristic polynomial (Exercise 12). This is used to define the characteristic polynomial of an operator independently of the basis choice.
True or false: Every linear operator on an n-dimensional vector space has n distinct eigenvalues. A: False. It may have fewer, or (over R) none at all.
True or false: If a real matrix has one eigenvector, then it has infinitely many eigenvectors. A: True. Any nonzero scalar multiple of an eigenvector is also an eigenvector.
True or false: Similar matrices always have the same eigenvectors. A: False. They have the same eigenvalues, but generally different eigenvectors.
Fill in the blank: λ is an eigenvalue of A if and only if det(______) = 0. A: det(A - λI) = 0.
Q: Find the eigenvalues of A = [[1, 1], [4, 1]].
A: det(A - tI) = (1-t)^2 - 4 = t^2 - 2t - 3 = (t-3)(t+1). Eigenvalues are λ = 3 and λ = -1.
Q: For the matrix A above, find the eigenvectors corresponding to each eigenvalue.
A: For λ = 3: solve (A - 3I)x = 0, giving [[-2, 1],[4, -2]]x = 0. Solutions: t[1, 2]^T, so eigenvectors are nonzero multiples of (1, 2). For λ = -1: solve (A + I)x = 0, giving [[2, 1],[4, 2]]x = 0. Solutions: t[1, -2]^T, so eigenvectors are nonzero multiples of (1, -2).
Q: Is the rotation of R^2 by π/2 diagonalizable over R? Explain.
A: No. The rotation maps every nonzero vector to a non-collinear vector, so no nonzero vector satisfies T(v) = λv for any real λ. No eigenvectors exist, hence no basis of eigenvectors exists.
Q: Let T on P_2(R) be defined by T(f(x)) = f(x) + (x+1)f'(x), with standard basis β. The matrix representation is A = [[1,1,0],[0,2,2],[0,0,3]]. Find all eigenvalues.
A: det(A - tI) = (1-t)(2-t)(3-t). Eigenvalues are λ = 1, 2, 3.
Q: Explain why det(T - tI) is well-defined for a linear operator T (i.e. independent of the choice of basis).
A: If β and γ are two ordered bases for V, then [T]_β and [T]_γ are similar matrices, and similar matrices have the same characteristic polynomial.
This section sets up Section 5.2 (Diagonalizability), which gives the complete test for when an operator can be diagonalized. The eigenvalue concept reappears in Section 5.3 (Matrix Limits and Markov Chains), where the condition that all eigenvalues lie inside or on the unit circle determines whether A^m converges. Eigenvalues also connect to determinants (det(A) = product of eigenvalues for triangular matrices) and to the trace (tr(A) = sum of eigenvalues, see Section 5.2 exercises).
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