Eigenvalues, eigenvectors and multiplicity

Objectives and prerequisites

Find eigenvalues and eigenspaces, separate algebraic from geometric multiplicity, and use trace, determinant and polynomial transformations as checks. Prerequisites: determinants, homogeneous systems and rank. A is square. State the field: a real matrix may require complex eigenvalues and vectors. Unless a question says otherwise, the examples here use real scalars.

1. The defining equation

An eigenpair satisfies Av=λv with v≠0. The scalar λ is an eigenvalue; v is an eigenvector. The image stays on the same line, though a negative λ reverses direction and λ=0 sends the vector to zero. Zero is a valid eigenvalue, but the zero vector is never an eigenvector. Every nonzero scalar multiple of v is another eigenvector for the same λ.

Rewrite as (A−λI)v=0. A nonzero solution exists exactly when A−λI is singular. We use the monic characteristic polynomial p(t)=det(tI−A). The alternative det(A−tI) differs by (−1)ⁿ and has the same roots. After finding a root, solve a homogeneous system; the determinant alone does not give the eigenvectors.

2. A complete calculation

Let A=[[2,1],[0,3]]. Then p(t)=(t−2)(t−3), so the eigenvalues are 2 and 3. For λ=2, (A−2I)v=0 gives y=0; the eigenspace is span{(1,0)ᵀ}. Its eigenvectors are (c,0)ᵀ with c≠0. For λ=3, the equation is −x+y=0, so the eigenspace is span{(1,1)ᵀ}. Check directly: A(1,1)ᵀ=(3,3)ᵀ.

The eigenspace Eλ=ker(A−λI) includes zero and is a vector space; the collection of eigenvectors alone excludes zero and is not a vector space. Do not mix these two descriptions. For a triangular matrix, the determinant of tI−A is the product of its diagonal entries, so its eigenvalues are its diagonal entries, including repetitions. Its eigenvectors need not be coordinate vectors.

3. Repeated roots and two multiplicities

Algebraic multiplicity is the number of times λ occurs as a root of p(t). Geometric multiplicity is dim Eλ=n−rank(A−λI). For each eigenvalue, 1≤geometric multiplicity≤algebraic multiplicity. Repetition does not determine the number of independent eigenvectors.

For J=[[4,1],[0,4]], p(t)=(t−4)², but (J−4I)(x,y)ᵀ=(y,0)ᵀ. Thus y=0 and E₄ has dimension one although the algebraic multiplicity is two. In contrast, 4I₂ has the same characteristic polynomial but E₄ is the entire two-dimensional space. Every nonzero vector is its eigenvector. This comparison motivates the next lesson on diagonalization.

Eigenvectors belonging to distinct eigenvalues are linearly independent. For two vectors, if v₂=cv₁ with c≠0, applying A would force λ₂=λ₁. The general result follows by applying A−λI to a proposed dependence and reducing the number of terms. Multiple eigenvectors for one eigenvalue may be dependent.

4. Fast invariants and transformations

Over C, count all n eigenvalues with algebraic multiplicity. Their sum is tr(A), the sum of diagonal entries, and their product is det(A). Hence A is invertible exactly when zero is not an eigenvalue. These checks can reject an arithmetic error, but trace and determinant alone do not determine the spectrum of a general matrix of size three or more.

If Av=λv, then Aᵏv=λᵏv for positive integer k; (A+cI)v=(λ+c)v; and for any polynomial f, f(A)v=f(λ)v. If A is invertible, A⁻¹v=λ⁻¹v. The same v works in each statement. For the example with eigenvalues 2 and 3, A²−3A+I has corresponding eigenvalues −1 and 1. Do not add eigenvalues of unrelated matrices to infer eigenvalues of their sum; a shared eigenvector is needed for that step.

A and Aᵀ have the same characteristic polynomial because determinants are unchanged by transpose. Their eigenvectors can differ. Similar matrices P⁻¹AP also share the characteristic polynomial, since det(tI−P⁻¹AP)=det(P⁻¹(tI−A)P)=det(tI−A).

5. The field matters

For R=[[0,−1],[1,0]], p(t)=t²+1. Over R there are no eigenvalues or nonzero eigenvectors; over C the eigenvalues are i and −i. A real matrix is not guaranteed real eigenvalues. Real symmetric matrices do have real eigenvalues, a special result used in the next lesson.

Exam workflow and pitfalls

Write p(t), factor it with multiplicities, solve each (A−λI)v=0, and specify a basis for each eigenspace. Check trace and determinant and substitute a representative eigenvector. Never include zero as an eigenvector, treat multiplicity as dimension without solving, or divide by a parameter without checking its zero case. These methods work without assuming that the matrix is diagonalizable.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

LE01. Which condition is required for v to be an eigenvector of A?

A. v=0 B. v≠0 and Av=λv for some scalar λ C. Av=0 only D. All coordinates of v are nonzero

Answer: B

A: The zero vector is excluded by definition.

B: An eigenvector is a nonzero vector whose image is a scalar multiple of itself.

C: Av=0 also admits v=0, which is not an eigenvector. Only its nonzero solutions are eigenvectors for eigenvalue zero; eigenvectors may also have other eigenvalues.

D: Some coordinates may be zero; the whole vector must be nonzero.

LE02. For A=[[2,1],[0,3]], which vector is an eigenvector for λ=3?

A. (1,0)ᵀ B. (0,1)ᵀ C. (1,1)ᵀ D. (1,−1)ᵀ

Answer: C

A: Its image is twice itself, so its eigenvalue is 2.

B: Its image is (1,3)ᵀ, not (0,3)ᵀ.

C: A(1,1)ᵀ=(3,3)ᵀ=3(1,1)ᵀ.

D: Its image is (1,−3)ᵀ, not (3,−3)ᵀ.

LE03. A 3×3 matrix has eigenvalues 2,2,−1 counted with algebraic multiplicity. What is det(A)?

A. 3 B. −4 C. 4 D. −1

Answer: B

A: 3 is the trace, the sum of eigenvalues.

B: The determinant is their product: 2×2×(−1)=−4.

C: This loses the negative sign.

D: This ignores the two factors of 2.

LE04. For J=[[4,1],[0,4]], what is dim ker(J−4I)?

A. 0 B. 1 C. 2 D. 4

Answer: B

A: The equation has nonzero solutions.

B: (J−4I)(x,y)ᵀ=(y,0)ᵀ=0 forces y=0; x is free.

C: Algebraic multiplicity two does not imply two independent eigenvectors.

D: An eigenspace in two coordinates cannot have dimension four.

LE05. If Av=2v with v≠0, what is (A²−3A+I)v?

A. −v B. 0 C. v D. 3v

Answer: A

A: Substitute the eigenvalue into the polynomial: 4−6+1=−1.

B: The polynomial does not vanish at 2.

C: The sign of 4−6+1 is negative.

D: This does not evaluate the polynomial at 2 correctly.

LE06. For real A=[[0,−1],[1,0]], which statement is correct?

A. It has real eigenvalues 1 and −1 B. It has eigenvalue 0 C. It has no eigenvalues over C D. It has no real eigenvalues, but complex eigenvalues i and −i

Answer: D

A: Its characteristic polynomial is t²+1, not t²−1.

B: The determinant is 1, so zero is not an eigenvalue.

C: Over C, t²+1 has roots i and −i.

D: The equation t²+1=0 has no real root and exactly those two complex roots.

Source and scope

Directly supports the official eigenvalues and eigenvectors heading. Multiplicity and polynomial transformations develop that heading; all practice is original.

Official GATE 2026 CS syllabus (verified 2026-10-14).