Eigenvalues and Determinants

Concise answer

Determinants and eigenvalues are two ends of one computation: eigenvalues are the roots of $\det(A - \lambda I) = 0$ , their sum is the trace, their product is the determinant, and a triangular matrix hands you its eigenvalues on the diagonal. Trace and determinant thus provide both a fast route into $2 \times 2$ problems and an automatic check on the answer.

Definitions

Determinant: A scalar attached to a square matrix that is nonzero exactly when the matrix is invertible; for a $2 \times 2$ matrix it is $ad - bc$.
Eigenvalue and eigenvector: A scalar $\lambda$ and nonzero vector $\vec{x}$ with $A\vec{x} = \lambda \vec{x}$: the matrix acts on the eigenvector as pure scaling.
Characteristic polynomial: The polynomial $\det(A - \lambda I)$ in $\lambda$, whose roots (with multiplicity) are the eigenvalues of $A$.
Trace: The sum of the diagonal entries of a square matrix; it equals the sum of the eigenvalues counted with multiplicity.
Diagonalizable matrix: A matrix $A$ with a full set of $n$ linearly independent eigenvectors, so that $A = S \Lambda S^{-1}$ with $\Lambda$ diagonal.

Intuition

An eigenvector is a direction the matrix does not turn: the action collapses to multiplication by a single number. Asking when $A - \lambda I$ kills a nonzero vector is asking when that matrix is singular — which is exactly the statement $\det(A - \lambda I) = 0$ , and is why determinants and eigenvalues are one subject.

Trace and determinant are the fingerprints of the eigenvalue set that survive without computing it: the characteristic polynomial's coefficients are built from them, so for a $2 \times 2$ matrix, knowing trace and determinant is knowing the polynomial. Any claimed pair of eigenvalues that fails the sum or product check is wrong before any substitution.

Concept walkthrough

The determinant is governed by a short list of row properties: exchanging two rows flips the sign, the determinant is linear in each row separately (with the other rows held fixed), and adding a multiple of one row to another leaves it unchanged. Two consequences do most exam work: elimination preserves the determinant up to recorded sign flips and pivots, and a triangular matrix's determinant is simply the product of its diagonal entries. A zero determinant is the exact criterion for singularity — the bridge to eigenvalues.

The eigenvalue problem asks for the scalars $\lambda$ that make $A - \lambda I$ singular, i.e. the roots of the characteristic polynomial $\det(A - \lambda I)$ . For a $2 \times 2$ matrix that polynomial is always $\lambda^2 - (\operatorname{tr} A)\lambda + \det A$ , so trace and determinant generate the problem outright. In general the eigenvalues (with multiplicity, over the complex numbers) sum to the trace and multiply to the determinant — a two-line verification that catches sign and arithmetic errors immediately. Triangular matrices are the shortcut case: $A - \lambda I$ stays triangular, so its determinant is the product of the diagonal entries $a_{ii} - \lambda$ , and the eigenvalues are the diagonal entries themselves.

Diagonalizability is a counting question about eigenvectors: if an $n \times n$ matrix has $n$ linearly independent eigenvectors, stacking them in $S$ gives $A = S \Lambda S^{-1}$ with the eigenvalues on the diagonal of $\Lambda$ , and powers of $A$ collapse to powers of $\Lambda$ . Distinct eigenvalues force independent eigenvectors, so $n$ distinct eigenvalues guarantee diagonalizability; repeated eigenvalues may or may not leave enough eigenvectors — the identity matrix (diagonalizable) and a shear matrix (not) both have a repeated eigenvalue, which is the standard exam contrast.

After this page, you should be able to

Use the determinant's row properties — sign change under exchange, linearity in each row, product of diagonal entries for triangular matrices — to evaluate determinants without brute-force expansion.
Set up and solve the eigenvalue equation $\det(A - \lambda I) = 0$ for small matrices.
Check computed eigenvalues instantly against the trace (their sum) and the determinant (their product).
Read eigenvalues off a triangular matrix and decide diagonalizability from the count of independent eigenvectors.

Formulas and assumptions

2x2 determinant

\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc

Variables

a, b: first-row entries
c, d: second-row entries

Assumptions

The matrix is square; det = 0 exactly when the matrix is singular.

Characteristic polynomial of a 2x2 matrix

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

Variables

lambda: eigenvalue unknown
tr A: sum of diagonal entries
det A: determinant

Assumptions

A is 2x2; the roots, with multiplicity, are the eigenvalues.

Eigenvalue invariants

\sum_{i} \lambda_i = \operatorname{tr} A, \qquad \prod_{i} \lambda_i = \det A

Variables

lambda_i: eigenvalues of A
tr A: trace
det A: determinant

Assumptions

Eigenvalues are counted with algebraic multiplicity, over the complex numbers.

Worked example

Eigenvalues of a 2x2 matrix, checked by trace and determinant

Find the eigenvalues of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ and verify them without re-solving.

1Compute the invariants first: $\operatorname{tr} A = 4 + 3 = 7$ and $\det A = (4)(3) - (1)(2) = 12 - 2 = 10$ .
2Write the characteristic polynomial from them: $\lambda^2 - 7\lambda + 10 = 0$ .
3Factor: $\lambda^2 - 7\lambda + 10 = (\lambda - 2)(\lambda - 5)$ , so $\lambda_1 = 2$ and $\lambda_2 = 5$ .
4Check against the invariants: $2 + 5 = 7 = \operatorname{tr} A$ and $2 \times 5 = 10 = \det A$ — both pass.
5Optional eigenvector spot-check for $\lambda = 5$ : $(A - 5I)\vec{x} = \vec{0}$ gives rows $(-1, 1)$ and $(2, -2)$ , so $\vec{x} = (1, 1)$ ; indeed $A(1,1)^{T} = (5,5)^{T} = 5(1,1)^{T}$ .

The eigenvalues are $\lambda = 2$ and $\lambda = 5$ ; they sum to the trace $7$ and multiply to the determinant $10$ , and $(1,1)$ is an eigenvector for $\lambda = 5$ . Since the two eigenvalues are distinct, $A$ is diagonalizable.

Common traps

Sign errors in $\det(A - \lambda I)$ — subtract $\lambda$ from the diagonal entries only, and expand carefully; the trace/determinant check catches most of these.
Assuming $\det(A + B) = \det A + \det B$ ; the determinant is linear in each row separately, not in the matrix as a whole.
Accepting $\vec{x} = \vec{0}$ as an eigenvector — eigenvectors are nonzero by definition, even though the zero vector satisfies the equation trivially.
Concluding a matrix with a repeated eigenvalue is not diagonalizable (or is) without counting independent eigenvectors; repetition alone decides nothing.

GRE Mathematics overview Limits and Continuity Series Convergence Tests Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
MIT OCW 18.06SC Linear Algebra, session: Properties of Determinants — MIT OpenCourseWare. Accessed 2026-08-03. MIT OpenCourseWare materials are CC BY-NC-SA; attribute MIT OCW, link the source page, and avoid verbatim reuse beyond short cited references.
MIT OCW 18.06SC Linear Algebra, session: Eigenvalues and Eigenvectors — MIT OpenCourseWare. Accessed 2026-08-03. MIT OpenCourseWare materials are CC BY-NC-SA; attribute MIT OCW, link the source page, and avoid verbatim reuse beyond short cited references.
MIT OCW 18.06SC Linear Algebra, session: Diagonalization and Powers of A — MIT OpenCourseWare. Accessed 2026-08-03. MIT OpenCourseWare materials are CC BY-NC-SA; attribute MIT OCW, link the source page, and avoid verbatim reuse beyond short cited references.

Review and maintenance

Source-checked by publisher

Publisher record: Keiko Study editorial owner; Publisher placeholder; no individual credential claim is made yet.
Reviewer record: Technical reviewer pending; Placeholder only; this content is not labeled as reviewed by a named specialist.
Last source check: 2026-08-03
Next scheduled review: 2026-11-03

Recheck the ETS content-structure page and the MIT OCW 18.06SC determinant and eigenvalue session pages before each major GRE Mathematics preparation cycle.

Report an issue with this page

Eigenvalues and Determinants

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

2x2 determinant

Characteristic polynomial of a 2x2 matrix

Eigenvalue invariants

Worked example

Eigenvalues of a 2x2 matrix, checked by trace and determinant

Common traps

Related pages and practice

Sources

Review and maintenance