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Worked example

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Eigenvalues of a triangular matrix

Use a characteristic polynomial and a trace check rather than assuming all matrix entries are eigenvalues.

Question

What are the eigenvalues of A=(250−1)A=\begin{pmatrix}2&5\\0&-1\end{pmatrix}?

  1. a.2 and −1
  2. b.5 and 0
  3. c.1 and −2
  4. d.2 and 5

Correct answer

A. 2 and −1

Full reasoning

  1. 1Form A−λI=(2−λ50−1−λ)A-\lambda I=\begin{pmatrix}2-\lambda&5\\0&-1-\lambda\end{pmatrix}.
  2. 2Its determinant is (2−λ)(−1−λ)−5(0)(2-\lambda)(-1-\lambda)-5(0).
  3. 3Set the determinant to zero: λ = 2 or λ = −1.
  4. 4Check the sum against the trace (1) and the product against the determinant (−2). The diagonal shortcut works because this matrix is triangular; it is not a general rule for arbitrary matrices.

Why each choice is right or wrong

Choice A

Correct. The zero below the diagonal makes the characteristic polynomial (2 − λ)(−1 − λ), with roots 2 and −1.

Choice B

These are the off-diagonal entries; they are not roots of the characteristic polynomial.

Choice C

These have the right product but the wrong sum, so they fail the trace check.

Choice D

The entry 5 couples coordinates but is not an eigenvalue here.

Related formula

Characteristic equation

det(A − λI) = 0

Assumptions: The determinant must be computed for the whole matrix; reading diagonal entries alone requires triangular structure.

Related topic and practice

Eigenvalues and DeterminantsLinear Algebra Domain GuideDeterminants of Scaled, Transposed, and Inverted MatricesCreate a free account to practice

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Sources

Original authored exercise; no database question ID and no protected or official exam-bank content. Specialist review pending.

  1. 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.
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Sources last checked 2026-09-13

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