Eigenvalues and Eigenvectors (AI HL)
For a square matrix \(A\), an eigenvector is a non-zero vector \(\mathbf v\) whose direction is unchanged by \(A\) - it only gets stretched or shrunk by a scale factor \(\lambda\), the eigenvalue. This page covers how to find both for \(2\times2\) matrices, with worked examples and the mistakes that lose the most marks. It's part of the broader Matrices topic.
20 questions on this sub-topic.
Finding eigenvalues and eigenvectors
Covered under IB syllabus reference AHL1.15, restricted to \(2\times2\) matrices with distinct real eigenvalues. Neither formula below is in the formula booklet - they're definitions you apply directly.
Characteristic equation
\(\det(A - \lambda I) = 0\)
Expand this into a quadratic in \(\lambda\); its two roots are the eigenvalues of \(A\). A diagonal matrix is the one shortcut - its eigenvalues are simply its diagonal entries.
Eigenvector for a given \(\lambda\)
\((A - \lambda I)\mathbf v = \mathbf 0\)
Substitute one eigenvalue at a time and solve the resulting equations for \(\mathbf v\). Any non-zero scalar multiple of the answer is an equally valid eigenvector.
Need the full syllabus wording for matrices in general? See Matrices. GDC steps for eigenvalues live in the Matrices GDC section.
Worked examples
\(A = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}.\)
(a)(i) Find the eigenvalue with \(\lambda<3.5\).
(a)(ii) Find the eigenvalue with \(\lambda>3.5.\)
Worked solution
Solve \(\det(A - \lambda I) = 0\): \((4-\lambda)(3-\lambda) - 2 = 0.\) M1
\(\lambda^2 - 7\lambda + 10 = 0\) M1
\((\lambda-5)(\lambda-2) = 0\) A1
\(\Rightarrow \lambda = 5\) or \(2.\) A1 A1
For \(A = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}\) with eigenvalues \(5\) and \(2\), find an eigenvector for each.
(a)(i) Find an eigenvector for eigenvalue \(\lambda=5\).
(a)(ii) Find an eigenvector for eigenvalue \(\lambda=2\).
Worked solution
(a)(i) \(\lambda = 5\): \((A - 5I)\mathbf v = \mathbf 0\) M1
\(\Rightarrow -v_1 + v_2 = 0.\) A1
Eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}.\) A1
(a)(ii) \(\lambda = 2\): \((A - 2I)\mathbf v = \mathbf 0\) M1
\(\Rightarrow 2v_1 + v_2 = 0.\) A1
Eigenvector \(\begin{pmatrix}1\\-2\end{pmatrix}.\) A1
Common mistakes
- Forgetting to subtract \(\lambda I\) from the whole matrix. Only the diagonal entries of \(A\) change when forming \(A - \lambda I\) - the off-diagonal entries stay exactly as they were.
- Stopping after finding the eigenvalues. A question asking for eigenvectors needs you to substitute each \(\lambda\) back into \((A-\lambda I)\mathbf v = \mathbf 0\) and solve - the eigenvalues alone aren't the full answer.
- Assuming an eigenvector answer is unique. Any non-zero scalar multiple of a valid eigenvector, such as \(\begin{pmatrix}2\\2\end{pmatrix}\) instead of \(\begin{pmatrix}1\\1\end{pmatrix}\), is equally correct - don't second-guess a "different" answer that's really the same direction.
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20 eigenvalue and eigenvector questions, marked instantly like the real exam.
Quick answers
How do you find the eigenvalues of a 2x2 matrix?
Solve the characteristic equation \(\det(A-\lambda I) = 0\). This expands to a quadratic in \(\lambda\); its two roots are the eigenvalues.
How do you find an eigenvector once you know an eigenvalue?
Substitute the eigenvalue into \((A-\lambda I)\mathbf v = \mathbf 0\) and solve the resulting equations for \(\mathbf v\). Any non-zero scalar multiple of the resulting vector is a valid eigenvector.