Eigenvalues and eigenvectors

At a Glance

• Frequency: 10 sub-parts across 10 of 14 years (2013, 2014, 2015, 2016, 2017, 2021, 2022, 2024, 2025, 2026)
• Priority tier: T1
• Marks (count): 10 (2), 12 (2), 15 (1), 20 (1), 8 (3)
• Average solve time: ~9 min
• Difficulty mix: medium 6, easy 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Eigenvalues and eigenvectors appear in 10 of the last 14 years — a perfect run, placing this atom among the highest-frequency items in Paper 1. Questions split roughly 6:3 between “find eigenvalues and eigenvectors of a given matrix” (pure computation) and “prove a theorem about eigenvalues” (conceptual). The computation questions follow a single repeatable template; the theorem questions each have a one-paragraph proof. The 2024 paper raised the stakes with a 20-mark two-part question that asked for eigenvectors of $A^{-15}$ — demonstrating that the “spectral inheritance” rule ($A^k$ has the same eigenvectors as $A$, eigenvalues $\lambda^k$) must be memorised explicitly.

Minimum Theory

Characteristic polynomial. For an $n\times n$ matrix $A$, the eigenvalues are the roots of the characteristic polynomial $p(\lambda)=\det(A-\lambda I)=0$. For a $3\times 3$ matrix this is a cubic; factor by trying integer divisors of the constant term. The algebraic multiplicity of $\lambda$ is its multiplicity as a root of $p$; the geometric multiplicity is $\dim\ker(A-\lambda I)=n-\operatorname{rank}(A-\lambda I)$. Always: geometric $\le$ algebraic multiplicity. $A$ is diagonalisable iff for every eigenvalue, geometric = algebraic multiplicity.

Key spectral facts:

• $\sum_i \lambda_i = \operatorname{tr}(A)$ and $\prod_i \lambda_i = \det(A)$ — use these to cross-check every computation.
• Real symmetric matrices have real eigenvalues and orthogonal eigenvectors for distinct eigenvalues.
• Hermitian matrices ($A^* = A$) have real eigenvalues (proof: $\bar\lambda\|x\|^2 = \overline{x^*Ax} = x^*A^*x = x^*Ax = \lambda\|x\|^2$, so $\bar\lambda = \lambda$).
• If $Av = \lambda v$ and $A$ is invertible, then $A^{-1}v = \lambda^{-1}v$; more generally $A^kv = \lambda^k v$ for any integer $k$.
• Eigenvectors for distinct eigenvalues are linearly independent.

Rotation in $\mathbb R^2$. The rotation matrix $T_\theta=\bigl(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\bigr)$ has characteristic polynomial $\lambda^2-2\cos\theta\,\lambda+1$. Discriminant $= 4\cos^2\theta-4 = -4\sin^2\theta \le 0$; real eigenvalues only when $\sin\theta=0$ (i.e. $\theta=0$ or $\pi$). For $\theta=\pi/2$: $p(\lambda)=\lambda^2+1$, eigenvalues $\pm i$.

Question Archetypes

Four patterns cover every eigenvalue question in the corpus.

eigen-computation (6 question(s); 2014, 2015, 2016, 2021, 2024, 2025)

Recognition Cues

• “Find the eigenvalues and eigenvectors of $A$.”
• “Hence find the eigenvalues (or eigenvectors) of $A^{-k}$, $A^k$, or $p(A)$” — the “hence” signals the spectral-inheritance rule; do not redo the computation from scratch.
• A $3\times3$ integer matrix; or a $2\times2$ complex (Hermitian) matrix.
• A block-diagonal or clearly structured matrix (factor the characteristic polynomial by inspection).

Solution Template

1. Write $\det(A-\lambda I)$ and expand (for $3\times3$: cofactor along the most sparse row/column).
2. Factor the cubic: find one integer root by testing divisors of the constant term; then divide to get a quadratic.
3. Cross-check: trace $=\sum\lambda_i$ and determinant $=\prod\lambda_i$.
4. For each $\lambda$: row-reduce $(A-\lambda I)$ to echelon form; identify the null space basis. Verify algebraic = geometric multiplicity at any repeated eigenvalue.
5. If the question asks for $A^{-k}$: state “eigenvectors of $A$ and $A^{-k}$ are identical; eigenvalues of $A^{-k}$ are $\lambda^{-k}$.” Do not re-compute.

Worked Example(s)

2014 Paper 1, 2014-P1-Q3c-i (8 marks)

$A=\bigl(\begin{smallmatrix}-2&2&-3\\2&1&-6\\-1&-2&0\end{smallmatrix}\bigr)$. Find eigenvalues and eigenvectors.

Characteristic polynomial. Expanding $\det(A-\lambda I)$ along row 1: $p(\lambda) = (\lambda+3)[-(\lambda+2)(\lambda-4)+7] = (\lambda+3)(\lambda+3)(\lambda-5) \cdot (-1)$ $\Rightarrow p(\lambda)=-(\lambda+3)^2(\lambda-5).$

Eigenvalues: $\lambda=5$ (simple) and $\lambda=-3$ (algebraic multiplicity 2).

Eigenvector for $\lambda=5$. Row-reduce $A-5I$: pivot equations yield $y=-2z$, $x=-z$. Take $z=-1$: $\boxed{\;v_1=(1,2,-1)\text{ for }\lambda=5.\;}$

Eigenvectors for $\lambda=-3$. $(A+3I)$ row-reduces to a single equation $x+2y-3z=0$ (rank 1 ⇒ geometric multiplicity 2). Two free variables: $\boxed{\;v_2=(-2,1,0),\quad v_3=(3,0,1)\text{ for }\lambda=-3.\;}$

Check: trace $=5+(-3)+(-3)=-1=-2+1+0$ ✓; det $=5\cdot9=45$: verify directly ✓.

2015 Paper 1, 2015-P1-Q2c (12 marks)

$A=\bigl(\begin{smallmatrix}1&1&3\\1&5&1\\3&1&1\end{smallmatrix}\bigr)$ (symmetric). Find eigenvalues and eigenvectors.

Characteristic polynomial. Expanding and collecting (note the $\lambda$ coefficient cancels — a reliable check): $p(\lambda) = -\lambda^3+7\lambda^2-36 = 0 \;\Longrightarrow\; \lambda^3-7\lambda^2+36=0.$

Trial root $\lambda=-2$: $-8-28+36=0$ ✓. Factor: $(\lambda+2)(\lambda-3)(\lambda-6)=0$.

Eigenvalues: $\lambda=-2,3,6$ (all distinct; trace $=7$ ✓).

Eigenvectors (each by 2-equation row reduction): $v_1=(1,0,-1)\ (\lambda=-2),\quad v_2=(1,-1,1)\ (\lambda=3),\quad v_3=(1,2,1)\ (\lambda=6).$

Orthogonality check ($A$ symmetric, distinct $\lambda$): $v_1\cdot v_2=0$, $v_1\cdot v_3=0$, $v_2\cdot v_3=0$ ✓.

2016 Paper 1, 2016-P1-Q2b-i (8 marks)

$A=\bigl(\begin{smallmatrix}1&1&0\\1&1&0\\0&0&1\end{smallmatrix}\bigr)$. Find eigenvalues and eigenvectors.

Block structure. $A$ is block-diagonal: $\bigl(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\bigr)\oplus[1]$. Characteristic polynomial: $p(\lambda)=[(1-\lambda)^2-1]\cdot(1-\lambda)=\lambda(\lambda-2)(1-\lambda).$

Eigenvalues: $\lambda=0,1,2$.

Eigenvectors: $\lambda=0$: $x+y=0,z=0$ → $(1,-1,0)$. $\lambda=1$: $x=y=0,z$ free → $(0,0,1)$. $\lambda=2$: $x=y,z=0$ → $(1,1,0)$.

2021 Paper 1, 2021-P1-Q4a-ii (10 marks)

$A=\bigl(\begin{smallmatrix}0&-i\\i&0\end{smallmatrix}\bigr)$ over $\mathbb C$. Find eigenvalues and eigenvectors.

Characteristic polynomial. $\det(A-\lambda I)=\lambda^2-(-i)(i)=\lambda^2-1$ (since $-i\cdot i=-i^2=1$). Eigenvalues: $\lambda=\pm1$ (real, as expected for a Hermitian matrix — this is the Pauli matrix $\sigma_y$).

Eigenvectors. For $\lambda=1$: $-v_1-iv_2=0\Rightarrow v_1=-iv_2$; take $v_2=1$: $v_1=(1,i)^T$. For $\lambda=-1$: $v_1-iv_2=0$; $v_2=(i,1)^T$.

Verify: $A(1,i)^T=(-i\cdot i,\,i\cdot 1)^T=(1,i)^T$ ✓. Orthogonality: $(1,-i)\cdot(i,1)^T=i-i=0$ ✓.

2024 Paper 1, 2024-P1-Q4a (20 marks)

$A=\bigl(\begin{smallmatrix}3&2&4\\2&0&2\\4&2&3\end{smallmatrix}\bigr)$. Find eigenvalues and eigenvectors of $A$; hence find those of $A^{-15}$.

Characteristic polynomial. Expanding $\det(A-\lambda I)$: $\lambda^3-6\lambda^2-15\lambda-8=0 \;\Longrightarrow\; (\lambda+1)^2(\lambda-8)=0.$

Eigenvalues: $\lambda=8$ (simple), $\lambda=-1$ (algebraic multiplicity 2). Trace $=6=8-1-1$ ✓; det $=8=8\cdot1\cdot1$ ✓.

Eigenvectors. For $\lambda=8$: row-reduce $A-8I$; get $v_1=2v_2$, $v_3=2v_2$ → $(2,1,2)^T$. For $\lambda=-1$: all rows of $A+I$ collapse to $2v_1+v_2+2v_3=0$ (rank 1, geometric multiplicity 2); basis $(1,-2,0)^T$ and $(0,-2,1)^T$.

Eigenvalues of $A^{-15}$. Eigenvectors are unchanged. Eigenvalues become $\lambda^{-15}$: $8^{-15}=2^{-45}$ for $(2,1,2)^T$ and $(-1)^{-15}=-1$ for $(1,-2,0)^T$ and $(0,-2,1)^T$.

$\boxed{\;\text{Same eigenvectors as }A;\quad \text{eigenvalues }2^{-45}\text{ and }-1\ (\text{mult. }2).\;}$

2025 Paper 1, 2025-P1-Q4c-i (12 marks)

$A=\bigl(\begin{smallmatrix}1&2&0\\2&1&-6\\2&-2&3\end{smallmatrix}\bigr)$. Find eigenvalues and eigenvectors.

Characteristic polynomial. Expanding along row 1 (third entry is 0 — one cofactor vanishes): $p(\lambda) = -\lambda^3+5\lambda^2+9\lambda-45 = 0 \;\Longrightarrow\; \lambda^3-5\lambda^2-9\lambda+45=0.$

Group: $\lambda^2(\lambda-5)-9(\lambda-5)=(\lambda-5)(\lambda^2-9)=(\lambda-5)(\lambda-3)(\lambda+3)=0$.

Eigenvalues: $\lambda=3,5,-3$.

Eigenvectors. For $\lambda=3$: $y=x$, $z=0$ → $(1,1,0)^T$. For $\lambda=5$: $y=2x$, $z=-x$ → $(1,2,-1)^T$. For $\lambda=-3$: $y=-2x$, $z=-x$ → $(1,-2,-1)^T$.

Common Traps

• Always verify the trace and determinant after computing eigenvalues — this catches arithmetic errors with almost no extra work.
• At a repeated eigenvalue, compute the rank of $(A-\lambda I)$ explicitly to confirm geometric multiplicity. If rank is $n-1$, nullity is 1 (not 2 even if algebraic multiplicity is 2) and $A$ is defective.
• The “hence” keyword in questions about $A^{-k}$ means inherit the eigenstructure: state “if $Av=\lambda v$ then $A^{-1}v=\lambda^{-1}v$” in one sentence, then read off the new eigenvalues. Never redo the characteristic polynomial.
• For complex/Hermitian matrices: the characteristic polynomial has real eigenvalues; the eigenvectors are complex. Check by substituting back: $Av=\lambda v$.

eigenvalue-bound (1 question(s); 2013)

Recognition Cues

• A matrix built from roots of unity ($\omega$, $\omega^2$, $1+\omega+\omega^2=0$) or a DFT-type Vandermonde structure.
• The question asks to “show $|\lambda_1|+|\lambda_2|+|\lambda_3| \le K$” rather than to compute eigenvalues explicitly.
• Key signal: you are meant to find $A^*A$ (or $AA^*$), recognise it as a scalar multiple of $I$, and invoke the spectral theorem for normal matrices.

Solution Template

1. Compute $A^*A$ entry-by-entry (dot products of rows of $\overline A$ with columns of $A$); use $1+\omega+\omega^2=0$ and $|\omega|=1$ to collapse each entry to zero (off-diagonal) or $n$ (diagonal).
2. Conclude $A^*A = cI$ (scalar matrix); singular values of $A$ are all $\sqrt c$.
3. For $A^2$: $(A^2)^*(A^2) = A^{**}A^*AA = A^*(cI)A = c\,A^*A = c^2 I$, so singular values of $A^2$ are all $c$.
4. $A^2$ is normal ($A^*A = AA^* \Rightarrow (A^2)^*A^2 = A^2(A^2)^*$). For normal matrices the spectral theorem gives $|\lambda_i| = \sigma_i$. Therefore each $|\lambda_i(A^2)| = c$, and the sum is $nc \le K$.

Worked Example(s)

2013 Paper 1, 2013-P1-Q2b-i (8 marks)

$A=\bigl(\begin{smallmatrix}1&1&1\\1&\omega^2&\omega\\1&\omega&\omega^2\end{smallmatrix}\bigr)$, $\omega\ne1$, $\omega^3=1$. Show $|\lambda_1|+|\lambda_2|+|\lambda_3|\le 9$.

Compute $A^*A$. Note $A$ is symmetric so $A^* = \overline A$. The $(i,j)$-entry of $\overline A \cdot A$ is the inner product of row $i$ of $\overline A$ with column $j$ of $A$. All off-diagonal entries involve $1+\omega+\omega^2=0$ or $1+\omega^3+\omega^3=3$ — using $\omega^3=1$, off-diagonal entries vanish and diagonal entries equal 3: $A^*A = 3I.$

Singular values of $A^2$. $(A^2)^*(A^2) = A^*(A^*A)A = A^*(3I)A = 3A^*A = 9I$. So singular values of $A^2$ are all $3$.

Spectral theorem. $A^2$ is normal (the same calculation shows $AA^*=3I$ as well, so $A^*A=AA^*$, and thus $(A^2)^*A^2=A^2(A^2)^*=9I$). For a normal matrix, $|\lambda_i|=\sigma_i$, so $|\lambda_i(A^2)|=3$ for all $i$. $|\lambda_1|+|\lambda_2|+|\lambda_3|=9\le 9.\ \blacksquare$

The bound is attained exactly (not strict).

Common Traps

• $A$ is symmetric, not Hermitian: $A^* = \overline A \ne A$ (entries are complex). So $A^*A\ne A^2$.
• The identity $1+\omega+\omega^2=0$ applies to any primitive cube root of unity $\omega\ne1$; use it for every off-diagonal computation.
• The inequality $\le 9$ looks like a strict bound but equals $9$ — the question is technically asking you to demonstrate equality.

eigenvector-independence-proof (1 question(s); 2017)

Recognition Cues

• “Prove that eigenvectors corresponding to distinct eigenvalues are linearly independent.”
• Can also appear as “show the eigenvector matrix is non-singular” for a Hermitian matrix with distinct eigenvalues.
• The proof is pure linear algebra; no matrix arithmetic required.

Solution Template

Proof by induction on the number of eigenvectors $k$:

1. Base case ($k=1$): a single non-zero vector is trivially independent.
2. Inductive step: assume any $k-1$ eigenvectors for distinct eigenvalues are independent. Suppose $\sum_{i=1}^k c_i v_i = \mathbf0$.
3. Apply $A$: $\sum c_i\lambda_i v_i = \mathbf0$. Subtract $\lambda_k$ times the original relation: $\sum_{i=1}^{k-1}c_i(\lambda_i-\lambda_k)v_i=\mathbf0$.
4. By hypothesis, $c_i(\lambda_i-\lambda_k)=0$ for $i<k$. Since $\lambda_i\ne\lambda_k$, we get $c_i=0$ for $i<k$. Then $c_kv_k=\mathbf0$ and $v_k\ne\mathbf0$ gives $c_k=0$.

Worked Example(s)

2017 Paper 1, 2017-P1-Q3b (10 marks)

Prove distinct non-zero eigenvectors of a matrix are linearly independent.

Setup. The correct statement: eigenvectors $v_1,\ldots,v_k$ belonging to distinct eigenvalues $\lambda_1,\ldots,\lambda_k$ are linearly independent. State this clearly — the problem’s wording is slightly imprecise.

Suppose $c_1v_1+\cdots+c_kv_k=\mathbf0$. $\qquad(1)$

Apply $A$: $c_1\lambda_1v_1+\cdots+c_k\lambda_kv_k=\mathbf0.\qquad(2)$

Subtract $\lambda_k\times$(1) from (2): $\sum_{i=1}^{k-1}c_i(\lambda_i-\lambda_k)v_i=\mathbf0.$

By the inductive hypothesis (applied to $v_1,\ldots,v_{k-1}$ for $k-1$ distinct eigenvalues): each $c_i(\lambda_i-\lambda_k)=0$. Since $\lambda_i\ne\lambda_k$, we get $c_i=0$ for all $i<k$. Then (1) gives $c_kv_k=\mathbf0$, and $v_k\ne\mathbf0$ forces $c_k=0$.

$\boxed{\;\text{All }c_i=0\text{; hence }v_1,\ldots,v_k\text{ are linearly independent.}\ \blacksquare\;}$

Common Traps

• The hypothesis is distinct eigenvalues, not just distinct vectors. Two eigenvectors for the same eigenvalue can be dependent. State the exact hypothesis at the start.
• After showing $c_1=\cdots=c_{k-1}=0$, an explicit appeal to $v_k\ne\mathbf 0$ is needed to conclude $c_k=0$ — examiners check this step.

no-real-eigenvalue (1 question(s); 2022)

Recognition Cues

• A rotation matrix $T_\theta$ (or a $90°$ rotation) and the question asks to “show it has no real eigenvalue.”
• Or a skew-symmetric/anti-Hermitian operator and the question asks about the nature of eigenvalues.

Solution Template

1. Write the standard rotation matrix $T_\theta = \bigl(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\bigr)$.
2. Compute the characteristic polynomial $p(\lambda)=\lambda^2-2\cos\theta\,\lambda+1$.
3. Discriminant $= 4\cos^2\theta-4=-4\sin^2\theta$. Real eigenvalues iff $\sin\theta=0$ (i.e. $\theta=0$ or $\pi$).
4. For $\theta=\pi/2$: $p(\lambda)=\lambda^2+1=0$, roots $\lambda=\pm i$. No real solutions.
5. Geometric conclusion: a $90°$ rotation maps every vector to a perpendicular direction; no non-zero vector can be parallel to its image, so no real eigenvector exists.

Worked Example(s)

2022 Paper 1, 2022-P1-Q4a (15 marks)

Rotation $T:\mathbb R^2\to\mathbb R^2$ by angle $\theta$; find the linear map; show $T_{\pi/2}$ has no real eigenvalue.

Linear map. The rotation by $\theta$ counter-clockwise acts as: $T_\theta(x,y) = (x\cos\theta-y\sin\theta,\ x\sin\theta+y\cos\theta),\qquad T_\theta=\begin{pmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{pmatrix}.$

Linearity: matrix multiplication is linear. Verify: $T_\theta(1,0)^T=(\cos\theta,\sin\theta)^T$ ✓.

No real eigenvalue at $\theta=\pi/2$. $T_{\pi/2}=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad p(\lambda)=\lambda^2+1.$ $\lambda^2=-1$ has no real solutions. $\boxed{\;T_{\pi/2}\text{ has eigenvalues }\pm i;\text{ no real eigenvalue.}\ \blacksquare\;}$

Common Traps

• A real eigenvalue $\lambda$ would require a non-zero vector $v$ with $T_\theta v = \lambda v$, i.e. a fixed direction (up to scaling). A rotation by $\theta\ne 0,\pi$ can never fix any real direction; state this as an intuition check.
• Real eigenvalues of $T_\theta$ exist only at $\theta=0$ ($\lambda=1$, identity) and $\theta=\pi$ ($\lambda=-1$, reversal). The discriminant formula $-4\sin^2\theta\le 0$ makes this precise.

Marks-Aware Writing

8-mark questions (2013-Q2b-i, 2014-Q3c-i, 2016-Q2b-i): Three steps — characteristic polynomial, eigenvalues (with trace/det check), eigenvectors. Write the row-reduction briefly; one sentence per pivot. No lengthy justification.

10–12-mark questions (2021-Q4a-ii, 2025-Q4c-i, 2017-Q3b): For computation: same three steps but show the cubic factorisation explicitly. For proofs: write the induction setup in one sentence, then the two displayed equations and the conclusion.

15-mark questions (2022-Q4a): Two parts — state the matrix (with linearity verified), then the characteristic polynomial and conclusion. Geometric interpretation earns the last method mark.

20-mark questions (2024-Q4a): Full computation for $A$, then a short bridge paragraph stating the spectral-inheritance rule, then a table of $\lambda^{-15}$ values and corresponding eigenvectors. The bridge paragraph is worth 4–6 marks; do not skip it.

Practice Set