Eigenvalues and Eigenvectors

Most directions get rotated and stretched into completely new directions when you multiply by a matrix — but for almost every matrix there are a few special directions that survive unturned, merely stretched or squashed (and possibly flipped) along their own line. Those directions are the eigenvectors, and the stretch factor along each is the eigenvalue. This is arguably the deepest idea in the matrices course and one of the most powerful in all of mathematics: eigentheory is how we diagonalise matrices (Lesson 9), solve coupled differential equations and recurrence relations, analyse the stability of dynamical systems, run Google's PageRank, and find the natural frequencies of a vibrating structure. The defining relation is disarmingly simple — $A\mathbf{v} = \lambda\mathbf{v}$ — but unpacking it gives a complete computational procedure (characteristic equation for the $\lambda$ 's; a homogeneous system for the $\mathbf{v}$ 's) and a rich geometric meaning (eigenvectors are the invariant lines through the origin from Lesson 6). This lesson builds all of it rigorously, with full $2\times2$ and $3\times3$ worked examples, the trace/determinant checks, the treatment of complex and repeated eigenvalues, and the exam technique examiners reward.

1. Where this sits in AQA 7367

Eigenvalues and eigenvectors are compulsory pure content for Papers 1 and 2, and they are the gateway to diagonalisation (Lesson 9) and the Cayley–Hamilton theorem (Lesson 10). Forming and solving the characteristic equation is AO1; using the trace/determinant relations as checks, or interpreting an eigenvector as an invariant line, is AO2; and multi-step problems — a $3\times3$ eigenproblem, a parameter that makes a matrix singular, a proof about eigenvalues of $A^2$ — are AO3. The topic draws directly on solving homogeneous systems (Lesson 5: $(A-\lambda I)\mathbf{v}=\mathbf{0}$ has non-trivial solutions exactly when $\det(A-\lambda I)=0$ ), on determinants (Lesson 3), and on the invariant-line idea (Lesson 6). Examiners particularly like questions that combine these strands — for instance, finding the eigenvalues of a matrix in terms of a parameter, then determining the value of that parameter for which the matrix is singular (a zero eigenvalue), or which makes two eigenvalues coincide (a repeated root, i.e. a zero discriminant of the characteristic quadratic). Recognising those triggers — "singular" means a zero eigenvalue, "repeated eigenvalue" means a zero discriminant — turns an apparently hard AO3 question into routine algebra.

2. Definitions and the geometric meaning

For a square matrix $A$ , a non-zero vector $\mathbf{v}$ is an eigenvector with eigenvalue $\lambda$ (a scalar) if

\boxed{\;A\mathbf{v} = \lambda\mathbf{v},\quad \mathbf{v}\neq\mathbf{0}.\;}

The condition $\mathbf{v}\neq\mathbf{0}$ is essential: $A\mathbf{0}=\lambda\mathbf{0}$ holds trivially for every $\lambda$ , so the zero vector is excluded by definition. Geometrically, $A$ maps $\mathbf{v}$ to a scalar multiple of itself — same line through the origin, scaled by $\lambda$ . So the line $\operatorname{span}\{\mathbf{v}\}$ is an invariant line of the transformation (exactly the invariant lines of Lesson 6), and $\lambda$ is the factor by which lengths along it are multiplied:

$\lambda > 1$ : stretched in the same direction; $0<\lambda<1$ : shrunk; $\lambda = 1$ : the direction is fixed (a line of invariant points);
$\lambda < 0$ : the direction is reversed (and scaled by $\lvert\lambda\rvert$ ); $\lambda = 0$ : the whole line is collapsed to the origin (so $A$ is singular).

This is the unifying picture: the eigenvectors mark out the "skeleton" of the transformation — the handful of directions along which the matrix acts as a pure scaling. Every other vector is a combination of eigenvectors, and the matrix acts on it by scaling each eigen-component independently. That is precisely why eigenvectors make matrix powers easy (Lesson 9) and why they reveal the long-term behaviour of repeated transformations: iterate $A$ many times and the eigen-component with the largest $\lvert\lambda\rvert$ dominates, which is the mathematics behind population models, Markov chains and PageRank.

The diagram below shows a matrix with eigenvectors along $\binom{1}{1}$ (eigenvalue $3$ , stretched) and $\binom{1}{-1}$ (eigenvalue $1$ , fixed): a vector on each dashed line stays on that line.

3. Finding eigenvalues: the characteristic equation

Rearrange the defining relation, writing $\lambda\mathbf{v} = \lambda I\mathbf{v}$ so both sides are matrix–vector products:

A\mathbf{v} = \lambda\mathbf{v} \;\iff\; A\mathbf{v} - \lambda I\mathbf{v} = \mathbf{0} \;\iff\; (A - \lambda I)\mathbf{v} = \mathbf{0}.

This is a homogeneous system. From Lesson 5 it has a non-zero solution $\mathbf{v}$ precisely when its coefficient matrix is singular:

\boxed{\;\det(A - \lambda I) = 0.\;}

This is the characteristic equation; for an $n\times n$ matrix it is a degree- $n$ polynomial in $\lambda$ (the characteristic polynomial), whose roots are the eigenvalues. The procedure is therefore always two-stage: (1) solve $\det(A-\lambda I)=0$ for the eigenvalues; (2) for each eigenvalue, solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$ for the eigenvectors.

For a $2\times2$ matrix this collapses to a memorable form. If $A=\binom{a\ \ b}{c\ \ d}$ then

\det(A-\lambda I) = \begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix} = \lambda^2 - (a+d)\lambda + (ad-bc) = \lambda^2 - (\operatorname{tr}A)\lambda + \det A.

So the $2\times2$ characteristic equation is $\lambda^2 - (\operatorname{tr}A)\lambda + \det A = 0$ — worth knowing by heart.

4. Worked Example 1 — a 2×2 eigenproblem (with mark scheme)

Find the eigenvalues and corresponding eigenvectors of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ .

Step 1 — characteristic equation. $\operatorname{tr}A = 7$ , $\det A = 12-2 = 10$ , so

\lambda^2 - 7\lambda + 10 = 0 \;\Rightarrow\; (\lambda-5)(\lambda-2) = 0 \;\Rightarrow\; \lambda_1 = 5,\ \lambda_2 = 2.

(Check: sum $5+2 = 7 = \operatorname{tr}A$ ✓; product $5\times2 = 10 = \det A$ ✓.)

Step 2 — eigenvectors. For $\lambda_1 = 5$ : $(A-5I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\binom{x}{y} = \mathbf{0}$ . Both rows give $-x+y=0$ , i.e. $y=x$ , so $\mathbf{v}_1 = \binom{1}{1}$ (any non-zero multiple).

For $\lambda_2 = 2$ : $(A-2I)\mathbf{v} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\binom{x}{y} = \mathbf{0}$ . Both rows give $2x+y=0$ , i.e. $y=-2x$ , so $\mathbf{v}_2 = \binom{1}{-2}$ .

Mark scheme: M1 form $\det(A-\lambda I)$ ; A1 characteristic equation $\lambda^2-7\lambda+10=0$ ; A1 $\lambda = 5, 2$ ; M1 substitute one $\lambda$ into $(A-\lambda I)\mathbf{v}=\mathbf{0}$ ; A1 $\mathbf{v}_1 = \binom{1}{1}$ ; A1 $\mathbf{v}_2 = \binom{1}{-2}$ . The trace/determinant check is examiner-approved insurance on the eigenvalues.

5. Worked Example 2 — a 3×3 eigenproblem (with mark scheme)

Find the eigenvalues and eigenvectors of $A = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 3 & 0 \\ -1 & 0 & 2 \end{pmatrix}$ .

This matrix is lower triangular, so its eigenvalues are the diagonal entries: $\lambda = 2, 3, 2$ — note $\lambda = 2$ is repeated. (Check: sum $2+3+2 = 7 = \operatorname{tr}A$ ✓.)

For $\lambda = 3$ : $(A-3I)\mathbf{v} = \begin{pmatrix} -1 & 0 & 0 \\ 1 & 0 & 0 \\ -1 & 0 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ . Row 1 gives $x=0$ ; row 3 then gives $z=0$ ; $y$ is free. Eigenvector $\mathbf{v}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ .

For $\lambda = 2$ (repeated): $(A-2I)\mathbf{v} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\mathbf{v} = \mathbf{0}$ . Row 3 gives $x=0$ ; row 2 then gives $y=0$ ; $z$ is free. So there is only one independent eigenvector $\mathbf{v}=\begin{pmatrix}0\\0\\1\end{pmatrix}$ for the repeated eigenvalue — a defective eigenvalue (algebraic multiplicity 2, geometric multiplicity 1). This will matter for diagonalisation (Lesson 9).

Mark scheme: B1 eigenvalues $2,3,2$ from the triangular structure (or M1 A1 via the characteristic polynomial); M1 solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$ ; A1 eigenvector for $\lambda=3$ ; A1 eigenvector for $\lambda=2$ ; A1 note only one independent eigenvector for the repeated eigenvalue.

Key fact: for any triangular matrix (upper or lower), the eigenvalues are exactly the diagonal entries — no characteristic-equation work needed.

Worked Example (parameter) — when is a matrix singular or defective? (with mark scheme)

The matrix $A = \begin{pmatrix} k & 2 \\ 1 & 3 \end{pmatrix}$ depends on a constant $k$ . (a) Find its eigenvalues in terms of $k$ . (b) For what value of $k$ is $A$ singular? (c) For what value of $k$ does $A$ have a repeated eigenvalue?

(a) $\operatorname{tr}A = k+3$ , $\det A = 3k - 2$ , so the characteristic equation is $\lambda^2 - (k+3)\lambda + (3k-2) = 0$ , giving $\lambda = \dfrac{(k+3) \pm \sqrt{(k+3)^2 - 4(3k-2)}}{2}$ .

(b) $A$ is singular when $0$ is an eigenvalue, i.e. $\det A = 0$ : $3k-2 = 0\Rightarrow k = \tfrac23$ .

(c) A repeated eigenvalue means the discriminant of the characteristic quadratic vanishes: $(k+3)^2 - 4(3k-2) = 0$ . Expanding, $k^2 + 6k + 9 - 12k + 8 = k^2 - 6k + 17 = 0$ , whose discriminant is $36 - 68 < 0$ — so there is no real $k$ giving a repeated eigenvalue. (The eigenvalues stay distinct for every real $k$ .)

Mark scheme: M1 form the characteristic equation in $k$ ; A1 correct quadratic; M1 set $\det A = 0$ for singularity; A1 $k = \tfrac23$ ; M1 set the discriminant to $0$ for a repeated eigenvalue; A1 conclude no real $k$ . The two triggers — "singular $\Rightarrow\det = 0$ ", "repeated $\Rightarrow$ discriminant $= 0$ " — are exactly what the question rewards.

6. Properties of eigenvalues — the toolkit and the checks

These relations are both exam shortcuts and error-catchers. Let $A$ be $n\times n$ with eigenvalues $\lambda_1,\dots,\lambda_n$ .

Property	Statement	Why it matters
Sum	$\lambda_1+\cdots+\lambda_n = \operatorname{tr}A$	quick check on the eigenvalues
Product	$\lambda_1\cdots\lambda_n = \det A$	quick check; ties to singularity
Singularity	$A$ singular $\iff 0$ is an eigenvalue	a zero eigenvalue means $\det A=0$
Powers	$A^k$ has eigenvalues $\lambda_i^{\,k}$ (same eigenvectors)	powers of $A$
Inverse	$A^{-1}$ has eigenvalues $1/\lambda_i$ (same eigenvectors)	needs all $\lambda_i\neq0$
Shift	$A+kI$ has eigenvalues $\lambda_i + k$ (same eigenvectors)	shifting the spectrum
Scalar	$cA$ has eigenvalues $c\lambda_i$	scaling

The "same eigenvectors" remark is the key insight for several of these: if $A\mathbf{v}=\lambda\mathbf{v}$ , then $A^2\mathbf{v}=A(\lambda\mathbf{v})=\lambda A\mathbf{v}=\lambda^2\mathbf{v}$ (so $\lambda^2$ is an eigenvalue of $A^2$ , with the same $\mathbf{v}$ ); similarly $(A+kI)\mathbf{v}=(\lambda+k)\mathbf{v}$ . These one-line proofs are popular "show that" questions.

Worked Example 3 — using the property toolkit (with mark scheme)

The matrix $A$ has eigenvalues $\lambda = 3$ and $\lambda = -1$ , with eigenvectors $\binom{2}{1}$ and $\binom{1}{-1}$ respectively. Without finding $A$ explicitly, state the eigenvalues and eigenvectors of (a) $A^3$ , (b) $A^{-1}$ , (c) $2A + 3I$ , and (d) write down $\det A$ and $\operatorname{tr}A$ .

(a) $A^3$ has eigenvalues $3^3 = 27$ and $(-1)^3 = -1$ , with the same eigenvectors $\binom{2}{1}$ , $\binom{1}{-1}$ .

(b) $A^{-1}$ exists (no zero eigenvalue) with eigenvalues $\tfrac13$ and $-1$ , same eigenvectors.

(d) $\det A = (3)(-1) = -3$ ; $\operatorname{tr}A = 3 + (-1) = 2$ .

Eigenvalues and Eigenvectors

Eigenvalues and Eigenvectors

1. Where this sits in AQA 7367

2. Definitions and the geometric meaning

3. Finding eigenvalues: the characteristic equation

4. Worked Example 1 — a 2×2 eigenproblem (with mark scheme)

5. Worked Example 2 — a 3×3 eigenproblem (with mark scheme)

Worked Example (parameter) — when is a matrix singular or defective? (with mark scheme)

6. Properties of eigenvalues — the toolkit and the checks

Worked Example 3 — using the property toolkit (with mark scheme)

More in Mathematics