HookThe twenty-five-billion-dollar eigenvector

In 1998 two Stanford PhD students, Sergey Brin and Larry Page, modelled the entire web as one enormous matrix: a row and column for every page, entries recording who links to whom. Their insight was that a page's importance is not a vote count — it is a fixed point. Important pages are the ones linked to by important pages, a circular definition that becomes solvable the moment you write it as $\mathbf{M}\mathbf{x}=\lambda\mathbf{x}$: importance is an eigenvector of the link matrix. Their algorithm, PageRank, found it by the crudest method imaginable — multiply a starting vector by the matrix, again and again, and watch it settle. When a 2006 SIAM paper set out the mathematics for students, its title needed no exaggeration: 'The \$25,000,000,000 Eigenvector'.

That is the destination of this section. The route to it is the entire matrix toolkit AQA examines: an algebra where multiplication stops commuting; matrices as geometric machines that rotate, reflect and stretch the plane; the determinant as a volume dial that can hit zero; inverses and the three-planes geometry of simultaneous equations; and finally eigenvalues, eigenvectors and diagonalisation — the trick that turns computing $\mathbf{M}^n$ from $n$ multiplications into three. Every layer feeds the next, and the last layer runs the modern world's search engines, population models and Markov chains.

ModelThe algebra — where familiar rules bend

Matrices add, subtract and scale exactly as you would hope: entry by entry, provided the two matrices have the same order. Multiplication is the different beast. The product $\mathbf{AB}$ exists only when the matrices are conformable — $\mathbf{A}$ is $n\times m$ and $\mathbf{B}$ is $m\times p$, inner dimensions matching — and the result is $n\times p$, each entry a row of $\mathbf{A}$ dotted with a column of $\mathbf{B}$.

Two old habits must die here. First, multiplication does not commute: $\mathbf{AB}\ne\mathbf{BA}$ in general — sometimes $\mathbf{BA}$ does not even exist, and even for square matrices the two products usually differ. Every time you multiply, you must know which side you are multiplying on. Second, you cannot cancel: $\mathbf{AB}=\mathbf{0}$ does not force $\mathbf{A}=\mathbf{0}$ or $\mathbf{B}=\mathbf{0}$ — try squaring $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and you get the zero matrix from a non-zero one.

The two special matrices anchor the algebra. The zero matrix $\mathbf{0}$ is the additive identity: $\mathbf{A}+\mathbf{0}=\mathbf{A}$. The identity matrix $\mathbf{I}$ — ones on the diagonal, zeros elsewhere — is the multiplicative one: $\mathbf{AI}=\mathbf{IA}=\mathbf{A}$. $\mathbf{I}$ plays the role of the number 1 throughout the topic: it defines inverses ($\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}$), appears in the characteristic equation as $\mathbf{M}-\lambda\mathbf{I}$, and represents the do-nothing transformation.

Worked example

Let $\mathbf{A}=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $\mathbf{B}=\begin{pmatrix}1&0\\3&1\end{pmatrix}$. Compute $\mathbf{AB}$ and $\mathbf{BA}$.

Step 1: $\mathbf{AB}=\begin{pmatrix}1\cdot 1+2\cdot 3&1\cdot 0+2\cdot 1\\0\cdot 1+1\cdot 3&0\cdot 0+1\cdot 1\end{pmatrix}=\begin{pmatrix}7&2\\3&1\end{pmatrix}$.

Step 2: $\mathbf{BA}=\begin{pmatrix}1\cdot 1+0\cdot 0&1\cdot 2+0\cdot 1\\3\cdot 1+1\cdot 0&3\cdot 2+1\cdot 1\end{pmatrix}=\begin{pmatrix}1&2\\3&7\end{pmatrix}$.

Step 3: $\mathbf{AB}\ne\mathbf{BA}$ — same two matrices, different products. Order is information, and every composite-transformation question in the next block trades on exactly this fact.

ModelMatrices as motion — transformations and what they leave alone

The fastest way to read any transformation matrix: its columns are the images of the basis vectors. Whatever $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ map to, write those images as columns — done. That one fact generates the whole catalogue: rotation by $\theta$ anticlockwise about the origin is $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$; reflections in the axes and in $y=\pm x$ swap or negate the columns accordingly; a stretch factor $k$ parallel to the $x$-axis is $\begin{pmatrix}k&0\\0&1\end{pmatrix}$; an enlargement is $k\mathbf{I}$. In 3D, AQA keeps to single reflections in the coordinate planes and rotations about the coordinate axes — rotation about the $x$-axis, for instance, fixes $x$ and rotates the $yz$-plane: $\begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}$.

Composition is multiplication read right to left: 'transformation $\mathbf{A}$ then $\mathbf{B}$' is the single matrix $\mathbf{BA}$, because $\mathbf{B}(\mathbf{A}\mathbf{x})=(\mathbf{BA})\mathbf{x}$. Get the order wrong and — as the previous block showed — you usually get a different transformation entirely.

What a transformation leaves alone is the examiner's favourite probe. An invariant point satisfies $\mathbf{M}\mathbf{x}=\mathbf{x}$; solving $(\mathbf{M}-\mathbf{I})\mathbf{x}=\mathbf{0}$ finds them all, and the origin always qualifies. An invariant line is a line mapped onto itself — individual points may slide along it. For lines $y=mx+c$, take the general point $(x,\ mx+c)$, apply $\mathbf{M}$, and demand the image satisfies the same equation for all $x$: comparing coefficients yields a quadratic in $m$ and a condition on $c$.

Worked example

Find the invariant lines of $\mathbf{M}=\begin{pmatrix}1&2\\3&2\end{pmatrix}$.

Step 1: a general point of $y=mx+c$ is $(x,\ mx+c)$. Its image is $\begin{pmatrix}1&2\\3&2\end{pmatrix}\begin{pmatrix}x\\mx+c\end{pmatrix}=\begin{pmatrix}x+2mx+2c\\3x+2mx+2c\end{pmatrix}$.

Step 2: for the line to be invariant, the image must satisfy $y'=mx'+c$: $3x+2mx+2c=m(x+2mx+2c)+c$ for ALL $x$.

Step 3 (coefficients of x): $3+2m=m+2m^2$, so $2m^2-m-3=0$, giving $(2m-3)(m+1)=0$: $m=\frac{3}{2}$ or $m=-1$.

Step 4 (constant terms): $2c=2mc+c$, i.e. $c(1-2m)=0$. Neither value of $m$ equals $\frac{1}{2}$, so $c=0$ in both cases.

Conclusion: the invariant lines are $y=\frac{3}{2}x$ and $y=-x$, both through the origin. Hold that result — the eigenvector block will land on the same two lines from a completely different direction.

ModelDeterminants and inverses — the scale factor with a sign

For $\mathbf{M}=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\det\mathbf{M}=ad-bc$; for a $3\times 3$, expand along any row or column with the alternating sign pattern, each entry multiplying the determinant of the $2\times 2$ left when its row and column are deleted. But the number means something: $|\det\mathbf{M}|$ is the area scale factor of the transformation in 2D and the volume scale factor in 3D, and a negative determinant means orientation is reversed — the plane has been flipped, as every reflection (determinant $-1$) demonstrates. A unit square mapped by a matrix of determinant $-5$ becomes a parallelogram of area 5, traced the opposite way round.

$\det\mathbf{M}=0$ is the degenerate case: the transformation crushes the plane onto a line (or a point), area scale factor zero. Such a matrix is singular, and no inverse can exist — you cannot un-crush a line back into a plane, because infinitely many points shared each image. Non-singular matrices invert cleanly: for $2\times 2$, $\mathbf{M}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ — swap the diagonal, negate the off-diagonal, divide by the determinant. For $3\times 3$, the by-hand route is cofactors, transpose (the adjugate), divide by the determinant; AQA also permits the calculator on the calculator papers, which makes the by-hand skill a checking tool as much as a method.

Two laws worth engraving: $\det(\mathbf{AB})=\det\mathbf{A}\det\mathbf{B}$ — scale factors multiply, which gives $\det(\mathbf{M}^n)=(\det\mathbf{M})^n$ as a free arithmetic check on any matrix-power answer — and $(\mathbf{AB})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$: to undo 'A then B', you must undo B first, socks and shoes.

MechanismThree equations, three planes — solving and seeing

A system of three linear equations in three unknowns is one matrix sentence: $\mathbf{A}\mathbf{x}=\mathbf{b}$. If $\det\mathbf{A}\ne 0$, the inverse exists and the solution is immediate and unique: $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$. Note the order — $\mathbf{A}^{-1}$ multiplies on the left of both sides, because that is the side $\mathbf{A}$ sits on.

The geometry is where AQA earns its interpretation marks. Each equation is a plane in 3D. Determinant non-zero: the three planes meet in exactly one point. Determinant zero splits into consistent and inconsistent cases, and you must test consistency (eliminate variables and see whether the equations agree) to tell them apart. Consistent: the planes share a whole line — a sheaf, like pages of an open book sharing the spine — or are coincident. Inconsistent: no solution, either because two or more planes are parallel and distinct, or because the planes form a prism — each pair meets in a line, but the three lines are parallel and never concur.

The exam pattern is predictable: a system with a parameter $k$, part (a) 'find the values of $k$ for which a unique solution exists' (solve $\det\mathbf{A}\ne 0$), part (b) 'for this particular $k$, solve or interpret geometrically'. Determinant first, always — it is the switch that decides which world you are in.

Worked example

Solve, using an inverse matrix: $x+y+z=6$, $2x-y+z=3$, $x+2y-z=2$.

Step 1: $\mathbf{A}=\begin{pmatrix}1&1&1\\2&-1&1\\1&2&-1\end{pmatrix}$, $\mathbf{b}=\begin{pmatrix}6\\3\\2\end{pmatrix}$. Expanding along the top row: $\det\mathbf{A}=1(1-2)-1(-2-1)+1(4+1)=-1+3+5=7\ne 0$, so a unique solution exists.

Step 2 (adjugate route): computing all nine cofactors and transposing gives $\mathbf{A}^{-1}=\frac{1}{7}\begin{pmatrix}-1&3&2\\3&-2&1\\5&-1&-3\end{pmatrix}$.

Step 3: $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}=\frac{1}{7}\begin{pmatrix}-6+9+4\\18-6+2\\30-3-6\end{pmatrix}=\frac{1}{7}\begin{pmatrix}7\\14\\21\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}$.

Step 4 (check): substitute back: $1+2+3=6$ ✓, $2-2+3=3$ ✓, $1+4-3=2$ ✓. Geometrically: three planes meeting at the single point $(1,2,3)$.

MechanismFactorising determinants without the bloodbath

Some determinants — especially those stuffed with algebraic entries like $a$, $b$, $c$ — are set precisely so that brute-force expansion produces an unfactorisable swamp. The intended weapon is row and column operations: adding a multiple of one row (or column) to another leaves the determinant unchanged, and taking out a common factor from a whole row or column pulls it outside as a factor of the determinant. Swapping two rows just changes the sign.

The strategy is always the same: engineer zeros, then extract factors before expanding. Subtract columns to create entries like $b-a$ and $c-a$; the moment a whole row or column shares a factor, pull it out; only expand once the determinant left behind is small and clean. A factorised answer such as $(b-a)(c-a)(c-b)$ is also self-checking: set $b=a$ in the original determinant and two columns coincide, forcing the determinant to zero — so $(b-a)$ must be a factor. That symmetry argument is both a way to predict the answer and a way to verify it.

Worked example

Factorise $\Delta=\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}$.

Step 1 (create zeros): replace column 2 by (column 2 $-$ column 1) and column 3 by (column 3 $-$ column 1) — determinant unchanged: $\Delta=\begin{vmatrix}1&0&0\\a&b-a&c-a\\a^2&b^2-a^2&c^2-a^2\end{vmatrix}$.

Step 2 (expand along the top row): only the first entry survives: $\Delta=(b-a)(c^2-a^2)-(c-a)(b^2-a^2)$.

Step 3 (factor each difference of squares): $\Delta=(b-a)(c-a)(c+a)-(c-a)(b-a)(b+a)$.

Step 4 (common factor): $\Delta=(b-a)(c-a)\big[(c+a)-(b+a)\big]=(b-a)(c-a)(c-b)$.

Sanity check: $b=a$ makes columns 1 and 2 of the original equal, so $\Delta=0$ — consistent with the factor $(b-a)$.

CaseEigenvalues and diagonalisation — the power shortcut

An eigenvector of $\mathbf{M}$ is a non-zero vector whose direction survives the transformation: $\mathbf{M}\mathbf{x}=\lambda\mathbf{x}$, stretched by the eigenvalue $\lambda$. Rearranged, $(\mathbf{M}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}$ has a non-zero solution only when $\mathbf{M}-\lambda\mathbf{I}$ is singular, giving the characteristic equation $\det(\mathbf{M}-\lambda\mathbf{I})=0$ — a quadratic for a $2\times 2$, a cubic for a $3\times 3$. Solve it for the eigenvalues; substitute each back and solve for the eigenvector directions. Geometrically, eigenvectors mark the invariant lines through the origin, with $\lambda$ the stretch along each; an eigenvalue of 1 means a line of invariant points, and a negative eigenvalue flips vectors along its line.

Diagonalisation packages the eigen-structure as a change of viewpoint. Load the eigenvectors as columns of $\mathbf{U}$ and the eigenvalues, in the same order, into the diagonal matrix $\mathbf{D}$; then $\mathbf{M}=\mathbf{U}\mathbf{D}\mathbf{U}^{-1}$. In eigenvector coordinates the transformation is just independent stretches. The jackpot is powers: $\mathbf{M}^n=\mathbf{U}\mathbf{D}^n\mathbf{U}^{-1}$, because every internal $\mathbf{U}^{-1}\mathbf{U}$ collapses to $\mathbf{I}$ — and $\mathbf{D}^n$ just raises each diagonal entry to the $n$. A hundredth power costs one line.

This is PageRank's engine room. Repeatedly multiplying a vector by $\mathbf{M}$ amplifies the component along the eigenvector with the largest $|\lambda|$ fastest — write the vector in eigencoordinates and each component scales like $\lambda^n$ — so the iteration converges to the dominant eigenvector: the ranking. The same mathematics runs population models and Markov chains; the exam version is simply 'find $\mathbf{M}^n$'.

Worked example

Diagonalise $\mathbf{M}=\begin{pmatrix}1&2\\3&2\end{pmatrix}$ and hence find $\mathbf{M}^n$.

Step 1 (eigenvalues): $\det(\mathbf{M}-\lambda\mathbf{I})=(1-\lambda)(2-\lambda)-6=\lambda^2-3\lambda-4=(\lambda-4)(\lambda+1)=0$, so $\lambda=4$ and $\lambda=-1$.

Step 2 (eigenvectors): for $\lambda=4$: $-3x+2y=0$ gives direction $\begin{pmatrix}2\\3\end{pmatrix}$. For $\lambda=-1$: $2x+2y=0$ gives $\begin{pmatrix}1\\-1\end{pmatrix}$. (These span the lines $y=\frac{3}{2}x$ and $y=-x$ — the invariant lines found earlier, now with their stretch factors.)

Step 3 (assemble): $\mathbf{U}=\begin{pmatrix}2&1\\3&-1\end{pmatrix}$, $\mathbf{D}=\begin{pmatrix}4&0\\0&-1\end{pmatrix}$, $\det\mathbf{U}=-5$, so $\mathbf{U}^{-1}=\frac{1}{5}\begin{pmatrix}1&1\\3&-2\end{pmatrix}$.

Step 4 (power): $\mathbf{M}^n=\mathbf{U}\begin{pmatrix}4^n&0\\0&(-1)^n\end{pmatrix}\mathbf{U}^{-1}=\frac{1}{5}\begin{pmatrix}2\cdot 4^n+3(-1)^n&2\cdot 4^n-2(-1)^n\\3\cdot 4^n-3(-1)^n&3\cdot 4^n+2(-1)^n\end{pmatrix}$.

Check (n = 1): $\frac{1}{5}\begin{pmatrix}8-3&8+2\\12+3&12-2\end{pmatrix}=\begin{pmatrix}1&2\\3&2\end{pmatrix}=\mathbf{M}$ ✓ — and a closed form like this is exactly what a section-A induction question would then ask you to prove.

VocabularyKey terms the mark scheme pays for

Conformable

Matrices whose dimensions allow the operation: same order for addition; inner dimensions matching ($n\times m$ times $m\times p$) for multiplication.

Identity matrix $\mathbf{I}$

Ones on the diagonal, zeros elsewhere: $\mathbf{AI}=\mathbf{IA}=\mathbf{A}$. The do-nothing transformation, and the '1' in every matrix equation.

Determinant

$ad-bc$ for a $2\times 2$; the area (2D) or volume (3D) scale factor of the transformation, with a negative sign signalling orientation reversal.

Singular matrix

A matrix with determinant zero: it crushes the plane onto a line or point, so no inverse exists and $\mathbf{A}\mathbf{x}=\mathbf{b}$ loses its unique solution.

Invariant line

A line mapped onto itself by a transformation — points may slide along it. Invariant lines through the origin are exactly the eigenvector directions.

Eigenvalue and eigenvector

A non-zero vector with $\mathbf{M}\mathbf{x}=\lambda\mathbf{x}$: direction preserved, length scaled by $\lambda$. Found via $\det(\mathbf{M}-\lambda\mathbf{I})=0$.

Characteristic equation

$\det(\mathbf{M}-\lambda\mathbf{I})=0$ — the polynomial whose roots are the eigenvalues; quadratic for $2\times 2$ matrices, cubic for $3\times 3$.

Diagonalisation

Writing $\mathbf{M}=\mathbf{U}\mathbf{D}\mathbf{U}^{-1}$ with eigenvectors as the columns of $\mathbf{U}$ and eigenvalues along $\mathbf{D}$; powers collapse to $\mathbf{M}^n=\mathbf{U}\mathbf{D}^n\mathbf{U}^{-1}$.

Sheaf and prism

The two $\det=0$ geometries for three planes: a sheaf shares a common line (consistent, infinitely many solutions); a prism's pairwise intersection lines are parallel (inconsistent, none).

TrapsMisconceptions that cost marks

“Matrix multiplication works like number multiplication — order doesn't matter and you can cancel.”

Actually: $\mathbf{AB}\ne\mathbf{BA}$ in general, and $\mathbf{AB}=\mathbf{0}$ does not force either factor to be zero: $\begin{pmatrix}0&1\\0&0\end{pmatrix}^2=\mathbf{0}$. Cancelling $\mathbf{A}$ from $\mathbf{AB}=\mathbf{AC}$ is legal only if $\mathbf{A}$ is non-singular — you are really multiplying by $\mathbf{A}^{-1}$, on a stated side.

“det A = 0 means the system Ax = b has no solutions.”

Actually: It means no UNIQUE solution. The system may still have infinitely many (planes forming a sheaf, or coincident) or none (parallel planes, or a prism) — you must test consistency to decide which. The unique-solution question and the any-solution question are different questions.

“Each eigenvalue has one specific eigenvector you must find.”

Actually: Any non-zero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue — it is a whole invariant LINE of directions. $\begin{pmatrix}2\\3\end{pmatrix}$ and $\begin{pmatrix}-4\\-6\end{pmatrix}$ are the same answer, so choose the smallest integer form and expect the mark scheme to say 'or any multiple'.

ExamWhat examiners want

Order discipline earns quiet marks all through this topic. Composite transformations: '$\mathbf{A}$ then $\mathbf{B}$' is $\mathbf{BA}$, and writing the product the wrong way round changes the answer, not just the notation. Solving systems: state $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$ with the inverse on the left. Diagonalisation: the eigenvalue order in $\mathbf{D}$ must match the eigenvector column order in $\mathbf{U}$ — mismatching them is the classic way to lose the final accuracy mark with all the hard work done.

Cheap self-checks are everywhere, and examiners' reports repeatedly note that the candidates who use them score higher: verify an eigenvector by one multiplication ($\mathbf{M}\mathbf{x}$ should be visibly $\lambda\mathbf{x}$); test a computed inverse with $\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}$; check a closed form for $\mathbf{M}^n$ at $n=1$; and remember $\det(\mathbf{M}^n)=(\det\mathbf{M})^n$ as a one-line audit. In geometric-interpretation parts, name the configuration precisely — 'the planes form a sheaf: consistent, meeting in a common line' — because the noun carries the mark; 'no unique solution' on its own is the hedge that earns nothing. When determinants carry algebraic entries, factorise with row and column operations and state the operation you performed at each step ($C_2\to C_2-C_1$): unexplained determinant jumps are where method marks vanish.