Paper 2 · Further Pure 2 ✦

Eigenvalues & Eigenvectors

Every matrix has directions it does not turn — it only stretches them. Find those, and powers, inverses and whole systems of planes fall out almost for free.

Syllabus 2.2Eigenvalues: real & distinct only2 × 2 and 3 × 3 only

What the examiner expects you to do

Turn 3 linear equations in 3 unknowns into a matrix equation and back.
Understand consistency, link it to whether the matrix is singular, solve consistent systems, and interpret geometrically in terms of lines and planes.
Understand the terms characteristic equation, eigenvalue and eigenvector; find them for 2 × 2 and 3 × 3 matrices — restricted to real, distinct eigenvalues.
Write a square matrix as ${𝐐 𝐃 𝐐}^{- 1}$ and use it — in particular to compute powers.
Use the fact that a square matrix satisfies its own characteristic equation.

§1 Three equations, three planes

◆ The matrix equation and what its determinant tells you

The system $\begin{aligned} a_{1} x + b_{1} y + c_{1} z & = d_{1} \\ a_{2} x + b_{2} y + c_{2} z & = d_{2} \\ a_{3} x + b_{3} y + c_{3} z & = d_{3} \end{aligned}$ is $𝐀 𝐱 = 𝐛$ , where each row of $𝐀$ is the normal of one plane.

$\det 𝐀 \neq 0$ : $𝐀^{- 1}$ exists, so $𝐱 = 𝐀^{- 1} 𝐛$ is the unique solution — the three planes meet in a single point.
$\det 𝐀 = 0$ : the system is either inconsistent (no solutions) or has infinitely many. The determinant alone cannot tell you which — the right-hand side decides.

◇ The geometry when

\det 𝐀 = 0

What elimination gives	Solutions	The picture
a row of $0 = 0$	infinitely many, forming a line	a sheaf: three distinct planes through one common line
a row of $0 = k$ , $k \neq 0$	none	a triangular prism: each pair meets in a line, but the three lines are parallel and distinct — or two (or three) planes are parallel and distinct

So the routine is always: eliminate down to the last row, then look at what that row says.

✎ Worked example — the same matrix, two different fates

Consider the system $x + y + z = 6, x + 2 y + 3 z = 14, x + 3 y + 5 z = d$ .

Here $𝐀 = (\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{array})$ and

\det 𝐀 = 1 (10 - 9) - 1 (5 - 3) + 1 (3 - 2) = 1 - 2 + 1 = 0,

so $𝐀$ is singular whatever $d$ is. Subtracting adjacent equations:

(2) - (1) : y + 2 z = 8, (3) - (2) : y + 2 z = d - 14 .

$d = 22$ : the two reduce to the same thing, $0 = 0$ . Consistent, infinitely many solutions. Setting $z = t$ gives $y = 8 - 2 t$ and $x = 6 - y - z = - 2 + t$ , so $(\begin{array}{c} x \\ y \\ z \end{array}) = (\begin{array}{c} - 2 \\ 8 \\ 0 \end{array}) + t (\begin{array}{c} 1 \\ - 2 \\ 1 \end{array})$ — a line. The three planes form a sheaf. Check in (3): $(- 2 + t) + 3 (8 - 2 t) + 5 t = 22$ ✓ for every $t$ .
$d = 20$ : we get $y + 2 z = 8$ and $y + 2 z = 6$ , i.e. $0 = - 2$ . Inconsistent — no solutions. The three planes form a triangular prism: each pair meets in a line, but there is no point on all three.

Identical coefficient matrix, identical determinant, opposite conclusions. This is exactly why " $\det = 0$ " is never a complete answer.

✎ Worked example — the non-singular case

$x + y + z = 6, 2 x - y + z = 3, x + 2 y - z = 2$ . Expanding along the first row,

\det (\begin{array}{ccc} 1 & 1 & 1 \\ 2 & - 1 & 1 \\ 1 & 2 & - 1 \end{array}) = 1 (1 - 2) - 1 (- 2 - 1) + 1 (4 + 1) = - 1 + 3 + 5 = 7 \neq 0,

so there is exactly one solution. Eliminating gives $x = 1, y = 2, z = 3$ . Check all three: $1 + 2 + 3 = 6$ ✓, $2 - 2 + 3 = 3$ ✓, $1 + 4 - 3 = 2$ ✓ — the three planes meet at the single point $(1,2,3)$ .

§2 Eigenvalues and eigenvectors

◇ The definition — and why it forces a determinant

A non-zero vector $𝐞$ is an eigenvector of $𝐀$ , with eigenvalue $λ$ , when

𝐀 𝐞 = λ 𝐞, 𝐞 \neq 𝟎 .

Rearranged, $(𝐀 - λ 𝐈) 𝐞 = 𝟎$ . A non-zero $𝐞$ can only satisfy this if $𝐀 - λ 𝐈$ is singular — otherwise we could multiply by its inverse and force $𝐞 = 𝟎$ . Hence the characteristic equation

\det (𝐀 - λ 𝐈) = 0 .

In this syllabus you only meet cases where its roots are real and distinct.

✎ Worked example — a 2 × 2

Find the eigenvalues and eigenvectors of $𝐀 = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ .

\det (𝐀 - λ 𝐈) = | \begin{array}{cc} 4 - λ & 1 \\ 2 & 3 - λ \end{array} | = (4 - λ) (3 - λ) - 2 = λ^{2} - 7 λ + 10 = (λ - 2) (λ - 5) .

So $λ = 2$ and $λ = 5$ — real and distinct.

$λ = 5$ : $(𝐀 - 5 𝐈) 𝐞 = 𝟎$ gives $(\begin{array}{cc} - 1 & 1 \\ 2 & - 2 \end{array}) (\begin{array}{c} x \\ y \end{array}) = 𝟎$ , i.e. $y = x$ . Take $𝐞_{1} = (\begin{array}{c} 1 \\ 1 \end{array})$ .

$λ = 2$ : $(\begin{array}{cc} 2 & 1 \\ 2 & 1 \end{array}) (\begin{array}{c} x \\ y \end{array}) = 𝟎$ gives $y = - 2 x$ . Take $𝐞_{2} = (\begin{array}{c} 1 \\ - 2 \end{array})$ .

Always check: $𝐀 𝐞_{1} = (\begin{array}{c} 5 \\ 5 \end{array}) = 5 𝐞_{1}$ ✓ and $𝐀 𝐞_{2} = (\begin{array}{c} 2 \\ - 4 \end{array}) = 2 𝐞_{2}$ ✓ — two multiplications, and your answer is certain.

Both rows of $𝐀 - λ 𝐈$ must give the same line. If they do not, your $λ$ is wrong — a free error check built into the method.

★ The link back to FP1 — eigenvectors are the invariant directions

In Topic 4 the invariant lines $y = m x$ came from $b m^{2} + (a - d) m - c = 0$ . That is the same condition as $𝐀 𝐞 = λ 𝐞$ for $𝐞 = (1, m)$ : an eigenvector points along an invariant line through the origin, and $λ$ says how far points slide along it. A negative $λ$ flips them to the other side; $λ = 1$ leaves them fixed.

◆ Properties proved straight from

𝐀 𝐞 = λ 𝐞

$λ^{n}$ is an eigenvalue of $𝐀^{n}$ , with the same eigenvector: $𝐀^{2} 𝐞 = 𝐀 (λ 𝐞) = λ 𝐀 𝐞 = λ^{2} 𝐞$ , and repeating gives $𝐀^{n} 𝐞 = λ^{n} 𝐞$ .
If $𝐀$ is invertible, $λ^{- 1}$ is an eigenvalue of $𝐀^{- 1}$ : multiply $𝐀 𝐞 = λ 𝐞$ by $𝐀^{- 1}$ to get $𝐞 = λ 𝐀^{- 1} 𝐞$ , so $𝐀^{- 1} 𝐞 = \frac{1}{λ} 𝐞$ .
$𝐀$ is singular exactly when $0$ is an eigenvalue, since $\det (𝐀 - 0 𝐈) = \det 𝐀$ .

These one-line arguments are exactly the "simple properties" the syllabus asks you to prove.

§3 The 3 × 3 case

✎ Worked example — a full 3 × 3

Find the eigenvalues and eigenvectors of $𝐀 = (\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{array})$ .

Characteristic equation. Expand along the top row:

| \begin{array}{ccc} 1 - λ & 1 & 0 \\ 1 & 2 - λ & 1 \\ 0 & 1 & 1 - λ \end{array} | = (1 - λ) [(2 - λ) (1 - λ) - 1] - 1 [(1) (1 - λ) - 0] .

Both terms carry a factor $(1 - λ)$ — take it out rather than expanding into a cubic:

= (1 - λ) [(2 - λ) (1 - λ) - 1 - 1] = (1 - λ) (λ^{2} - 3 λ) = (1 - λ) λ (λ - 3) .

So $λ = 0, 1, 3$ — real and distinct. ( $λ = 0$ appearing tells us immediately that $\det 𝐀 = 0$ .)

$λ = 0$ : $𝐀 𝐞 = 𝟎$ reads $x + y = 0, x + 2 y + z = 0, y + z = 0$ . The first gives $y = - x$ , the third $z = - y = x$ , and the middle is then automatic. $𝐞 = (\begin{array}{c} 1 \\ - 1 \\ 1 \end{array})$ .

$λ = 1$ : $𝐀 - 𝐈 = (\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{array})$ gives $y = 0$ and $x + z = 0$ . $𝐞 = (\begin{array}{c} 1 \\ 0 \\ - 1 \end{array})$ .

$λ = 3$ : $𝐀 - 3 𝐈 = (\begin{array}{ccc} - 2 & 1 & 0 \\ 1 & - 1 & 1 \\ 0 & 1 & - 2 \end{array})$ gives $y = 2 x$ and $y = 2 z$ , so $z = x$ . $𝐞 = (\begin{array}{c} 1 \\ 2 \\ 1 \end{array})$ .

Check all three: $𝐀 (\begin{array}{c} 1 \\ - 1 \\ 1 \end{array}) = (\begin{array}{c} 0 \\ 0 \\ 0 \end{array})$ ✓, $𝐀 (\begin{array}{c} 1 \\ 0 \\ - 1 \end{array}) = (\begin{array}{c} 1 \\ 0 \\ - 1 \end{array})$ ✓, $𝐀 (\begin{array}{c} 1 \\ 2 \\ 1 \end{array}) = (\begin{array}{c} 3 \\ 6 \\ 3 \end{array}) = 3 (\begin{array}{c} 1 \\ 2 \\ 1 \end{array})$ ✓

★ A* tip — do not multiply the cubic out

A 3 × 3 characteristic determinant almost always has a visible common factor, or an obvious root such as $λ = 1$ or $λ = - 1$ to test. Factorise as you expand. Multiplying out to $- λ^{3} + \dots$ and then hunting for roots wastes minutes and invites sign errors.

§4 Diagonalisation: $𝐀 = {𝐐 𝐃 𝐐}^{- 1}$

◆ The construction

Put the eigenvectors in as the columns of $𝐐$ , and the matching eigenvalues down the diagonal of $𝐃$ in the same order. Then

𝐀 = {𝐐 𝐃 𝐐}^{- 1} and, crucially, 𝐀^{n} = {𝐐 𝐃}^{n} 𝐐^{- 1},

because the middle terms telescope: $𝐀^{2} = {𝐐 𝐃 𝐐}^{- 1} {𝐐 𝐃 𝐐}^{- 1} = 𝐐 𝐃 (𝐐^{- 1} 𝐐) {𝐃 𝐐}^{- 1} = {𝐐 𝐃}^{2} 𝐐^{- 1}$ . And $𝐃^{n}$ is trivial — just raise each diagonal entry to the power $n$ .

✎ Worked example — a formula for

𝐀^{n}

For $𝐀 = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ we found $λ = 5$ with $(\begin{array}{c} 1 \\ 1 \end{array})$ , and $λ = 2$ with $(\begin{array}{c} 1 \\ - 2 \end{array})$ . So

𝐐 = (\begin{array}{cc} 1 & 1 \\ 1 & - 2 \end{array}), 𝐃 = (\begin{array}{cc} 5 & 0 \\ 0 & 2 \end{array}), 𝐐^{- 1} = \frac{1}{- 3} (\begin{array}{cc} - 2 & - 1 \\ - 1 & 1 \end{array}) = \frac{1}{3} (\begin{array}{cc} 2 & 1 \\ 1 & - 1 \end{array}) .

Then $𝐀^{n} = {𝐐 𝐃}^{n} 𝐐^{- 1}$ :

𝐀^{n} = \frac{1}{3} (\begin{array}{cc} 1 & 1 \\ 1 & - 2 \end{array}) (\begin{array}{cc} 5^{n} & 0 \\ 0 & 2^{n} \end{array}) (\begin{array}{cc} 2 & 1 \\ 1 & - 1 \end{array}) = \frac{1}{3} (\begin{array}{cc} 2 \cdot 5^{n} + 2^{n} & 5^{n} - 2^{n} \\ 2 \cdot 5^{n} - 2 \cdot 2^{n} & 5^{n} + 2 \cdot 2^{n} \end{array}) .

Check $n = 1$ : $\frac{1}{3} (\begin{array}{cc} 12 & 3 \\ 6 & 9 \end{array}) = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) = 𝐀$ ✓ Check $n = 2$ : $\frac{1}{3} (\begin{array}{cc} 54 & 21 \\ 42 & 33 \end{array}) = (\begin{array}{cc} 18 & 7 \\ 14 & 11 \end{array})$ , and $𝐀^{2} = (\begin{array}{cc} 18 & 7 \\ 14 & 11 \end{array})$ ✓

Testing $n = 1$ costs ten seconds and catches a swapped column of $𝐐$ instantly.

⚠ The order must match, and eigenvectors may be scaled

If $𝐞_{1}$ is the first column of $𝐐$ , then $λ_{1}$ must be the first diagonal entry of $𝐃$ . Swap one and not the other and everything collapses.

Any non-zero multiple of an eigenvector is also an eigenvector, so $(\begin{array}{c} 2 \\ 2 \end{array})$ would do just as well as $(\begin{array}{c} 1 \\ 1 \end{array})$ ; $𝐐$ changes but ${𝐐 𝐃 𝐐}^{- 1}$ does not. Do not panic if your $𝐐$ differs from the mark scheme by a scale factor.

§5 Cayley–Hamilton

◆ A matrix satisfies its own characteristic equation

If the characteristic equation of $𝐀$ is $λ^{2} - p λ + q = 0$ , then

𝐀^{2} - p 𝐀 + q 𝐈 = 𝟎,

where — and this is where marks go — the constant becomes $q 𝐈$ , not the scalar $q$ . For a 2 × 2, $p = a + d$ (the sum of the diagonal) and $q = \det 𝐀$ .

✎ Worked example — two standard uses

For $𝐀 = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ the characteristic equation is $λ^{2} - 7 λ + 10 = 0$ , so

𝐀^{2} - 7 𝐀 + 10 𝐈 = 𝟎 .

(a) Reducing powers. $𝐀^{2} = 7 𝐀 - 10 𝐈$ , and multiplying by $𝐀$ again,

𝐀^{3} = 7 𝐀^{2} - 10 𝐀 = 7 (7 𝐀 - 10 𝐈) - 10 𝐀 = 39 𝐀 - 70 𝐈 .

Check: $39 (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) - 70 𝐈 = (\begin{array}{cc} 86 & 39 \\ 78 & 47 \end{array})$ , and multiplying $𝐀^{2} 𝐀 = (\begin{array}{cc} 18 & 7 \\ 14 & 11 \end{array}) (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) = (\begin{array}{cc} 86 & 39 \\ 78 & 47 \end{array})$ ✓

(b) Finding the inverse. Rearrange to isolate $𝐈$ and multiply through by $𝐀^{- 1}$ :

10 𝐈 = 7 𝐀 - 𝐀^{2} ⟹ 10 𝐀^{- 1} = 7 𝐈 - 𝐀 ⟹ 𝐀^{- 1} = \frac{1}{10} (7 𝐈 - 𝐀) = \frac{1}{10} (\begin{array}{cc} 3 & - 1 \\ - 2 & 4 \end{array}) .

Check against the standard 2 × 2 formula: $\det 𝐀 = 10$ , so $𝐀^{- 1} = \frac{1}{10} (\begin{array}{cc} 3 & - 1 \\ - 2 & 4 \end{array})$ ✓ — the same. For a 3 × 3, where the standard route means nine cofactors, this shortcut is a genuine time-saver.

★ A* tip — why the inverse trick always works

The constant term of the characteristic equation is $\pm \det 𝐀$ . So it is non-zero exactly when $𝐀$ is invertible — which is exactly when you are allowed to divide by it. If that constant is $0$ , the matrix is singular and there is no inverse to find.

§6 Interactive: watch a vector line up with its eigenvector

Spin the blue vector $𝐯$ and watch its image $𝐀 𝐯$ . For most directions the two point different ways; along an eigendirection they line up exactly, and $λ$ is the stretch factor. The panel checks $𝐀 𝐞 = λ 𝐞$ , ${𝐐 𝐃 𝐐}^{- 1} = 𝐀$ , ${𝐐 𝐃}^{n} 𝐐^{- 1} = 𝐀^{n}$ , and Cayley–Hamilton — all numerically.

Direction of

𝐯

30°

Power

n

to test 5

v Av eigendirections image of the unit circle

line them up ✎

§7 Examiner traps & A* checklist

⚠ The eight most common mark-losers

Concluding "no solutions" from $\det 𝐀 = 0$ alone. It could equally be infinitely many — eliminate and look at the last row.
Writing $\det (𝐀 - λ) = 0$ instead of $\det (𝐀 - λ 𝐈) = 0$ . You cannot subtract a scalar from a matrix.
In Cayley–Hamilton, writing $𝐀^{2} - p 𝐀 + q = 𝟎$ . The constant must be $q 𝐈$ .
Mismatching the order of the columns of $𝐐$ with the diagonal of $𝐃$ .
Giving $𝐞 = 𝟎$ as an eigenvector. The zero vector is excluded by definition — it would satisfy the equation for every $λ$ .
Only using one row of $𝐀 - λ 𝐈$ for a 3 × 3 and stopping too early: you need two independent rows to pin the direction down.
Not stating the geometry when the question says "interpret geometrically" — name the picture: single point, sheaf through a common line, or triangular prism / parallel planes.
Forgetting that any scalar multiple of an eigenvector is also an eigenvector, and wasting time trying to match the book exactly.

★ Before you turn the page

Did I verify each eigenvector by multiplying $𝐀 𝐞$ and comparing with $λ 𝐞$ ?
Do my eigenvalues sum to the sum of the diagonal entries of $𝐀$ , and multiply to $\det 𝐀$ ? (Both follow from the characteristic equation and catch errors instantly.)
Did I test any $𝐀^{n}$ formula at $n = 1$ ?

§8 Video explainers (curated)

▶Three planes & consistency
Curated search — the geometric cases and how to spot them.↗ ▶Eigenvalues & eigenvectors
Curated search — characteristic equation, 2 × 2 and 3 × 3.↗ ▶Diagonalisation & Cayley–Hamilton
Curated search — powers of a matrix, and the inverse shortcut.↗

§9 Check yourself

Score 0 / 5

Q1 · characteristic equation

The characteristic equation of $(\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ is:

Solution

$(4 - λ) (3 - λ) - (1) (2) = λ^{2} - 7 λ + 12 - 2 = λ^{2} - 7 λ + 10$ . Instant check: for a 2 × 2 the equation is $λ^{2} - (sum of the diagonal) λ + \det = 0$ ; here that is $λ^{2} - 7 λ + 10$ ✓ The last option swaps those two numbers over.

Q2 · eigenvector

An eigenvector of $(\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ corresponding to $λ = 5$ is:

Solution

$(\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) (\begin{array}{c} 1 \\ 1 \end{array}) = (\begin{array}{c} 5 \\ 5 \end{array}) = 5 (\begin{array}{c} 1 \\ 1 \end{array})$ ✓ The second option is the eigenvector for $λ = 2$ ; the last is excluded by definition — $𝟎$ satisfies $𝐀 𝐞 = λ 𝐞$ for every $λ$ , so it is never an eigenvector.

Q3 · geometry of three planes

Three planes have a singular coefficient matrix, and elimination reduces the system to $0 = 3$ . Geometrically the planes:

Solution

$0 = 3$ is impossible, so the system is inconsistent — there is no common point. The picture is a triangular prism (or a pair of parallel planes). A row of $0 = 0$ instead would have given a sheaf with a common line; a single point requires $\det \neq 0$ .

Q4 · Cayley–Hamilton

Given $𝐀 = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ satisfies $𝐀^{2} - 7 𝐀 + 10 𝐈 = 𝟎$ , it follows that $𝐀^{- 1}$ equals:

Solution

Multiply through by $𝐀^{- 1}$ : $𝐀 - 7 𝐈 + 10 𝐀^{- 1} = 𝟎$ , so $𝐀^{- 1} = \frac{1}{10} (7 𝐈 - 𝐀) = \frac{1}{10} (\begin{array}{cc} 7 - 4 & - 1 \\ - 2 & 7 - 3 \end{array}) = \frac{1}{10} (\begin{array}{cc} 3 & - 1 \\ - 2 & 4 \end{array})$ . Check: $(\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) (\begin{array}{cc} 3 & - 1 \\ - 2 & 4 \end{array}) = (\begin{array}{cc} 10 & 0 \\ 0 & 10 \end{array})$ ✓

Q5 · full method — try it on paper first

The matrix $𝐀 = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array})$ has eigenvalues $5$ and $2$ .

(i) Find corresponding eigenvectors and write down $𝐐$ and $𝐃$ with $𝐀 = {𝐐 𝐃 𝐐}^{- 1}$ . (ii) Hence find a formula for $𝐀^{n}$ . (iii) Verify your formula at $n = 2$ .

Solution

(i) For $λ = 5$ , $(𝐀 - 5 𝐈) 𝐞 = 𝟎$ gives $- x + y = 0$ , so $𝐞_{1} = (\begin{array}{c} 1 \\ 1 \end{array})$ . For $λ = 2$ , $2 x + y = 0$ gives $𝐞_{2} = (\begin{array}{c} 1 \\ - 2 \end{array})$ . Taking the eigenvectors as columns in the matching order,

𝐐 = (\begin{array}{cc} 1 & 1 \\ 1 & - 2 \end{array}), 𝐃 = (\begin{array}{cc} 5 & 0 \\ 0 & 2 \end{array}) .

Since $\det 𝐐 = - 2 - 1 = - 3$ , $𝐐^{- 1} = \frac{1}{- 3} (\begin{array}{cc} - 2 & - 1 \\ - 1 & 1 \end{array}) = \frac{1}{3} (\begin{array}{cc} 2 & 1 \\ 1 & - 1 \end{array})$ .

(ii) $𝐀^{n} = {𝐐 𝐃}^{n} 𝐐^{- 1}$ . First ${𝐐 𝐃}^{n} = (\begin{array}{cc} 5^{n} & 2^{n} \\ 5^{n} & - 2 \cdot 2^{n} \end{array})$ , and then

𝐀^{n} = \frac{1}{3} (\begin{array}{cc} 5^{n} & 2^{n} \\ 5^{n} & - 2 \cdot 2^{n} \end{array}) (\begin{array}{cc} 2 & 1 \\ 1 & - 1 \end{array}) = \frac{1}{3} (\begin{array}{cc} 2 \cdot 5^{n} + 2^{n} & 5^{n} - 2^{n} \\ 2 \cdot 5^{n} - 2 \cdot 2^{n} & 5^{n} + 2 \cdot 2^{n} \end{array}) .

(iii) At $n = 2$ the formula gives

\frac{1}{3} (\begin{array}{cc} 50 + 4 & 25 - 4 \\ 50 - 8 & 25 + 8 \end{array}) = \frac{1}{3} (\begin{array}{cc} 54 & 21 \\ 42 & 33 \end{array}) = (\begin{array}{cc} 18 & 7 \\ 14 & 11 \end{array}),

and multiplying directly, $𝐀^{2} = (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) (\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}) = (\begin{array}{cc} 18 & 7 \\ 14 & 11 \end{array})$ ✓ The two agree. $■$

(Testing $n = 1$ as well is worth the ten seconds: it gives $\frac{1}{3} (\begin{array}{cc} 12 & 3 \\ 6 & 9 \end{array}) = 𝐀$ ✓)

FP2 · Topic 9 · Matrices II

Eigenvalues & Eigenvectors

What the examiner expects you to do

§1 Three equations, three planes

§2 Eigenvalues and eigenvectors

§3 The 3 × 3 case

§4 Diagonalisation: 𝐀=𝐐𝐃𝐐−1

§5 Cayley–Hamilton

§6 Interactive: watch a vector line up with its eigenvector

§7 Examiner traps & A* checklist

§8 Video explainers (curated)

§9 Check yourself

§4 Diagonalisation: $𝐀 = {𝐐 𝐃 𝐐}^{- 1}$