Diagonalisation

If eigenvectors are the "skeleton" of a matrix (Lesson 8), diagonalisation is the act of changing to a coordinate system built from that skeleton — and in those coordinates the matrix becomes utterly simple: a diagonal matrix that merely scales each axis. Formally, a matrix $A$A with enough independent eigenvectors can be written $A = PDP^{-1}$A=PDP−1, where $D$D carries the eigenvalues down its diagonal and $P$P carries the eigenvectors as its columns. This factorisation is one of the most useful in the whole subject, because it makes powers trivial: $A^n = PD^nP^{-1}$An=PDnP−1, and raising a diagonal matrix to a power just raises each diagonal entry to that power. That single idea lets you compute $A^{100}$A100 without a hundred multiplications, solve linear recurrence systems in closed form, integrate systems of differential equations, and read off the long-term behaviour of a repeated transformation at a glance. This lesson develops the factorisation rigorously, gives full $2\times2$2×2 and $3\times3$3×3 worked examples with verification, derives the $A^n = PD^nP^{-1}$An=PDnP−1 result, and pins down exactly when diagonalisation is — and is not — possible.

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1. Where this sits in AQA 7367

Diagonalisation is compulsory pure content for Papers 1 and 2, building directly on eigenvalues/eigenvectors (Lesson 8) and feeding into Cayley–Hamilton (Lesson 10). Constructing $P$P and $D$D and verifying $A = PDP^{-1}$A=PDP−1 is AO1/AO2; computing $A^n$An via $PD^nP^{-1}$PDnP−1 is AO1; deciding whether a matrix is diagonalisable, and justifying it through the count of independent eigenvectors, is AO2/AO3; and applying diagonalisation to a recurrence or a "find $A^n$An in closed form" problem is AO3. The mechanics lean on inverses (Lesson 4: you need $P^{-1}$P−1) and matrix multiplication (Lesson 2). A typical exam item gives a $2\times2$2×2 or $3\times3$3×3 matrix and asks you to find $P$P and $D$D, then "hence" find $A^n$An or solve a recurrence — so fluency in the whole pipeline (eigenvalues $\to$→ eigenvectors $\to$→ $P,D$P,D $\to$→ $P^{-1}$P−1 $\to$→ $PD^nP^{-1}$PDnP−1) is what earns the marks, and an error anywhere propagates, which is why the verification step is so valuable.

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2. The diagonalisation factorisation

Suppose the $n\times n$n×n matrix $A$A has $n$n linearly independent eigenvectors $\mathbf{v}_1,\dots,\mathbf{v}_n$v1​,…,vn​ with eigenvalues $\lambda_1,\dots,\lambda_n$λ1​,…,λn​. Form

$$P = \bigl(\,\mathbf{v}_1\ \mid\ \mathbf{v}_2\ \mid\ \cdots\ \mid\ \mathbf{v}_n\,\bigr), \qquad D = \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}.$$P=(v1​ ∣ v2​ ∣ ⋯ ∣ vn​),D=​λ1​​⋱​λn​​​.

Then $AP = PD$AP=PD, and since the independence of the eigenvectors makes $P$P invertible,

$$\boxed{\;A = PDP^{-1},\qquad\text{equivalently}\qquad D = P^{-1}AP.\;}$$A=PDP−1,equivalentlyD=P−1AP.​

Why $AP = PD$AP=PD? The $j$j-th column of $AP$AP is $A\mathbf{v}_j = \lambda_j\mathbf{v}_j$Avj​=λj​vj​; the $j$j-th column of $PD$PD is $P$P times the $j$j-th column of $D$D, which is $\lambda_j\mathbf{v}_j$λj​vj​. The columns agree, so $AP = PD$AP=PD. This is the entire derivation, and it makes the rule for building $P$P and $D$D unforgettable: eigenvectors as the columns of $P$P, their eigenvalues in the same order down $D$D. The order is free, but $P$P and $D$D must use the same order — column $j$j of $P$P must be the eigenvector for the $j$j-th diagonal entry of $D$D.

There is a deeper way to read $A = PDP^{-1}$A=PDP−1 that makes every later result obvious. The matrix $P^{-1}$P−1 translates a vector from standard coordinates into eigen-coordinates (its components along the eigenvectors); $D$D then scales each eigen-coordinate by the corresponding eigenvalue; and $P$P translates back to standard coordinates. So $A$A "really is" a diagonal scaling — it only looks complicated because we insist on writing it in the standard basis. This change-of-basis viewpoint is why diagonalisation is sometimes called finding the natural coordinates of a transformation, and it is the reason the eigenbasis decouples everything: in those coordinates the axes are exactly the invariant directions, so the transformation cannot mix them.

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3. When is a matrix diagonalisable?

A matrix is diagonalisable if and only if it has $n$n linearly independent eigenvectors. Two sufficient conditions cover almost every exam case:

• $n$n distinct eigenvalues $\Rightarrow$⇒ diagonalisable. (Eigenvectors for distinct eigenvalues are automatically independent — proved in Lesson 8.)
• $A$A real and symmetric $\Rightarrow$⇒ diagonalisable, with real eigenvalues and orthogonal eigenvectors (the spectral theorem).

Diagonalisation fails only when a repeated eigenvalue fails to supply as many independent eigenvectors as its multiplicity — a defective matrix. The standard example is the shear $\binom{1\ 1}{0\ 1}$(0 11 1​): eigenvalue $1$1 twice, but only one eigendirection, so it is not diagonalisable. The logic is always: distinct eigenvalues are safe; at a repeated eigenvalue, count the independent eigenvectors before claiming diagonalisability.

The precise bookkeeping uses two numbers for each eigenvalue. The algebraic multiplicity is how many times $\lambda$λ appears as a root of the characteristic polynomial; the geometric multiplicity is the number of independent eigenvectors it produces — the dimension of the eigenspace $\ker(A-\lambda I)$ker(A−λI). It is always true that $1 \le \text{geometric} \le \text{algebraic}$1≤geometric≤algebraic. A matrix is diagonalisable exactly when every eigenvalue has geometric multiplicity equal to its algebraic multiplicity (so the eigenspaces together fill all $n$n dimensions). A simple eigenvalue (algebraic multiplicity $1$1) automatically satisfies this — which is why distinct eigenvalues always give a diagonalisable matrix. The only thing that can go wrong is a repeated eigenvalue whose eigenspace is too small, as in the shear, where eigenvalue $1$1 has algebraic multiplicity $2$2 but geometric multiplicity $1$1.

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

Diagonalise $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$A=(42​13​), and verify $PDP^{-1} = A$PDP−1=A.

From Lesson 8: eigenvalues $\lambda_1 = 5,\ \lambda_2 = 2$λ1​=5, λ2​=2, with eigenvectors $\mathbf{v}_1 = \binom{1}{1},\ \mathbf{v}_2 = \binom{1}{-2}$v1​=(11​), v2​=(−21​). Two distinct eigenvalues $\Rightarrow$⇒ diagonalisable. Take

$$P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \qquad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}.$$P=(11​1−2​),D=(50​02​).

Compute $P^{-1}$P−1: $\det P = (1)(-2) - (1)(1) = -3$detP=(1)(−2)−(1)(1)=−3, so $P^{-1} = \dfrac{1}{-3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}$P−1=−31​(−2−1​−11​)=(2/31/3​1/3−1/3​).

Verify. $PD = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}$PD=(11​1−2​)(50​02​)=(55​2−4​), then

$$PDP^{-1} = \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}\begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix} = \begin{pmatrix} 12/3 & 3/3 \\ 6/3 & 9/3 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A.\ ✓$$PDP−1=(55​2−4​)(2/31/3​1/3−1/3​)=(12/36/3​3/39/3​)=(42​13​)=A. ✓

Mark scheme: B1 eigenvalues/eigenvectors quoted or found; M1 form $P$P and $D$D consistently (same order); M1 compute $P^{-1}$P−1; A1 correct $P^{-1}$P−1; M1 multiply $PDP^{-1}$PDP−1; A1 obtain $A$A, confirming. Mismatched ordering of $P$P and $D$D is the classic error and forfeits the verification.

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5. Powers via diagonalisation: $A^n = PD^nP^{-1}$An=PDnP−1

Here is the pay-off. If $A = PDP^{-1}$A=PDP−1 then

$$A^2 = (PDP^{-1})(PDP^{-1}) = PD\underbrace{(P^{-1}P)}_{I}DP^{-1} = PD^2P^{-1},$$A2=(PDP−1)(PDP−1)=PDI(P−1P)​​DP−1=PD2P−1,

and by the same telescoping (the inner $P^{-1}P = I$P−1P=I cancels at every join),

$$\boxed{\;A^n = PD^nP^{-1},\qquad D^n = \begin{pmatrix} \lambda_1^{\,n} & & \\ & \ddots & \\ & & \lambda_n^{\,n} \end{pmatrix}.\;}$$An=PDnP−1,Dn=​λ1n​​⋱​λnn​​​.​

Raising the diagonal matrix to the $n$n-th power is trivial — just power each entry — so the only real work is one matrix inverse and two products, regardless of how large $n$n is.

Worked Example 2 — a clean closed-form power (with mark scheme)

Find $A^n$An in closed form for $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$A=(30​12​).

$A$A is upper triangular, so eigenvalues are $\lambda = 3, 2$λ=3,2 (distinct $\Rightarrow$⇒ diagonalisable). Eigenvectors: for $\lambda=3$λ=3, $(A-3I)\mathbf v = \binom{0\ \ 1}{0\,-1}\mathbf v = \mathbf 0\Rightarrow y=0$(A−3I)v=(0−10  1​)v=0⇒y=0, so $\binom{1}{0}$(01​); for $\lambda=2$λ=2, $(A-2I)\mathbf v = \binom{1\ 1}{0\ 0}\mathbf v = \mathbf 0\Rightarrow x=-y$(A−2I)v=(0 01 1​)v=0⇒x=−y, so $\binom{1}{-1}$(−11​). Thus

$$P = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix},\quad D = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix},\quad P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}$$P=(10​1−1​),D=(30​02​),P−1=(10​1−1​)

(note $\det P = -1$detP=−1, and this $P$P happens to be self-inverse). Then

$$A^n = P\begin{pmatrix} 3^n & 0 \\ 0 & 2^n \end{pmatrix}P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 3^n & 0 \\ 0 & 2^n \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 3^n & 3^n - 2^n \\ 0 & 2^n \end{pmatrix}.$$An=P(3n0​02n​)P−1=(10​1−1​)(3n0​02n​)(10​1−1​)=(3n0​3n−2n2n​).

(Check $n=1$n=1: $\binom{3\ \,1}{0\ \,2}$(0 23 1​) ✓; $n=2$n=2: $\binom{9\ \,5}{0\ \,4}$(0 49 5​), and direct squaring gives the same.)

Mark scheme: M1 eigenvalues; M1 eigenvectors; A1 $P, D, P^{-1}$P,D,P−1; M1 form $PD^nP^{-1}$PDnP−1; A1 closed form $\binom{3^n\ \ 3^n-2^n}{0\ \ \ \ 2^n}$(0    2n3n  3n−2n​). Verifying at $n=1$n=1 secures the accuracy.

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6. The 3×3 case

Nothing changes in principle: find three eigenvalues and three independent eigenvectors, stack the eigenvectors as the columns of a $3\times3$3×3 matrix $P$P, put the eigenvalues in matching order down a $3\times3$3×3 diagonal $D$D, and $A = PDP^{-1}$A=PDP−1. The only extra labour is the $3\times3$3×3 inverse $P^{-1}$P−1 (Lesson 4). For example, if $A$A has eigenvalues $1, 2, 3$1,2,3 with eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3$v1​,v2​,v3​, then

$$A = P\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}P^{-1}, \qquad A^n = P\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 3^n \end{pmatrix}P^{-1},$$A=P​100​020​003​​P−1,An=P​100​02n0​003n​​P−1,

with $P = (\mathbf{v}_1\mid\mathbf{v}_2\mid\mathbf{v}_3)$P=(v1​∣v2​∣v3​). Three distinct eigenvalues guarantee the three eigenvectors are independent, so $P$P is invertible and the diagonalisation goes through.

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7. Applications — why diagonalisation matters

Application	How diagonalisation is used
Large powers $A^n$An	$A^n = PD^nP^{-1}$An=PDnP−1 — one inverse, then power the diagonal
Linear recurrences	$\mathbf{x}_{n+1} = A\mathbf{x}_n \Rightarrow \mathbf{x}_n = A^n\mathbf{x}_0 = PD^nP^{-1}\mathbf{x}_0$xn+1​=Axn​⇒xn​=Anx0​=PDnP−1x0​
Systems of ODEs	$\mathbf{x}' = A\mathbf{x}$x′=Ax decouples into $n$n scalar equations in eigen-coordinates
Stability / long-run behaviour	governed by the largest $\lvert\lambda_i\rvert$∣λi​∣; if all $\lvert\lambda_i\rvert<1$∣λi​∣<1, $A^n\to O$An→O

The common thread is that in the eigenbasis the transformation decouples: the messy coupled action of $A$A becomes $n$n independent one-dimensional scalings, which we can analyse, power, or integrate term by term, and then transform back via $P$P.

Worked Example 3 — a recurrence relation via diagonalisation (with mark scheme)

A sequence of pairs satisfies $\mathbf{x}_{n+1} = A\mathbf{x}_n$xn+1​=Axn​ with $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$A=(10​23​) and $\mathbf{x}_0 = \binom{1}{1}$x0​=(11​). Find a closed form for $\mathbf{x}_n$xn​.

The iteration gives $\mathbf{x}_n = A^n\mathbf{x}_0$xn​=Anx0​. Diagonalise $A$A: triangular, so eigenvalues $1, 3$1,3. For $\lambda=1$λ=1: $\binom{0\ 2}{0\ 2}\mathbf v=\mathbf0\Rightarrow y=0$(0 20 2​)v=0⇒y=0, $\binom{1}{0}$(01​); for $\lambda=3$λ=3: $\binom{-2\ \ 2}{0\ \ 0}\mathbf v=\mathbf0\Rightarrow x=y$(0  0−2  2​)v=0⇒x=y, $\binom{1}{1}$(11​). Hence $P = \binom{1\ 1}{0\ 1}$P=(0 11 1​), $D = \binom{1\ 0}{0\ 3}$D=(0 31 0​), $P^{-1} = \binom{1\,-1}{0\ \ 1}$P−1=(0  11−1​). Then

$$A^n = P\begin{pmatrix} 1 & 0 \\ 0 & 3^n \end{pmatrix}P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 3^n \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3^n - 1 \\ 0 & 3^n \end{pmatrix},$$An=P(10​03n​)P−1=(10​11​)(10​03n​)(10​−11​)=(10​3n−13n​),

so $\mathbf{x}_n = A^n\binom{1}{1} = \begin{pmatrix} 1 & 3^n-1 \\ 0 & 3^n \end{pmatrix}\binom{1}{1} = \binom{3^n}{3^n}$xn​=An(11​)=(10​3n−13n​)(11​)=(3n3n​). (Check $n=0$n=0: $\binom{1}{1} = \mathbf{x}_0$(11​)=x0​ ✓; $n=1$n=1: $A\binom{1}{1} = \binom{3}{3}$A(11​)=(33​) ✓.)

Mark scheme: M1 recognise $\mathbf{x}_n = A^n\mathbf{x}_0$xn​=Anx0​; M1 diagonalise; A1 $A^n$An; M1 multiply by $\mathbf{x}_0$x0​; A1 $\mathbf{x}_n = \binom{3^n}{3^n}$xn​=(3n3n​). Verifying at $n=0,1$n=0,1 protects the accuracy marks.