IB Maths AI HLEigenvalues & EigenvectorsPaper 1 & 2Diagonalisation & powers~8 min read
Diagonalisation & Powers of Matrices
A diagonal matrix is the friendliest kind of matrix — its powers are read off in one step. If M = PDP−1 for some diagonal D, then Mn = PDnP−1 for free. The trick uses the eigenvalues and eigenvectors from the previous note.
📘 What you need to know
A square matrix is diagonal if every off-diagonal entry is zero.
A 2×2 matrix is diagonalisable iff it has two real distinct eigenvalues.
To diagonalise: put the eigenvalues on the diagonal of D; the corresponding eigenvectors as columns of P, in the matching order.
M = PDP−1 ⇔ D = P−1MP.
Powers of a diagonal: ((a, 0), (0, b))n = ((an, 0), (0, bn)) — in the formula booklet.
Powers of M: Mn = PDnP−1 — in the formula booklet.
Diagonal and diagonalisable matrices
A diagonal matrix is square with zeros everywhere off the leading diagonal — e.g. ((1, 0, 0), (0, 3, 0), (0, 0, −1)). A general matrix M is diagonalisable if there exists a matrix P such that P−1MP is diagonal, or equivalently M = PDP−1 for some diagonal D.
Scope: in IB AI HL you only need to diagonalise 2×2 matrices with real distinct eigenvalues. If there is only one (repeated) eigenvalue the matrix is either already diagonal or cannot be diagonalised; complex eigenvalues are outside the syllabus.
Diagonalising a 2×2 matrix
For a 2×2 matrix M with real distinct eigenvalues λ1, λ2 and eigenvectors v1, v2, the diagonalisation is built directly from them: D takes the eigenvalues along its diagonal, and P takes the eigenvectors as columns, in the same order.
Different orderings give different (but equally valid) diagonalisations — swapping the columns of P just swaps the entries of D. What matters is that column j of P pairs with diagonal entry j of D.
Powers via Mn = PDnP−1
Computing M5 directly takes five 2×2 multiplications — tedious. The diagonalisation shortcut collapses this to one easy step plus two cheap matrix products: take the eigenvalues to the nth power on the diagonal, then sandwich between P and P−1.
Direct multiplication: five 2×2 products for M5, more for higher powers. With diagonalisation, the exponent acts only on the two diagonal entries of D.
Find eigenvalues λ1, λ2 and corresponding eigenvectors v1, v2.
Form D with the eigenvalues on the diagonal; form P with the eigenvectors as columns, in the matching order.
Find P−1 using the 2×2 formula (or GDC).
Raise D to the nth power: just raise each diagonal entry to n.
MultiplyMn = P · Dn · P−1.
Worked examples
WE 1
Identify diagonal matrices
Which of the following are diagonal? A = ((2, 0), (0, 5)), B = ((1, 0), (2, 3)), C = ((4, 0, 0), (0, −1, 0), (0, 0, 3)), D = ((1, 0, 0), (0, 0, 2), (0, 3, 0)).
check every off-diagonal entry is 0A: off-diagonal entries 0, 0 ✓B: entry (2,1) = 2 ≠ 0 ✗C: all six off-diagonal entries 0 ✓D: entries (2,3) = 2 and (3,2) = 3 ≠ 0 ✗A and C are diagonaldiagonal matrices must be square; the ‘leading diagonal’ is the top-left to bottom-right.
WE 2
Powers of a diagonal matrix
Find D4 for D = ((2, 0), (0, −3)).
raise each diagonal entry to the 4th power2⁴ = 16(−3)⁴ = 81D⁴ = ((16, 0), (0, 81))even powers wipe out the sign; odd powers keep it.
WE 3
Assemble a diagonalisation
A matrix N has eigenvalues λ1 = 2 and λ2 = 5 with eigenvectors (1, −1)T and (2, 1)T respectively. Find P, D and P−1 so that N = PDP−1.
eigenvalues on the diagonal of DD = ((2, 0), (0, 5))eigenvectors as columns of P, matching the orderP = ((1, 2), (−1, 1))P⁻¹: det P = 1·1 − 2·(−1) = 3P⁻¹ = (1/3)((1, −2), (1, 1))D = ((2,0),(0,5)) · P = ((1,2),(−1,1)) · P⁻¹ = ⅓((1,−2),(1,1))swapping the two columns of P swaps the diagonal of D; both are valid.
WE 4
Use diagonalisation to find M³
For M = ((4, 3), (−1, 0)) the eigenvalues are 1 and 3 with eigenvectors (1, −1)T and (−3, 1)T. Use the diagonalisation to find M3.
For M = ((3, −2), (−4, 1)) the eigenvalues are 5 and −1 with eigenvectors (1, −1)T and (1, 2)T. (a) Write down P, D, P−1. (b) Hence find M3.
(a) eigenvalues on D, eigenvectors as columns of PD = ((5, 0), (0, −1)); P = ((1, 1), (−1, 2))det P = 2 + 1 = 3P⁻¹ = (1/3)((2, −1), (1, 1))(b) D³ = ((125, 0), (0, −1))PD³ = ((125, −1), (−125, −2))M³ = (1/3)((125, −1), (−125, −2))((2, −1), (1, 1))= (1/3)((249, −126), (−252, 123))M³ = ((83, −42), (−84, 41))verify with M·M·M on the GDC.
💡 Top tips
Order matters: column j of P must be the eigenvector for diagonal entry j of D.
Both M = PDP−1 and Mn = PDnP−1 are in the formula booklet — no need to derive.
Any non-zero scalar multiple of an eigenvector works as a column of P; pick the cleanest.
For high powers use the diagonalisation; for M2 or M3 direct multiplication is often quicker.
Always cross-check with the GDC if available.
âš Common mistakes
Eigenvectors as rows of P instead of columns — the formula won’t work.
Mismatched order: D = diag(λ1, λ2) but P‘s first column is the λ2-eigenvector.
Computing Dn as nD — powers are entry-wise on the diagonal.
Forgetting to invert P or computing det P incorrectly.
Trying to diagonalise a matrix with a repeated eigenvalue — this often fails.
That completes Eigenvalues & Eigenvectors. From the characteristic polynomial through eigenvectors to diagonalisation and powers, you have the full AI HL toolkit for 2×2 matrix dynamics.
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