A special direction the matrix only stretches: Most vectors get knocked sideways when you multiply by a matrix A. But a few special directions come out pointing the same way — just longer or shorter.
Eigenvalue and eigenvector
- For those directions, Av = λv: the matrix acts like plain multiplication by a number λ.
- That number is the eigenvalue.
- The direction v is the eigenvector.
Why do we care?: In a population or network model the matrix is applied over and over. The eigenvalue λ = 1 picks out the mix that never changes — the long-run steady state.
Where the equation comes from: Rearrange Av = λv to Av − λv = 0, i.e. (A − λI)v = 0 (the I keeps it a matrix subtraction).
We want a non-zero v, so the matrix A − λI must be singular — its determinant is zero.
That single condition, det(A − λI) = 0, is the characteristic equation. Solve it for λ.
IB-style question — find the eigenvalues
A city splits people between Downtown (D) and Suburbs (S). Each year the population is updated by
A = [[0.8, 0.3], [0.2, 0.7]].
Find the eigenvalues of A.
Step by step
- Write A − λI by subtracting λ down the diagonal.
- Set the determinant (top-left × bottom-right − the other diagonal) to 0.
- Expand: 0.56 − 1.5λ + λ² − 0.06 = 0, which tidies to the characteristic equation.
- Factorise (or use the GDC's polynomial solver).
- Read off the two eigenvalues.
Final answer
Eigenvalues λ = 1 and λ = 0.5. (The λ = 1 here is the giveaway that A is a transition matrix — its columns each add to 1.)
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One eigenvector per eigenvalue, then bundle them up: For each eigenvalue λ, the eigenvector is any non-zero v solving (A − λI)v = 0.
Pick the simplest: The two rows always give the same relationship between x and y, so just pick the simplest whole-number vector.
Diagonalise
Find both eigenvectors
Once you have both eigenvectors, diagonalise.
Build P
Put the eigenvectors as the columns of P.
Build D
Put the matching eigenvalues down the diagonal of D.
Write A
Then A = PDP⁻¹.
Eigenvectors in P, eigenvalues in D.
Gold for powers: Applying A many times becomes Aⁿ = PDⁿP⁻¹, and raising the diagonal D to a power is just raising each entry to that power.
IB-style question — find the eigenvectors
For the same A = [[0.8, 0.3], [0.2, 0.7]] (eigenvalues 1 and 0.5), find an eigenvector for each eigenvalue.
Step by step
- λ = 1: solve (A − I)v = 0.
- Both rows say −0.2x + 0.3y = 0, i.e. y = (2/3)x. Choose x = 3.
- λ = 0.5: solve (A − 0.5I)v = 0.
- Both rows say x + y = 0, i.e. y = −x. Choose x = 1.
Final answer
Eigenvector (3, 2) for λ = 1; eigenvector (1, −1) for λ = 0.5. Any non-zero multiple is also correct.
IB-style question — diagonalise
Using those eigenvectors, write down matrices P and D so that A = PDP⁻¹.
Step by step
- P holds the eigenvectors as columns (same order as the eigenvalues in D).
- D holds the eigenvalues on the diagonal (λ = 1 first, matching the first column).
- The order must match: column 1 of P (for λ = 1) lines up with the 1 in D.
Final answer
P = [[3, 1], [2, −1]], D = [[1, 0], [0, 0.5]], with A = PDP⁻¹. (Swap both an eigenvalue and its column together and it still works.)