Matrix powers
With A = P D P^(-1), the power A^n equals P D^n P^(-1). Raise each diagonal entry of D to the n-th power.
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Analogy
Repeating a simple stretch many times is much easier than repeating a messy one.
Example
v is an eigenvector with L = 5.A v = 5v, A^2 v = 25v.A^3 v = 125 v.Result (125, 125).
In practice
Predicts the long-run behaviour of repeated processes, from populations to web page rankings.
Practice
Flashcards
- How do you compute A^n using diagonalisation?A^n = P D^n P^(-1).
Interview questions
- Explain how diagonalisation helps compute high powers of a matrix.
- A = P D P^(-1)
- A^n = P D^n P^(-1)
- D^n only needs each diagonal entry raised to n
- Largest eigenvalue dominates the long run
Questions people ask
What happens to A to the n as n gets large?
The eigenvalue of largest size dominates. Eigenvalues bigger than 1 in size make components grow, those smaller than 1 make them shrink to zero. For eigenvalues 5 and 2, the 5 part wins easily over time, since 5 to the tenth is far larger than 2 to the tenth.
Why is computing A to the n directly hard?
Multiplying a matrix by itself n times is slow and the numbers grow large. With the diagonal form you only raise the diagonal numbers to the power n, then do two matrix multiplications. It makes huge powers practical. A power like 50 or 100 takes almost no extra effort.