Characteristic equation
The eigenvalues of A are the solutions of det(A - lambda I) = 0. This gives a polynomial in lambda.
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Analogy
A lock whose combination is the set of stretch factors that make the matrix collapse.
Example
det = (4 - L)(3 - L) - (1)(2).= L^2 - 7L + 10.= (L - 5)(L - 2).Eigenvalues L = 5 and L = 2.
In practice
The first step in finding eigenvalues by hand.
Practice
Flashcards
- How do you find eigenvalues?Solve det(A - lambda I) = 0.
- What are the eigenvalues of [[4,1],[2,3]]?5 and 2.
Interview questions
- Find the eigenvalues and eigenvectors of A = [[4,1],[2,3]].
- Characteristic equation L^2 - 7L + 10 = 0
- Eigenvalues 5 and 2
- For 5: (A - 5I)v = 0 gives v = (1,1)
- For 2: gives v = (1,-2)
- Check using trace 7 and determinant 10
Questions people ask
Why does det(A minus lambda I) equal zero?
We need A v equals lambda v for a non-zero v, which means A minus lambda I sends v to zero. A matrix that kills a non-zero vector has no inverse, so its determinant must be zero. Solving this finds each allowed lambda.
How many eigenvalues does an n by n matrix have?
Up to n, counting repeats, and allowing complex numbers. The characteristic polynomial has degree n, so it has n roots in the complex numbers. A 2 by 2 matrix has two eigenvalues, though they may be equal or complex. For example, a rotation has two complex eigenvalues.