Eigenvalues and Eigenvectors. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Transcription

1 Eigenvalues and Eigenvectors

2 Consider the equation A x = λ x, where A is an nxn matrix. We call x (must be non-zero) an eigenvector of A if this equation can be solved for some value of λ. We call λ an eigenvalue of matrix A, and we call x an eigenvector corresponding to λ.

3 Consider the equation A x = λ x, where A is an nxn matrix. We call x (must be non-zero) an eigenvector of A if this equation can be solved for some value of λ. We call λ an eigenvalue of matrix A, and we call x an eigenvector corresponding to λ. We can rearrange the equation to see how to solve for λ and x: λ is an eigenvalue of A if, and only if the equation A λi x = 0 has a nontrivial solution. So the set of solutions is just the nullspace of (A λi). This is a subspace of R n and we call it the eigenspace corresponding to the eigenvalue λ. nxn identity matrix

4 Example: Consider the matrix A = An eigenvalue of this matrix is λ = 2. Find the associated eigenvectors.

5 Example: Consider the matrix A = An eigenvalue of this matrix is λ = 2. Find the associated eigenvectors. We need to find the nullspace of A λi : A 2I = = Nullspace will be vectors satisfying x + 2x 2 = 0 x = 2 is an eigenvector. The eigenspace is the -dimensional subspace of R 2 consisting of all multiples of 2.

6 Example: Consider the matrix A = An eigenvalue of this matrix is λ = 2. Find the associated eigenvectors. We need to find the nullspace of A λi : A 2I = = Nullspace will be vectors satisfying x + 2x 2 = 0 x = 2 is an eigenvector. The eigenspace is the -dimensional subspace of R 2 consisting of all multiples of 2. We can check that the eigenvector equation A x = λ x is satisfied: = 4 2 = 2 2

7 We know how to find eigenvectors, but only if we know what eigenvalue to use. How do we find eigenvalues? Here is the equation we need to solve, and we want to find nontrivial solutions: A λi x = 0

8 We know how to find eigenvectors, but only if we know what eigenvalue to use. How do we find eigenvalues? Here is the equation we need to solve, and we want to find nontrivial solutions: A λi x = 0 The only way to get non-trivial solutions is if the determinant is 0. This gives us an equation we can solve for λ. A scalar λ is an eigenvalue of an nxn matrix A if and only if λ satisfies the characteristic equation det A λi = 0.

9 We know how to find eigenvectors, but only if we know what eigenvalue to use. How do we find eigenvalues? Here is the equation we need to solve, and we want to find nontrivial solutions: A λi x = 0 The only way to get non-trivial solutions is if the determinant is 0. This gives us an equation we can solve for λ. A scalar λ is an eigenvalue of an nxn matrix A if and only if λ satisfies the characteristic equation det A λi = 0. Solving the characteristic equation will give all the eigenvalues of the matrix. They could be real (possibly repeated) or complex. We will look at several examples.

10 Example : Consider the matrix A =

11 Example : Consider the matrix A = First we solve the characteristic equation to find the e-values: 3 λ λ = 0 3 λ 8 λ 6 = 0 λ2 λ + 8 = 0 λ 2 λ 9 = 0 λ = 2 or λ = 9

12 Example : Consider the matrix A = First we solve the characteristic equation to find the e-values: 3 λ λ = 0 3 λ 8 λ 6 = 0 λ2 λ + 8 = 0 λ 2 λ 9 = 0 λ = 2 or λ = 9 Here we have 2 distinct, real eigenvalues. We expect to get a -dimensional eigenspace for each of them. We already found the e-vectors for λ = 2. For λ = 9 we need to find the nullspace of (A-9I): x x 2 = 0 x = 3.

13 Example : Consider the matrix A = Here are our results: Eigenvalue Eigenspace λ = 2 2 span λ 2 = 9 span 3

14 Example : Consider the matrix A = Here are our results: Eigenvalue Eigenspace λ = 2 2 span λ 2 = 9 span 3 We can give this a geometrical interpretation. The matrix A represents a linear transformation from R 2 R 2, and the eigenspaces are preserved under that transformation. i.e. any vector that is in an eigenspace is mapped to another vector in that eigenspace (scaled by the eigenvalue).

15 Example : Consider the matrix A = Here are our results: Eigenvalue Eigenspace λ = 2 2 span λ 2 = 9 span This vector is 9 times as long 0 e-vector for λ=2 e-vector for λ= This vector is twice as long

16 2 2 Example 2: Consider the matrix A =

17 2 2 Example 2: Consider the matrix A = First we find the characteristic equation: 2 λ 2 det 3 λ = λ

18 2 2 Example 2: Consider the matrix A = First we find the characteristic equation: 2 λ 2 det 3 λ = λ Cofactor expansion along the first column: 2 λ 3 λ 2 2 λ λ λ = 0 2 λ 3 λ 2 λ λ λ = 0 2 λ λ 2 5λ λ 3 = 0 2 λ λ )(λ 4 + 3(λ ) = 0 λ λ λ 5 = 0 λ = 5 or λ = repeated We say the eigenvalue λ= has multiplicity 2

19 2 2 Example 2: Consider the matrix A = Find eigenvectors corresponding to each eigenvalue: For λ = 5, find nullspace of x = For λ =, find nullspace of x = or x = 0

20 2 2 Example 2: Consider the matrix A = Here are our results: Eigenvalue λ = (multiplicity 2) span x = λ = 5 Eigenspace 2 0 span x 3 =, x 2 = 0 This time we have a repeated e-value, and we get a 2-dimensional eigenspace (won t always happen). So any vector that starts in the plane formed by x and x 2 will stay in that plane. In fact, since the e-value is, that vector won t change at all. The other e-value has multiplicity, so we get a -dimensional eigenspace (a line) just as in the previous example. Any vector starting on that line will stay the same direction, but be transformed to a vector 5 times longer. The diagram on the next slide shows these transformations.

21 2 2 Example 2: Consider the matrix A = x 3 5 x 3 x 2 x 2 x x Original e-vectors (zoomed in so you can see them better) Transformed e-vectors

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