Tutorial 6. Eigenvalues & Eigenvectors.

2 1. Reminder: Eigenvectors A vector x invariant up to a scaling by λ to a multiplication by matrix A is called an eigenvector of A with eigenvalue λ : The eigenvectors can be found via: This equation a has nontrivial (nonzero) solutions only for λ satisfying:

3 Reminder (continued) For an n-by-n matrix, this is a polynomial of order n. It is known as the characteristic polynomial. This poynomial has exactly n roots (possibly with multiplicities) λ 1, …, λ n. The eigenvector x j, corresponding to λ j, is found via solution of

4 Symmetric Matrices - Theorem Let A be a symmetric real n-by-n matrix. then: 1.All the eigenvalues are real λ 1, …, λ n. 2.Eigenvectors with different eigenvalues are orthogonal. 3.There is an orthonormal basis consisting of eigenvectors of A. Proof First, let us note that Indeed,

5 Symmetric Matrices (Continued) 1. In particular, for an eigenvector v, 2. For a pair of eigenvectors, v and w with different eigenvalues:

6 3. (There is an orthonormal basis consisting of eigenvectors of A) For the case d=1, the proof is trivial. We proceed by induction. Consider the general case d=n>1. The matrix A has at least one eigenvalue (solution, possibly multiple of characteristic equation) for which we have an eigenvector, and it is real due to (1). Let v 1 be the associated eigenvector. Let The dimension of is n-1, and for a vector w from it, the transformation by A keeps it within : Symmetric Matrices (Continued)

7 (3) In the subspace an orthonormal basis y 1,…y n-1 can be constructed. The linear transformation produced by A in is described by a n-1·n-1 matrix B in the basis of y 1,…y n-1. Consider the n·n-1 matrix Y, describing the transition from coordinates y 1,…y n-1, to x 1,…x n. Then, B=Y T AY, and therefore B is symmetric, and real. Therefore, B has at least one eigenvector, v B. Then for this eigenvector: We have shown that Yv B is another eigenvector of A, orthogonal to v 1. From here the proof continue by induction via d=n-2 to d=1. Symmetric Matrices (Continued)

8 Spectral Factorization In the basis of eigenvectors of matrix A, the action of A on arbitrary vector x is very simple: Now, consider an action of orthonormal matrix Q, built of eigenvectors of A, on an arbitrary vector x: Therefore,

9 Spectral Factorization (continued) Since for any vector x, Example: Perform a spectral factorization of the following matrix: First, let’s find the eigenvalues of A:

10 First, let’s find the eigenvalues of A: Now we can find corresponding eigenvectors: Spectral Factorization (continued)

11 Spectral Factorization (continued)

12 Spectral Factorization (continued)