University of Debrecen
Diagonalization and eigenvalues Diagonalization We have seen that if A is an n n square matrix, then A is diagonalizable if and only if for all λ eigenvalues of A we have dim(u λ ) = mult(λ). the sum of the geometric multiplicities of the eigenvalues is n. Or equivalently, A is diagonalizable if and only if it has n linearly independent eigenvectors.
Example Consider the matrix 2 4 A = 1 2 2 4 1 3 6 1 Computing det(a λi 4 ), we get λ 2 ( 1 λ) 2, thus the eigenvalues of A are λ 1 = 1 and λ 2 = with mult( 1) = 2 and mult() = 2.
Example The eigenvectors corresponding to λ 1 = 1 are v 1 = 1 and v 2 =. 1 However, the only eigenvector we find for λ 2 = is 2 v 3 = 1. What happens now?
Jordan Normal Form Jordan Normal Form For a given n n square matrix A, there exists an invertible matrix S, such that S 1 AS = diag(j 1,... J p ), where J i is the square matrix λ i 1... λ i 1... J i =......... λ i and is called the Jordan-block corresponding to the ith eigenvalue.
Example So for the matrix A in our previous example, we should have 1 S 1 AS = 1 1 Now, we just have to find a suitable matrix, S.
Generalized eigenvectors Definition Suppose that λ is an eigenvalue of the n n square matrix A, with multiplicity k 1. Then the vectors v R n are called generalized eigenvectors of A, if (A λi n ) k v =. An important property If mult(λ) = k, then there are exactly k generalize eigenvectors corresponding to λ. Another important property If the columns of S are the generalized eigenvectors of A, then S 1 AS will have Jordan form.
How can we find the generalized eigenvectors? From our previous example, we see that AS = SJ, where J is the Jordan form of A. If S = (v 1 v 2 v 3 v 4 ), then this means, that 1 A(v 1 v 2 v 3 v 4 ) = (v 1 v 2 v 3 v 4 ) 1 1 So, we have (A + 1I 4 )v 1 = (A + 1I 4 )v 2 = (A + I 4 )v 3 = (A + I 4 )v 4 = v 3 So, (A + I 4 ) 2 v 4 = (A + I 4 )v 3 =.
How can we find the generalized eigenvectors? So, we have to compute (A + I 4 ) 2, and solve the linear equation Doing so yields v t 1 1 2 3 (A + I 4 ) 2 v =. + t 2 1 4 6, t 1, t 2 R. Now, we pick a vector v 4 from the solution set, for which (A I 4 )v 4. Then, we can choose v 3 = (A I 4 )v 4.
How can we find the generalized eigenvectors? In this case, pick and then v 4 = 1 2 3, 2 v 3 = 1.
In general Chains Suppose that λ is an eigenvalue of A of multiplicity k 2, and suppose, that we have found a generalized eigenvector, v k of λ. (Or, in other words, solved the equation (A λi n ) k v =.) Then we have the following Jordan chain. v k 1 = (A λi n )v k,..., v i 1 = (A λi n )v i,... v 1 = (A λi n )v 2. The vectors v 1, v 2,..., v k are the generalized eigenvalues corresponding to λ, and they generate the generalized eigenspace corresponding to λ.
Summing it up For our example A, we can choose 2 1 2 4 1 S = 1 1 2, so S 1 = 3 6 1 1. 1 3 1 2 And this way 1 S 1 AS = 1 1
Shur Decomposition Since the Jordan form is numerically unstable, in computations usually Shur decomposition is used. Shur Decomposition If A is an n n square matrix, than A can be expressed as A = QUQ 1, where Q is an orthogonal matrix (i.e. Q t Q = I n ), and U is an upper triangular matrix. Shur decomposition can be calculated for example with the QR-algorithm.
Singular Decomposition Singular Values Decomposition If A is and m n real matrix, then the singluar value decomposition of A is the product A = USV t, where U is an m m orthogonal matrix, S is an m n rectangular diagonal matrix, and V t is an n n orthogonal matrix. Nomenclature The diagonal entries of S are called singular values of A. The columns of U are called left-singular vectors of A. The columns of V are called right-singular vectors of A.
How to compute SVD? Computation The computation of the singular value decomposition is done through eigenvector and eigenvalue calculations. Namely The non-zero singular values of A (i.e. the diagonal entries of S) are the square roots of the non-zero eigenvectors of both A t A and AA t. The left-singular vectors of A (i.e. the columns of U) are the eigenvectors of AA t. The right singular vectors of A (i.e. the columns of V ) are the eigenvectors of A t A