The Jordan Normal Form and its Applications
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1 The and its Applications Jeremy IMPACT Brigham Young University
2 A square matrix A is a linear operator on {R, C} n. A is diagonalizable if and only if it has n linearly independent eigenvectors. What happens if A does not have n linearly independent eigenvectors? When does this happen? What general form can we obtain in this case?
3 The The is one decomposition of a matrix, A = P 1 JP where J is the normal form. It has the advantage of corresponding to the eigenspaces and of being as close to diagonal as possible. More specifically, if a matrix is diagonal then its Jordan Normal Form is the diagonalization.
4 Complementary Subspaces Definition Two subspaces U, W of a vector space V are complementary if U W = {0} and for all v V, there exist u U, w W such that v = u + w. In fact, u, w are the unique vectors that satisfy this property. We denote this V = U W.
5 Complementary Subspaces Remark This idea extends to finite collections: V = W 1 W 2 W m. Remark If U, W are subspaces of V with dim U + dim W = dim V and U W = {0} then it can be shown that U W = V.
6 The Index of a Matrix Recall and N(A) N(A 2 ) N(A 3 )... R(A) R(A 2 ) R(A 3 )...
7 The Index of a Matrix Definition The index of a matrix is the smallest nonnegative integer k = Ind(A) such that N(A k ) = N(A k+1 ) =... R(A k ) = R(A k+1 ) =... where A 0 = I. Note that Ind(A) = 0 if A is invertible.
8 The Index of a Matrix Theorem Let A be a square matrix and let k = Ind(A). Then V = N(A k ) R(A k ).
9 The Index of a Matrix Proof. Suppose x N(A k ) R(A k ). Then A k x = 0 and there exists y such that x = A k y. Therefore, A k A k y = A 2k y = 0 so that y N(A 2k ). But N(A 2k ) = N(A k ) so that x = A k y = 0. The rank-nullity theorem implies that so V = N(A k ) R(A k ). dim N(A k ) + dim R(A k ) = n = dim(v )
10 Invariant Subspaces Definition A subspace W V is said to be invariant (with respect to a matrix A) if AW W.
11 Invariant Subspaces Example Notice that for any matrix A, the range R(A) is invariant since for x R(A), Ax R(A) by definition. It follows that R(A k ) is invariant for any k. Also, N(A) is invariant since Ax = 0 N(A). So is N(A k ). Another example is an eigenspae N(A λi ) because any vector satisfies Ax = λx N(A λi ).
12 Decomposing a matrix If V = U W and U and W are A-invariant subspaces then there exists an invertible matrix P such that [ ] A = P 1 AU 0 P. 0 A W In fact, P = [p 1,..., p r, p r+1,... p n ] where {p 1,..., p r } is a basis for U and {p r+1,..., p n } is a basis for W. Furthermore, A U = A U is the restriction of A to the subspace U.
13 Matrix Diagonalization When we diagonalize A, we are simply using complementary invariant spaces. These are the eigenspaces: V = N(A λ 1 I ) N(A λ 2 I ) N(A λ r I ). The matrix that diagonalizes A is P containing bases for the eigenspaces (the columns are eigenvectors) and the blocks A λi are diagonal because on the space N(A λi ), the action of A is simply that of λi.
14 Matrix Diagonalization How do we know that V = n N(A λ i I )? i=1
15 Matrix Diagonalization Example Consider the matrix A = [ ] Since A is upper diagonal, its only eigenvalue is 2. What are the eigenvectors?
16 Matrix Diagonlization Clearly V N(A 2I ) because the dimensions do not match. This matrix cannot be diagonalized because it doesn t have a full set of linearly independent eigenvectors.
17 Generalized Eigenspaces Notice that we had repeated eigenvalues. Remember that if we have n distinct eigenvalues we know there are n linearly independent eigenvectors. This problem only occurs when we have repeated eigenvalues.
18 Generalized Eigenspaces What if we could make N(A 2I ) bigger so that it covered all of V? Let s try N(A 2I ) 2 for example. It is easy to show that V = N(A 2I ) 2. Notice that ([ ] [ ]) ( ) 0 = 1 ( ) 1 0 So we found a vector not in N(A 2I ) such that (A 2I )x N(A 2I ). This is called an generalized eigenvector of second order.
19 Generalized Eigenspaces This can be repeated. In fact, we can show that if λ 1,..., λ r are the distinct eigenvalues of A and k i = Ind(A λ i I ) then V = N(A λ 1 I ) k 1 N(A λ r I ) kr. N(A λ i I ) k i is called the generalized eigenspace of A corresponding to λ i.
20 Diagonalization Revisited If we can t diagonalize a matrix, how do we choose a basis that gets us close? Remember that if x is a generalized eigenvector of order k then (A λi )x is a generalized eigenvector of order k 1. Repeating we may obtain a sequence x 1, x 2,..., x k such that 0 = (A λi )x 1 x 1 = (A λi )x 2. x k 1 = (A λi )x k
21 Diagonalization Revisited What is the action of A on the space spanned by {x 1,..., x k }? Well, we know what A λi looks like relative to this basis: A λi =
22 Diagonlization Revisited So then A must be λ λ A = λ
23 Diagonalization Revisited Of course, {x 1,..., x k } may not span all of N(A λi ) k. So, to get a basis for N(A λi ) k we follows this same idea. Take a basis {x 1,... x d1 } for N(A λi ). Extend this to a basis for N(A λi ) 2 so that {x 1,..., x d1, x d1 +1,..., x d2 } (A λi ){x d1 +1,..., x d2 } = {x 1,..., x d2 d 1 }. Then the portion of P corresponding to N(A λi ) is be [x 1, x d1 +1, x d2 +1,..., x 2, x d1 +2,..., x d1 ].
24 If we choose our basis this way, we can decompose A into the following form: J(λ 1 ) J(λ 2 )... 0 A = J(λ r ) The block J(λ i ) is called a Jordan segment for λ i.
25 A Jordan segment is a matrix of the form J 1 (λ i ) J 2 (λ i )... 0 J(λ i ) = J ri (λ i ) Each J l (λ i ) is called a Jordan block for λ i.
26 A Jordan block for λ i is a matrix λ i λ i J l (λ i ) = λ i
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