Lecture 13: Proof of existence of upper-triangular matrices for complex linear transformations; invariant subspaces and block upper-triangular matrices for real linear transformations (1) Travis Schedler Thurs, Oct 27, 2011 (version: Thurs, Oct 27, 1:00 PM)
Goals (2) Theorem on existence of upper-triangular matrices for F = C, when V is f.d. and nonzero Theorem on existence of block upper-triangular matrices for F = R, with diagonal blocks of size 2 or 1, when V is f.d. and nonzero Corollary: existence of eigenvalues when V is odd-dimensional Read Chapter 5, and do PS 6. Then, start on Chapter 6 and PS 7.
Warm-up exercise (3) Let A Mat(n, n, F) be an upper-triangular (square) matrix. Show that A is invertible if and only if its diagonal entries are nonzero (use Gaussian elimination!) Conclude that A λi is not injective if and only if λ appears on the diagonal of A. Conclude that the eigenvalues of A (i.e., λ such that Av = λv has a nonzero solution v Mat(n, 1, F)) are precisely the diagonal entries. Remark: A somewhat different proof, without using Gaussian elimination, is given in Axler, Proposition 5.16.
Solution to warm-up exercise (4) If A has nonzero diagonal entries, then we can rescale all the rows so that the diagonal entries are one. This is in row-echelon form with no nonzero rows, so the rank of the matrix is equal to n, i.e., A is invertible. Conversely, if A has some diagonal entries that are zero, then performing Gaussian elimination, the entries that are zero will yield free columns (and the nonzero entries will yield pivot columns). So not all columns will be pivot columns, and the number of these is the rank of A which is then strictly less than n. So A is not invertible.
Upper-triangular matrices (5) In general one has no eigenbasis, i.e., there is no diagonal matrix M(T ). However much more generally:
Upper-triangular matrices (5) In general one has no eigenbasis, i.e., there is no diagonal matrix M(T ). However much more generally: Proposition (Proposition 5.12) The following are equivalent for a basis (v 1,..., v n ) of V and T L(V ): (a) M(T ) is upper-triangular; (b) Tv k Span(v 1,..., v k ) for all k; (c) Span(v 1,..., v k ) is T -invariant for all k.
Upper-triangular matrices (5) In general one has no eigenbasis, i.e., there is no diagonal matrix M(T ). However much more generally: Proposition (Proposition 5.12) The following are equivalent for a basis (v 1,..., v n ) of V and T L(V ): (a) M(T ) is upper-triangular; (b) Tv k Span(v 1,..., v k ) for all k; (c) Span(v 1,..., v k ) is T -invariant for all k. Proof (on board): Just write out M(T ).
Upper-triangular matrices (5) In general one has no eigenbasis, i.e., there is no diagonal matrix M(T ). However much more generally: Proposition (Proposition 5.12) The following are equivalent for a basis (v 1,..., v n ) of V and T L(V ): (a) M(T ) is upper-triangular; (b) Tv k Span(v 1,..., v k ) for all k; (c) Span(v 1,..., v k ) is T -invariant for all k. Proof (on board): Just write out M(T ). Theorem (Theorem 5.13) If F = C and V is finite-dimensional, then there exists a basis such that M(T ) is upper-triangular.
Upper-triangular matrices (5) In general one has no eigenbasis, i.e., there is no diagonal matrix M(T ). However much more generally: Proposition (Proposition 5.12) The following are equivalent for a basis (v 1,..., v n ) of V and T L(V ): (a) M(T ) is upper-triangular; (b) Tv k Span(v 1,..., v k ) for all k; (c) Span(v 1,..., v k ) is T -invariant for all k. Proof (on board): Just write out M(T ). Theorem (Theorem 5.13) If F = C and V is finite-dimensional, then there exists a basis such that M(T ) is upper-triangular. Proof (on board; see next slide.)
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0).
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector.
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector. Write V = Span(v 1 ) U for some U. Let S L(U) be S(u) = P U,Span(v1 )T (u) for all u.
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector. Write V = Span(v 1 ) U for some U. Let S L(U) be S(u) = P U,Span(v1 )T (u) for all u. By induction, there is a basis (v 2,..., v n ) of U in which M(S) is upper-triangular.
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector. Write V = Span(v 1 ) U for some U. Let S L(U) be S(u) = P U,Span(v1 )T (u) for all u. By induction, there is a basis (v 2,..., v n ) of U in which M(S) is upper-triangular. Now P U,Span(v1 )T (Span(v 2,..., v k )) Span(v 2,..., v k ). This implies T (Span(v 2,..., v k )) Span(v 1, v 2,..., v k ).
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector. Write V = Span(v 1 ) U for some U. Let S L(U) be S(u) = P U,Span(v1 )T (u) for all u. By induction, there is a basis (v 2,..., v n ) of U in which M(S) is upper-triangular. Now P U,Span(v1 )T (Span(v 2,..., v k )) Span(v 2,..., v k ). This implies T (Span(v 2,..., v k )) Span(v 1, v 2,..., v k ). Hence Span(v 1,..., v k ) is T -invariant for all k.
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector. Write V = Span(v 1 ) U for some U. Let S L(U) be S(u) = P U,Span(v1 )T (u) for all u. By induction, there is a basis (v 2,..., v n ) of U in which M(S) is upper-triangular. Now P U,Span(v1 )T (Span(v 2,..., v k )) Span(v 2,..., v k ). This implies T (Span(v 2,..., v k )) Span(v 1, v 2,..., v k ). Hence Span(v 1,..., v k ) is T -invariant for all k. Note that (v 1,..., v n ) is a basis since it is put together from bases of Span(v 1 ) and U.
Proof of Theorem 5.13 (on board) (6) Proof: by induction on dim V. Base case: dim V = 1 (or 0). Inductive step: let v 1 =nonzero eigenvector. Write V = Span(v 1 ) U for some U. Let S L(U) be S(u) = P U,Span(v1 )T (u) for all u. By induction, there is a basis (v 2,..., v n ) of U in which M(S) is upper-triangular. Now P U,Span(v1 )T (Span(v 2,..., v k )) Span(v 2,..., v k ). This implies T (Span(v 2,..., v k )) Span(v 1, v 2,..., v k ). Hence Span(v 1,..., v k ) is T -invariant for all k. Note that (v 1,..., v n ) is a basis since it is put together from bases of Span(v 1 ) and U. By the proposition, M(T, (v 1,..., v n )) is upper-triangular.
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries.
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries. Proposition (Proposition 5.18) The eigenvalues of an upper-triangular matrix are exactly the diagonal entries.
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries. Proposition (Proposition 5.18) The eigenvalues of an upper-triangular matrix are exactly the diagonal entries. ( ) ( ) ( ) 2 1 1 1 Ex: A = : eigenvals 2, 1; eigenvecs and. 0 1 0 1
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries. Proposition (Proposition 5.18) The eigenvalues of an upper-triangular matrix are exactly the diagonal entries. ( ) ( ) ( ) 2 1 1 1 Ex: A = : eigenvals 2, 1; eigenvecs and. 0 1 0 1 If not upper-triangular, not true: e.g., 90 rotation: no eigenvalue of zero; complex eigenvalues are ±i.
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries. Proposition (Proposition 5.18) The eigenvalues of an upper-triangular matrix are exactly the diagonal entries. ( ) ( ) ( ) 2 1 1 1 Ex: A = : eigenvals 2, 1; eigenvecs and. 0 1 0 1 If not upper-triangular, not true: e.g., 90 rotation: no eigenvalue of zero; complex eigenvalues are ±i. Proof (from the warm-up!): The eigenvalues are exactly λ F such that null(a λi ) {0}.
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries. Proposition (Proposition 5.18) The eigenvalues of an upper-triangular matrix are exactly the diagonal entries. ( ) ( ) ( ) 2 1 1 1 Ex: A = : eigenvals 2, 1; eigenvecs and. 0 1 0 1 If not upper-triangular, not true: e.g., 90 rotation: no eigenvalue of zero; complex eigenvalues are ±i. Proof (from the warm-up!): The eigenvalues are exactly λ F such that null(a λi ) {0}. So we have to show: an upper-triangular matrix is invertible if and only if its diagonal entries are nonzero (Prop. 5.16).
Eigenvalues and upper-triangular matrices (7) Recall that the eigenvalues of a diagonal matrix are the diagonal entries. Proposition (Proposition 5.18) The eigenvalues of an upper-triangular matrix are exactly the diagonal entries. ( ) ( ) ( ) 2 1 1 1 Ex: A = : eigenvals 2, 1; eigenvecs and. 0 1 0 1 If not upper-triangular, not true: e.g., 90 rotation: no eigenvalue of zero; complex eigenvalues are ±i. Proof (from the warm-up!): The eigenvalues are exactly λ F such that null(a λi ) {0}. So we have to show: an upper-triangular matrix is invertible if and only if its diagonal entries are nonzero (Prop. 5.16). We proved this in the warm-up using Gaussian elimination.
The real case (8) Now let F := R. Then T need not have an eigenvalue. Example: rotation matrices!
The real case (8) Now let F := R. Then T need not have an eigenvalue. Example: rotation matrices! Theorem (Theorem 5.24) V 0 admits a T -invariant subspace of dimension 1 or 2.
The real case (8) Now let F := R. Then T need not have an eigenvalue. Example: rotation matrices! Theorem (Theorem 5.24) V 0 admits a T -invariant subspace of dimension 1 or 2. Proof. As we proved, there is a polynomial f (x) such that f (T ) = 0.
The real case (8) Now let F := R. Then T need not have an eigenvalue. Example: rotation matrices! Theorem (Theorem 5.24) V 0 admits a T -invariant subspace of dimension 1 or 2. Proof. As we proved, there is a polynomial f (x) such that f (T ) = 0. Factor f (x) (over R) using: Theorem (Thereom 4.14) Every polynomial with real coefficients factors into linear and quadratic factors.
The real case (8) Now let F := R. Then T need not have an eigenvalue. Example: rotation matrices! Theorem (Theorem 5.24) V 0 admits a T -invariant subspace of dimension 1 or 2. Proof. As we proved, there is a polynomial f (x) such that f (T ) = 0. Factor f (x) (over R) using: Theorem (Thereom 4.14) Every polynomial with real coefficients factors into linear and quadratic factors. Thus, for some linear or quadratic factor g(x), we must have null g(t ) 0.
The real case (8) Now let F := R. Then T need not have an eigenvalue. Example: rotation matrices! Theorem (Theorem 5.24) V 0 admits a T -invariant subspace of dimension 1 or 2. Proof. As we proved, there is a polynomial f (x) such that f (T ) = 0. Factor f (x) (over R) using: Theorem (Thereom 4.14) Every polynomial with real coefficients factors into linear and quadratic factors. Thus, for some linear or quadratic factor g(x), we must have null g(t ) 0. Now, say v null g(t ) is nonzero. Then, Span(v, Tv) is an invariant subspace of dimension 2 (or 1).
Real case: block upper-triangular form (9) Theorem Let F = R and T L(V ). Then, in some basis, M(T ) has the A 1 0 A 2 block upper-triangular form......, where each A j 0 0 A n is either 1 1 or 2 2; in the latter case, A j has no real eigenvalues.
Real case: block upper-triangular form (9) Theorem Let F = R and T L(V ). Then, in some basis, M(T ) has the A 1 0 A 2 block upper-triangular form......, where each A j 0 0 A n is either 1 1 or 2 2; in the latter case, A j has no real eigenvalues. By Gaussian elimination, A λi is noninjective if and only if A i λi is for some i (even if λ C!) So the real eigenvalues are the entries of the 1 1 matrices A i.
Real case: block upper-triangular form (9) Theorem Let F = R and T L(V ). Then, in some basis, M(T ) has the A 1 0 A 2 block upper-triangular form......, where each A j 0 0 A n is either 1 1 or 2 2; in the latter case, A j has no real eigenvalues. By Gaussian elimination, A λi is noninjective if and only if A i λi is for some i (even if λ C!) So the real eigenvalues are the entries of the 1 1 matrices A i. The non-real complex eigenvalues are the eigenvalues of the 2 2 A i. Using PS 6, #10 you see that the complex eigenvalues of each of these are of the form a ± bi for some a, b R. (This is because the eigenvalues of A i are fixed by complex conjugation, since A i itself is).
Proof of the theorem (on board) (10) We prove this by induction on dim V, beginning with zero.
Proof of the theorem (on board) (10) We prove this by induction on dim V, beginning with zero. For V 0, we showed there exists an invariant subspace U V of dimension 1 or 2. Let this be 1 if possible; otherwise let it be 2.
Proof of the theorem (on board) (10) We prove this by induction on dim V, beginning with zero. For V 0, we showed there exists an invariant subspace U V of dimension 1 or 2. Let this be 1 if possible; otherwise let it be 2. Pick a basis of U. Then M(T U ) is either 1 1 or, if it is 2 2, it has no real eigenvalue.
Proof of the theorem (on board) (10) We prove this by induction on dim V, beginning with zero. For V 0, we showed there exists an invariant subspace U V of dimension 1 or 2. Let this be 1 if possible; otherwise let it be 2. Pick a basis of U. Then M(T U ) is either 1 1 or, if it is 2 2, it has no real eigenvalue. Write V = U U for some complement U V.
Proof of the theorem (on board) (10) We prove this by induction on dim V, beginning with zero. For V 0, we showed there exists an invariant subspace U V of dimension 1 or 2. Let this be 1 if possible; otherwise let it be 2. Pick a basis of U. Then M(T U ) is either 1 1 or, if it is 2 2, it has no real eigenvalue. Write V = U U for some complement U V. Inductively, pick a basis of U so that S := P U,U T U, A 2 viewed as a map U U, has the form...... 0 A n
Proof of the theorem (on board) (10) We prove this by induction on dim V, beginning with zero. For V 0, we showed there exists an invariant subspace U V of dimension 1 or 2. Let this be 1 if possible; otherwise let it be 2. Pick a basis of U. Then M(T U ) is either 1 1 or, if it is 2 2, it has no real eigenvalue. Write V = U U for some complement U V. Inductively, pick a basis of U so that S := P U,U T U, A 2 viewed as a map U U, has the form...... 0 A n Put this together with our basis of U to get a basis of V. Then M(T ) is as desired.
Operators on odd-dimensional vector spaces have eigenvalues (11) Corollary (Theorem 5.26) Every operator on an odd-dimensional real vector space has an eigenvalue.
Operators on odd-dimensional vector spaces have eigenvalues (11) Corollary (Theorem 5.26) Every operator on an odd-dimensional real vector space has an eigenvalue. Proof. Pick a basis so that the matrix M(T ) is in the above block upper-triangular form.
Operators on odd-dimensional vector spaces have eigenvalues (11) Corollary (Theorem 5.26) Every operator on an odd-dimensional real vector space has an eigenvalue. Proof. Pick a basis so that the matrix M(T ) is in the above block upper-triangular form. Then, at least one block must be (λ j ) for λ j R.
Operators on odd-dimensional vector spaces have eigenvalues (11) Corollary (Theorem 5.26) Every operator on an odd-dimensional real vector space has an eigenvalue. Proof. Pick a basis so that the matrix M(T ) is in the above block upper-triangular form. Then, at least one block must be (λ j ) for λ j R. So λ j is an eigenvalue.
Operators on odd-dimensional vector spaces have eigenvalues (11) Corollary (Theorem 5.26) Every operator on an odd-dimensional real vector space has an eigenvalue. Proof. Pick a basis so that the matrix M(T ) is in the above block upper-triangular form. Then, at least one block must be (λ j ) for λ j R. So λ j is an eigenvalue. Remark: we will give another proof much later: the eigenvalues are the roots of the characteristic polynomial det(xi A), which has real coefficients. Its degree equals dim V, and if this is odd, it must have a real root.
Similarity of matrices (12) Recall from class and the homework: Two matrices A and B can be the matrix of the same linear transformation if and only if they are conjugate: A = CBC 1.
Similarity of matrices (12) Recall from class and the homework: Two matrices A and B can be the matrix of the same linear transformation if and only if they are conjugate: A = CBC 1. This is because if ( v 1 v n) = ( v1 v n ) C, then M(T, (v 1,..., v n )) = CM(T, (v 1,..., v n))c 1.
Similarity of matrices (12) Recall from class and the homework: Two matrices A and B can be the matrix of the same linear transformation if and only if they are conjugate: A = CBC 1. This is because if ( v 1 v n) = ( v1 v n ) C, then M(T, (v 1,..., v n )) = CM(T, (v 1,..., v n))c 1. Example: if (v 1,..., v n ) is the standard basis of V = Mat(n, 1, F), then C = (v 1 v n) is the vectors v 1,..., v n put together.
Corollaries of upper-triangular theorems (13) Corollary If A is a square complex matrix, then there exists an invertible complex matrix C such that B = C 1 AC is upper-triangular. The eigenvalues of A (or B) are then exactly the diagonal entries.
Corollaries of upper-triangular theorems (13) Corollary If A is a square complex matrix, then there exists an invertible complex matrix C such that B = C 1 AC is upper-triangular. The eigenvalues of A (or B) are then exactly the diagonal entries. Corollary If A is a square real matrix, then there exists an invertible real matrix C such that B = C 1 AC is block upper-triangular with diagonal blocks 1 1 or 2 2; in the latter case the block has no real eigenvalues. The real eigenvalues of A (or B) are the entries of the 1 1-diagonal blocks of B, and the nonreal complex eigenvalues are the eigenvalues of the 2 2-diagonal blocks of B (which are always of the form a ± bi, a, b R).