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• Linear Algebra Written Examinations Study Guide

Eduardo Corona Other Authors as they Join In

November 2, 2008

Contents

1 Vector Spaces and Matrix Operations 2

2 Linear Operators 2

3 Diagonalizable Operators 3 3.1 The Rayleigh Quotient and the Min-Max Theorem . . . . . . . . 4 3.2 Gershgorins Discs Theorem . . . . . . . . . . . . . . . . . . . . . 4

4 Hilbert Space Theory: Interior Product, Orthogonal Projection and Adjoint Operators 5 4.1 Orthogonal Projection . . . . . . . . . . . . . . . . . . . . . . . . 7 4.2 The Gram-Schmidt Process and QR Factorization . . . . . . . . 8 4.3 Riesz Representation Theorem and The Adjoint Operator . . . . 9

5 Normal and Self-Adjoint Operators: Spectral Theorems and Related Results 11 5.1 Unitary Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 14 5.2 Positive Operators and Square Roots . . . . . . . . . . . . . . . . 15

6 Singular Value Decomposition and the Moore-Penrose Gener- alized Inverse 16 6.1 Singular Value Decomposition . . . . . . . . . . . . . . . . . . . . 16 6.2 The Moore-Penrose Generalized Inverse . . . . . . . . . . . . . . 18 6.3 The Polar Decomposition . . . . . . . . . . . . . . . . . . . . . . 19

7 Matrix Norms and Low Rank Approximation 19 7.1 The Frobenius Norm . . . . . . . . . . . . . . . . . . . . . . . . 19 7.2 Operator Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 7.3 Low Rank Matrix Approximation: . . . . . . . . . . . . . . . . . 21

1

• 8 Generalized Eigenvalues, the Jordan Canonical Form and eA 22 8.1 The Generalized Eigenspace K� . . . . . . . . . . . . . . . . . . . 23 8.2 A method to compute the Jordan Form: The points diagram . . 24 8.3 Applications: Matrix Powers and Power Series . . . . . . . . . . . 24

9 Nilpotent Operators 24

10 Other Important Matrix Factorizations 24

11 Other Topics (which appear in past exams) 24

12 Yet more Topics I can think of 25

1 Vector Spaces and Matrix Operations

2 Linear Operators

Denition 1 Let U ,V be vector spaces =F (Usually F = R or C). Then L(U; V ) = fT : U ! V j T is linearg: In particular, L(U;U) = L(U) is the space of linear operators of U , and L(U;F ) = U� is its algebraic dual.

Denition 2 Important Subspaces: Given W � V (subspace), T�1(W ) � U: In particular, we are interested in T�1(f0g) = Ker(T ): Also, if S � U; then T (S) � U: We are most interested in T (U) = Ran(T ):

Theorem 3 U; V ev=F , dim(U) = n; dim(V ) = m: Given B = fu1; :::; ung basis of U and B0 = fv1; :::; vmg basis of V; to each T 2 L(U; V ) we can associate a matrix [T ]B

0

B such that:

Tui = a1iv1 + :::+ amivm 8i 2 f1; ::;mg [T ]B

0

B = (aij) in Mm�n(F )

T U �! V

B l l B0 Fn �! Fm

[T ]B 0

B

Conversely, given a matrix A 2Mm�n(F ); there is a unique TA 2 L(U; V ) such that A = [T ]B

0

B :

Proposition 4 Given T 2 L(U); there exist basis B and B0 of U such that:

[T ]B 0

B =

� I 0 0 0

� B is constructed as an extension for a basis for Ker(T ); and B0 as an extension for fT (u)gu2B :

2

• Theorem 5 (Rank and Nullity) dim(U) = dim(Ker(T ))+dim(Ran(T )): �(T ) = dim(Ker(T )) is known as the nullity of T; and r(T ) = dim(Ran(T )) as the rank of T:

Change of Basis: U; V ev=F; B and � basis of U; B0,�0 basis of V; there exists P invertible such that:

[T ]B 0

B = P [T ] �0

� P �1

P is a matrix that performs a change of coordinates. This means that, if two matrices are similar, they represent the same linear operator using a di¤erent basis. This further justies that key properties of matrices are preserved under similarity.

3 Diagonalizable Operators

If U = V (T;is a linear operator), it is natural to impose that both basis B and B0 are also the same. In this case, it is no longer generally true that we can nd a basis B such that the corresponding matrix is diagonal. However, if there exists a basis B such that [T ]B

0

B = � diagonal matrix, we say T is diagonalizable.

Denition 6 V ev=F; T 2 L(V ). � 2 F is an eigenvalue of T if 9 v 2 V a nonzero vector such that Tv = �v: All nonzero vectors such that this holds are known as eigenvectors of T:

We can immediately derive, from this denition, that the existence of the eigenpair (�; v) (eigenvalue � and corresponding eigenvector v) is equivalent to the existence of a nonzero solution v to

(T � �I)v = 0

This in turn tells us that the eigenvalues of T are those such that the operator T � �I is not invertible. After selecting a basis B for V; this also means: u

det([T ]BB � �I) = 0

Which is called the characteristic equation of T . We notice this equation does not depend on the choice of basis B; since it is invariant under similarity:

det(PAP�1 � �I) = det(P (A� �I)P�1) = det(A� �I)

This equation nally is equivalent to nding the complex roots of a polyno- mial in �:We know this to be a really hard problem for n � 5, and a numerically ill-posed problem at that.

3

• Denition 7 V ev=F; T 2 L(V ), � an eigenvector of T: Then E� = fv 2 V j Tv = �vg is the eigenspace for �:

Theorem 8 V ev=F of nite dimension; T 2 L(V ): The following are equiva- lent: i) T is diagonalizable ii) V has a basis of eigenvectors of T iii) There exist subspaces W1; :::;Wn such that dim(Wi) = 1; T (Wi) � Wi and V =

Ln i=1Wi

iv) V = Lk

i=1E�k with f�1; :::; �kg eigenvalues of T v) Pk

i=1 dim(E�i) = dim(V )

Proposition 9 V ev=C; T 2 L(V ) then T has at least one eigenvalue (this is a corollary of the Fundamental Theorem of Algebra, applied to the characteristic equation).

Theorem 10 (Schurs factorization) V ev=C; T 2 L(V ): There always exists a basis B such that [T ]B is upper triangular.

3.1 The Rayleigh Quotient and the Min-Max Theorem

3.2 Gershgorins Discs Theorem

Although calculating eigenvalues of a big matrix is a very di¢ cult problem (computationally and analytically), it is very easy to come up with regions on the complex plane where all the eigenvalues of a particular operator T must lie. This technique was rst devised by the russian mathematician Semyon Aranovich Gershgorin (1901� 1933):

Theorem 11 (Gershgorin, 1931) Let A = (aij) 2Mn(C): For each i 2 f1; ::; ng; we dene the ith "radius of A" as ri(A) =

P j 6=i jaij j and the ith Gershgorin

disc as Di(A) = fz 2 C j jz � aiij < ri(A)

Then, if we dene �(A) = f� j � is an eigenvalue of Ag; it follows that:

�(A) � n[ i=1

Di(A)

That is, all eigenvalues of A must lie inside one or more Gershgorin discs.

4

• Proof. Let � be an eigenvalue of A; v an associated eigenvector. We x i as the ith coordinate of v with maximum modulus, that is, jvij � jvkj 8k. Necessarily, jvij 6= 0. Then,

Av = �v =) �vi =

X aijvj

(�� aii)vi = X j 6=i

aijvj

j(�� aii)j jvij � X j 6=i

jaij j jvij

� 2 Di(A)

Now, we know that A represents a linear operator T 2 L(Rn), and that therefore its eigenvalues are invariant under transposition of A and under simi- larity. Therefore:

Corollary 12 Let A = (aij) 2Mn(C): Then, �(A) � \ f

n[ i=1

Di(PAP �1) j P

is invertibleg

Of course, if T is diagonalizable, one of this P 0s is the one such that PAP�1

is diagonal, and therefore the Gershgorin discs degenerate to the n points we are looking for. However, if we dont want to compute the eigenvalues, we can still use this to come up with a ne heuristic to reduce the region given by the union of the gershgorin discs: We can use permutation matrices or diagonal matrices as our P 0s to get a "reasonable region". This result also hints at the fact that, if we perturb a matrix A, the eigenvalues change continuously. The Gershgorin disc theorem is also a quick way to prove A is invertible if it

is diagonal dominant, and it also provides us with results when the eigenvalues of A are all distinct (namely, that there must be at least one eigenvalue per Gershgorin disc).

4 Hilbert Space Theory: Interior Product, Or- thogonal Projection and Adjoint Operators

Denition 13 Let V ev=F: An interior product on V is a function : V � V ! F such that: 1) hu+ v; wi = hu;wi+ hv; wi 8u; v; w 2 V 2) h�u;wi = � hu; vi 8u; v 2 V , � 2 F 3) hu; vi = hv; ui 4) hu; ui � 0 and hu; ui = 0 =) u = 0

5

• By denition, every interior product induces a natural norm for V , given by kvkV =

p hv; vi

Denition 14 We say u and v are orthogonal, or u?v, if hu; vi = 0

Some important identities:

1. Pythagoras Theorem: u?v () ku+ vk2 = kuk2 + kvk2

2. Cauchy-Bunyakowski-Schwarz: jhu; vij � kuk kvk 8u; v 2 V with equality () u = �v

3. Parallelogram: ku+ vk2 + ku� vk2 = 2(kuk2 + kvk2) 8u; v 2 V

4. Polarization:

hu; vi = 1 4 fku+ vk2 � ku� vk2g 8u; v 2 V if F = R

hu; vi = 1 4

4X k=1

ik

u+ ikv

2 8u; v 2 V if F = C

In fact, identities 3 and 4 (Parallelogram and Polarization) give us both necessary and su¢ cient conditions for a norm to be induced by some interior product. In this fashion, we can prove kk1 and kk1 are not induced by an interior product by showing paralellogram fails.

Denition 15 v 2 V is said to be of unit norm if kvk = 1

Denition 16 A

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