Linear Algebra: Graduate Level Problems and Solutions

Linear Algebra: Graduate Level Problems and Solutions

Igor Yanovsky

1

Linear Algebra

Igor Yanovsky, 2005

2

Disclaimer: This handbook is intended to assist graduate students with qualifying examination preparation. Please be aware, however, that the handbook might contain, and almost certainly contains, typos as well as incorrect or inaccurate solutions. I can not be made responsible for any inaccuracies contained in this handbook.

Linear Algebra

Igor Yanovsky, 2005

3

Contents

1 Basic Theory

4

1.1 Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Linear Maps as Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Dimension and Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Matrix Representations Redux . . . . . . . . . . . . . . . . . . . . . . . 6

1.5 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.6 Linear Maps and Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.7 Dimension Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.8 Matrix Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.9 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Inner Product Spaces

8

2.1 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2 Orthonormal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Gram-Schmidt procedure . . . . . . . . . . . . . . . . . . . . . . 9

2.2.2 QR Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Orthogonal Complements and Projections . . . . . . . . . . . . . . . . . 9

3 Linear Maps on Inner Product Spaces

11

3.1 Adjoint Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2 Self-Adjoint Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.3 Polarization and Isometries . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.4 Unitary and Orthogonal Operators . . . . . . . . . . . . . . . . . . . . . 14

3.5 Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.6 Normal Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.7 Unitary Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.8 Triangulability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Determinants

17

4.1 Characteristic Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5 Linear Operators

18

5.1 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5.2 Dual Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6 Problems

23

Linear Algebra

Igor Yanovsky, 2005

4

1 Basic Theory

1.1 Linear Maps

Lemma. If A M atmxn(F) and B M atnxm(F), then

tr(AB) = tr(BA).

Proof. Note that the (i, i) entry in AB is

m i=1

jiij

.

mn

Thus tr(AB) =

ij ji,

i=1 j=1

nm

tr(BA) =

jiij .

j=1 i=1

n j=1

ij

ji,

while

(j, j)

entry

in

BA

is

1.2 Linear Maps as Matrices

Example. Let Pn = {0 + 1t + ? ? ? + ntn : 0, 1, . . . , n F} be the space of polynomials of degree n and D : V V the differential map

D(0 + 1t + ? ? ? + ntn) = 1 + ? ? ? + nntn-1.

If we use the basis 1, t, . . . , tn for V then we see that D(tk) = ktk-1 and thus

the (n + 1)x(n + 1) matrix representation is computed via

0 1 0 ??? 0

[D(1) D(t) D(t2)

???

D(tn)]

=

[0 1 2t

???

ntn-1]

=

[1 t t2

???

tn]

0 0 ...

0 0 ...

2

0 ...

??? ...

...

0 0 n

0 0 0 ??? 0

1.3 Dimension and Isomorphism

A linear map L : V W is isomorphism if we can find K : W V such that LK = IW and KL = IV .

V ---L- W

IV

IW

V -K--- W

Linear Algebra

Igor Yanovsky, 2005

5

Theorem. V and W are isomorphic there is a bijective linear map L : V W .

Proof. If V and W are isomorphic we can find linear maps L : V W and K : W V so that LK = IW and KL = IV . Then for any y = IW (y) = L(K(y)) so we can let x = K(y), which means L is onto. If L(x1) = L(x2) then x1 = IV (x1) = KL(x1) = KL(x2) = IV (x2) = x2, which means L is 1 - 1. Assume L : V W is linear and a bijection. Then we have an inverse map L-1 which satisfies L L-1 = IW and L-1 L = IV . In order for this inverse map to be allowable as K we need to check that it is linear. Select 1, 2 F and y1, y2 W . Let xi = L-1(yi) so that L(xi) = yi. Then we have

L-1(1y1 + 2y2) = L-1(1L(x1) + 2L(x2)) = L-1(L(1x1 + 2x2)) = IV (1x1 + 2x2) = 1x1 + 2x2 = 1L-1(y1) + 2L-1(y2).

Theorem. If Fm and Fn are isomorphic over F, then n = m.

Proof. Suppose we have L : Fm Fn and K : Fn Fm such that LK = IFn and KL = IFm. L M atnxm(F) and K M atmxn(F). Thus

n = tr(IFn) = tr(LK) = tr(KL) = tr(IFm) = m.

Define the dimension of a vector space V over F as dimF V = n if V is isomorphic to Fn.

Remark. dimC C = 1, dimR C = 2, dimQ R = . The set of all linear maps {L : V W } over F is homomorphism, and is denoted

by homF(V, W ). Corollary. If V and W are finite dimensional vector spaces over F, then homF(V, W ) is also finite dimensional and

dimF homF(V, W ) = (dimF W ) ? (dimF V )

Proof. By choosing bases for V and W there is a natural mapping

homF(V, W ) M at(dimF W )?(dimF V )(F) F(dimF W )?(dimF V )

This map is both 1-1 and onto as the matrix represetation uniquely determines the linear map and every matrix yields a linear map.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download