Prof. Dr. Otto May 18, 2011

Linear Algebra II

Exercise Sheet no. 7

Summer term 2011

Prof. Dr. Otto May 18, 2011

Dr. Le Roux Dr. Linshaw

Exercise 1 (Warm up: the trace) Recall Exercise E4.3 about the trace.

LetV:=R^(n,n)be theR-vector space of all realn×nmatrices and letS⊆V be the subspace consisting of all symmetric matrices (i.e., all matricesAwithA^t=A). ForA,B∈V, we define

〈A,B〉:=Tr(AB), where thetraceTr(A)of a matrixA= (ai j)is defined as

Tr(A):=

n

X

i=1

a_ii.

(a) Show that〈. , .〉is bilinear.

(b) Show that〈. , .〉is a scalar product onS.

Solution:

a) LetA= (ai j),B= (bi j), andC= (ci j)be matrices andλ∈R. Since

〈A,B〉=

n

X

i,k=1

a_ikb_ki

it follows that

〈A+C,B〉=

n

X

i,k=1

(a_ik+c_ik)b_ki=

n

X

i,k=1

a_ikb_ki+

n

X

i,k=1

c_ikb_ki=〈A,B〉+〈C,B〉,

〈λA,B〉=

n

X

i,k=1

λa_ikb_ki=λ

n

X

i,k=1

a_ikb_ki=λ〈A,B〉.

In the same way, we show that〈A,B+C〉=〈A,B〉+〈A,C〉and〈A,λB〉=λ〈A,B〉. b) We have

〈A,A〉=

n

X

i,k=1

a_ika_ki=

n

X

i,k=1

(a_ik)²¾0 .

Furthermore, it follows that we have〈A,A〉=0if and only ifA=0.

Exercise 2 (Cauchy-Schwarz and triangle inequalities) (a) (Exercise 2.1.4 on page 60 of the notes)

Let(V,〈. , .〉)be a euclidean or unitary vector space. Show that equality holds in the Cauchy-Schwarz inequality, i.e., we havek〈v,w〉k=kvk · kwk, if, and only if,vandware linearly dependent.

1

(b) (Exercise 2.1.5 on page 60 of the notes)

Let u,v,w be pairwise distinct vectors in a euclidean or unitary vector space (V,〈. , .〉), and write a := v−u, b:=w−v. Show that equality holds in the triangle inequality

d(u,w) =d(u,v) +d(v,w), or, equivalently,ka+bk=kak+kbk,

if, and only if,aandbarepositive realscalar multiples of each other (geometrically: v=u+s(w−u)for some s∈(0, 1)⊆R).

Solution:

a) Without loss of generality we may assume thatv,w6=0.

Whenvandware linearly dependent, thenw=λvfor someλ. It follows that

k〈v,w〉k=k〈v,λv〉k=kλk k〈v,v〉k=kλkkvk²=kvkkλvk=kvkkwk.

Conversely, suppose that

k〈v,w〉k=kvk · kwk

and writeλ=^〈v,w〉_〈v,v〉. Then it follows, as in the proof of Proposition 2.1.10 (the Cauchy-Schwarz inequality) on page 59 of the notes, that

〈w−λv,w−λv〉=〈w,w〉 −〈w,v〉〈v,w〉

〈v,v〉 =kwk²−kvk²kwk² kvk² =0 . So by positive definiteness of the scalar product,w=λv.

b) Note thata,b6=0, becauseu,v,ware pairwise distinct.

Whenb=λawith0< λ∈R, then

ka+bk=ka+λak= (1+λ)kak=kak+λkak=kak+kλak=kak+kbk.

Conversely, whenka+bk=kak+kbk, also(ka+bk)²= (kak+kbk)². But

(ka+bk)²=〈a+b,a+b〉=〈a,a〉+〈a,b〉+〈b,a〉+〈b,b〉

¶kak²+2k〈a,b〉k+kbk²,and (kak+kbk)²=kak²+2kakkbk+kbk².

Thereforekakkbk¶k〈a,b〉k, andkakkbk=k〈a,b〉kby Cauchy-Schwarz. So we know thatb=λafor someλ∈C by Exercise (E3.2). Fromka+bk=kak+kbk, we deduce thatk1+λk=1+kλk, which implies thatλis a positive real.

Exercise 3 (Orthogonal matrices) We consider realn×nmatrices. Set

O(n):={A∈R^(n,n) |A^tA=E_n}. Show thatO(n)is a subgroup ofGL_n(R).

Solution:

We have to show thatE_n∈O(n)and thatO(n)is closed under multiplication and inverses.

SinceE_n^tE_n=E_n, we haveE_n∈O(n). Furthermore, forA,B∈O(n), we have (AB)^tAB=B^tA^tAB=B^tE_nB=B^tB=E_n.

Hence,AB∈O(n). Similarly, one can show thatA⁻¹∈O(n). For the inverse, we first note thatA^tA=E_nimpliesA^t=A⁻¹. Therefore, we have

(A⁻¹)^tA⁻¹= (A^t)^tA^t=AA^t= (A^tA)^t=E_n^t =E_n.

2

Exercise 4 (Orthogonal vectors)

LetV be a euclidean or unitary space andS={v1, . . . ,vn}be a set of non-null pairwise orthogonal vectors.

(a) Show thatSis linearly independent.

(b) Letu∈V. Show that the vector

w:=u−

n

X

i=1

〈v_i,u〉

〈v_i,v_i〉vi

is orthogonal toS. Note thatPn i=1〈vi,u〉

〈vi,v_i〉viis the orthogonal projection ofwonspan(S).

(c) [Parseval’s identity]Suppose thatV is finite dimensional and thatSis an othornormal basis ofV. Show that

〈v,w〉=

n

X

i=1

〈v,vi〉〈vi,w〉 for allv,w∈V.

(d) [Bessel’s inequality]Suppose thatV is euclidean andSis orthonormal. Show that

n

X

i=1

〈vi,u〉²≤ kuk² for allu∈V.

Solution:

a) Supposev₁, . . . ,v_nsatisfyPn

i=1λiv_i=0. We need to show that eachλiis zero. For each j=1, . . . ,n, we get that

0=〈v_j,0〉=

* v_j,

n

X

i=1

λiv_i +

=

n

X

i=1

λi〈v_j,v_i〉=λj〈v_j,v_j〉,

since〈v_j,v_i〉=0whenever j6=i. Butv_j6=0. So〈v_j,v_j〉 6=0since the scalar product is positive definite. Hence, λj=0for each j=1, . . . ,n. ThereforeSis linearly independent.

b) For each j=1, . . . ,n, we have that

〈v_j,w〉=

* v_j,u−

n

X

i=1

〈v_i,u〉

〈v_i,v_i〉v +

=〈v_j,u〉 −

n

X

i=1

〈v_j,u〉

〈v_i,v_i〉〈v_j,v_i〉

=〈vj,u〉 − 〈v_j,u〉

〈v_j,v_j〉〈vj,vj〉

=0 . c) By Lemma 2.3.2, we have thatw=Pn

i=1〈vi,w〉vi. Applying the operation〈v, .〉on both sides, we obtain the result.

d) Settingw:=u−Pn

i=1〈v_i,u〉v_iwe have

kwk²=〈w,w〉=

* u−

n

X

i=1

〈v_i,u〉v_i,u−

n

X

j=1

〈v_j,u〉v_j +

=〈u,u〉 −2

n

X

i=1

〈v_i,u〉²+

n

X

i,j=1

〈v_i,u〉〈v_j,u〉〈v_i,v_j〉

=〈u,u〉 −

n

X

i=1

〈v_i,u〉². Sincekwk²≥0, the inequality follows.

3

Exercise 5 (Jordan normal form for describing processes)

Suppose that we use vectorss_n∈R³to describe the state of a3-dimensional system at stepn∈N(for example, the position of a particle in space). The evolution of the system from stagenton+1is given by

s_n+1=As_n, where A=







−4 2 −1

−4 3 0

14 −5 5





.

(a) Use a transformation of the givenAinto Jordan normal form in order to get a feasible formula fors_n, as a function of the indexnand the initial states₀.

(b) Computes₁₀₀fors₀=





 1 3 1





. Solution:

(a) The characteristic polynomial of Ais p_A= (1−X)²(2−X), soλ1 =1and λ2= 2are the eigenvalues ofA. The corresponding eigenspaces are 1-dimensional, with generators





 1 2

−1





 forV_λ1 and





 1 4 2





 forV_λ2. So the Jordan normal form ofAhas two blocks, one of size 2 and one of size 1. As

(A−E₃)²=







3 −1 1

12 −4 4

6 −2 2





,

dim(ker(A−E₃)²) =2. Hence, the Jordan block of size 2 has entries 1 on the diagonal. Therefore the Jordan normal form ofAis

J=







1 1 0

0 1 0

0 0 2





.

To find a matrixSsuch thatA=SJ S⁻¹, we take as third columnu₃=





 1 4 2





, an eigenvector with eigenvalue 2, and as

second column an element ofker(A−E₃)²\ker(A−E₃), for exampleu₂=





 0 1 1





. The first column will then be

u₁= (A−E₃)u2=





 1 2

−1





. Hence, S=







1 0 1

2 1 4

−1 1 2





. We haves_n=Aⁿs₀=SJⁿS⁻¹s₀. Furthermore

Jⁿ=







1 n 0

0 1 0

0 0 2ⁿ





.

(b) Fors₀=





 1 3 1





=





 1 4 2





−





 0 1 1





, we have

s_n=2ⁿ





 1 4 2





−





 0 1 1





−n





 1 2

−1





. Hence, s₁₀₀=2¹⁰⁰





 1 4 2





−





 100 201

−99





.

4