MAT247 - Problem Set 1 Solution

Problem 1

a) It is straightforward to verify this is an inner product.

b) Again, easily veriﬁed using matrix multiplication.

c) Not an inner product. We have < x, x >=x2

1−4x2

2, take x1= 5 and

x2= 1, we see that <·,·>is not positive deﬁnite. Hence, not an inner

product.

d) Not an inner product. Take f(x) = x(x−i) then < f, f >= 0, but f6= 0,

hence not an inner product.

e) Easy to verify it is an inner product. For condition 4, we use the property

that <·,·>0is an inner product, hence we get positive deﬁniteness.

Problem 2

In this question, to be precise, we need to make sure the deﬁnition is well-

deﬁned. If x∈span{v1, . . . , vn}, with vi’s in β(i.e x=Pn

i=1 aivi), then for

any other basis vector w∈β, the only way to represent xas an element

of span{v1, . . . , vn, w}is by x=a1v1+. . . +anvn+ 0 ·w, since βis a

basis. Therefore, in computing < x, y >, taking any ﬁnite collection of

basis elements whose span contains both xand ywill give the same answer.

It is then straightforward to verify that we indeed get an inner product, and

in particular, in the case of Vis ﬁnite-dimensional one recovers the standard

inner product.

1

Problem 3

First, convince yourself that we’ve been given an inner product. Let u1=

(1,1,0), u2= (1,0,1), u3= (0,1,1). We apply the Gram-Schmidt algorithm:

take v1=u1, and set

v2=u2−< u2, v1>

||v1||2v1= (1,0,1) −3

5(1,1,0) = (2/5,−3/5,1)

v3=u3−< u3, v1>

||v1||2v1−< u3, v2>

||v2||2v2= (0,1,1) −2

5(1,1,0) −(−1/5)

(55/25)(2/5,−3/5,1)

=(−4/11,6/11,12/11)

and {v1, v2, v3}form an orthogonal basis. If v= (0,2,1), we can write

v=< v, v1>

||v1||2v1+< v, v2>

||v2||2v2+< v, v3>

||v3||2v3=4

5v1−v2+9

4v3

Problem 4

One approach would be to apply the Gram-Schmidt procedure to the basis

{1, x, x2}, which yields an orthonormal basis {1

√2,1

√2(x−1), x2+ 1}. Then, as

in problem 3 (or by eyeball), we can write

x2+ (1 + i)x+i= (x2+1)+(1+i)√2(x−1

√2)+2√2i(1

√2)

Of course, if you chose a diﬀerent basis your answers may vary.

Problem 5

Note that W=span{(1,0,√2,−1),(0,1,0, i)}, so applying the Gram-Schmidt

procedure to these two vectors (which are clearly independent) and then nor-

malizing yields a basis {1

2(1,0,√2,−1),1

2√7(i, 4,√2i, 3i)}.

Now W⊥={(x, y, z, w)∈C4|x+√2z−w= 0, y +iw = 0}, which is

spanned by (say) (√2,0,−1,0) and (0,−√2i, 1,√2). Applying Gram-Schmidt

and normalizing gives a basis {1

√3(√2,0,−1,0),1

√118 (√2/5,−√2i, 4/5,√2)}for

W⊥.

Problem 6

Let (tij ) be the matrix of Twith respect to the basis β={x1, . . . , xn}. For

any v=Pn

i=1 aixi∈V, we have that

< T (v), T (v)>=< T (

n

X

i=1

aixi), T (

n

X

j=1

ajxj)>=

n

X

i,j=1

aiaj< T (xi), T (xj)>

=

n

X

i,j=1

aiaj< T (xi),

n

X

k=1

tkj xk>

=

n

X

i,j,k=1

aiajtkj < T (xi), xk>= 0

2

since each < T (xi), xj>= 0. So for each v∈V,T(v) = 0; i.e., Tis the zero

operator.

Problem 7

a) Suppose Vis an n-dimensional complex inner product space, and Tis an

invertible linear operator on V. Let cT(x) = det(x·Id−T) be the characteristic

polynomial of T. This is a polynomial of degree nover the complex numbers,

hence it has a zero (Cis an algebraically closed ﬁeld - Appendix D in the

text has one proof of this fact). In other words, there is a λ∈Csuch that

det(λ·Id −T) = 0, and so there is some non-zero x∈Vsuch that T(x) = λx.

Applying T−1to this expression, we have x=λT −1(x), which implies λ6= 0,

since xis non-zero.

Thus,

< T (x), x >=< λx, x >=λ < x, x >=λ||x||26= 0

since both xand λare non-zero.

b) Now suppose Vis real, and has odd dimension. The characteristic

polynomial cT(x) = det(x·Id−T) = xn+. . . is a (real) polynomial in xof degree

n, which is odd, and so it has a (real) zero ( note that limx→+∞cT(x) = +∞

and limx→−∞ cT(x) = −∞, so being a continuous function cT(x) has to hit

the x-axis somewhere). So, as above, there is some λ∈Rand (non-zero)

x∈Vsuch that T(x) = λx. Applying T−1, we see that λ6= 0, and that

< T (x), x >=λ||x||26= 0.

c) Now suppose Vis real, and of dimension n= 2m. Changing notation

slightly, let β={x1, . . . xm, y1, . . . ym}be an orthonormal basis for V, and deﬁne

U:V→Von basis elements by U(xi) = −yi,U(yi) = xi(a linear operator is

clearly determined uniquely by where it sends basis elements). Let T:V→Vbe

deﬁned by T(xi) = yi,T(yi) = −xi. Then T◦Uand U◦Tare both the identity,

so T=U−1i.e. Uis invertible. However, for any x=Pm

i=1 aixi+biyi∈V, we

have

< U(x), x > =<

m

X

i=1

aiU(xi) + biU(yi),

m

X

j=1

ajxj+bjyj>

=<

m

X

i=1

ai(−yi) + bi(xi),

m

X

j=1

ajxj+bjyj>

=

m

X

i,j=1

(−aiaj< yi, xj>−aibj< yi, yj>+biaj< xi, xj>+bibj< xi, yj>)

=

m

X

i=1

(−aibi< yi, yi>+biai< xi, xi>)

=

m

X

i=1

(−aibi+aibi) = 0

where the fourth and ﬁfth lines follow from the assumption that our basis was

orthonormal.

3

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###### Document Summary

Problem 1: it is straightforward to verify this is an inner product, again, easily veri ed using matrix multiplication. 1 4x2: not an inner product. 2, take x1 = 5 and x2 = 1, we see that < , > is not positive de nite. Hence, not an inner product: not an inner product. Take f (x) = x(x i) then < f, f >= 0, but f (cid:54)= 0, hence not an inner product: easy to verify it is an inner product. For condition 4, we use the property that < , >(cid:48) is an inner product, hence we get positive de niteness. (cid:80)n. In this question, to be precise, we need to make sure the de nition is well- de ned. , vn}, with vi"s in (i. e x = i=1 aivi), then for any other basis vector w , the only way to represent x as an element of span{v1, .

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