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MA 1112: Linear Algebra II

Selected answers/solutions to the assignment for March 19, 2019

1.(a) Forx1=x2andy1=y2we get the value 2x1y1which also assumes negative values, so it is not a scalar product.

(b) Forx1=x2andy1=y2we get the valuex21which is nonnegative but vanishes for a nonzero vector µ0

1

, so it is not a scalar product.

(c) This formula is manifestly bilinear and symmetric, and forx1=x2and y1=y2 we getx21+7y12 which implies positivity, so it is a scalar product.

(d) Forx1=x2andy1=y2we obtainx12+2x1y1+y12=(x1+y1)2, and there are nonzero vectors for which this vanishes, so it is not a scalar product.

(e) It is not symmetric, so it is not a scalar product.

2.First we make this set into a set of orthogonal vectors. We put

e1=f1=

 2 1 0

,

e2=f2−(e1,f2) (e1,e1)e1=

−4/5 8/5

1

,

e3=f3−(e1,f3)

(e1,e1)e1−(e2,f3) (e2,e2)e2=

 1/7

−2/7 4/7

. To conclude, we normalise the vectors, obtaining the answer

p1 5

 2 1 0

, 1 p105

−4 8 5

, 1 p21

 1

−2 4

.

3.We first orthogonalise these vectors, noting thatR1

1f(t)d t is equal to 0 if f(t) is an odd function (this shows that our computations are actually quite easy, because even powers oft are automatically orthogonal to odd powers):

e1=1, e2=t−(1,t)

(1, 1)1=t, e3=t2−(1,t2)

(1, 1)1−(t,t2)

(t,t) t=t2−1 3, e4=t3−(1,t3)

(1, 1)1−(t,t3)

(t,t) t− (t213,t3)

(t213,t213)(t2−1

3)=t3−3 5, e5=t4−(1,t4)

(1, 1)1−(t,t4)

(t,t) t− (t213,t4)

(t213,t213)(t2−1

3)− (t335,t4)

(t335,t335)(t3−3

5)=t4−6 7t2+ 3

35. To conclude, we normalise these vectors, obtaining

p1 2,

p3t p2 ,

p5(3t2−1) 2p

2 ,

p7(5t3−3t) 2p

2 ,3(35t4−30t2+3) 8p

2

4.(a) The first two formulas are manifestly bilinear, the third one is not since

tr(A+B1+B2)=tr(A)+tr(B1)+tr(B2)6=tr(A)+tr(B1)+tr(A)+tr(B2)=tr(A+B1)+tr(A+B2),

(2)

the fourth one is not bilinear since det(2AB)=4 det(AB)6=2 det(AB).

(b) All of them are symmetric: the first one is because of the property tr(AB)=tr(B A) proved in the first semester, the second one becauseB AT =(ABT)T, the third one becauseA+B =B+A, the fourth one because det(AB)=det(A) det(B)=det(B) det(A)=det(B A).

(c) ForA=B, the first one becomes tr(A2) which is not always nonnegative, e.g. forA=

µ0 −1

1 0

¶ , the second one is the sum of squares of entries of A, so is positive, the third one is just tr(2A) so clearly is not positive, the fourth one is det(A2)=det(A)2which is nonnegative but vanishes for many nonzero matrices, e.g. forA=

µ1 0 0 0

¶ . 5.We have

|v+w|2=(v+w,v+w)=(v,v)+2(v,w)+(w,w), which is less than

(v,v)+2|v||w| +(w,w)=(|v| + |w|)2

by Cauchy–Schwartz inequality, so we get the statement of the problem after extracting square roots.

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