116 CHAPTER 5. LINEAR TRANSFORMATIONS

19. Let L be the linear transformation taking polynomials of degree at most three topolynomials of degree at most three given by

D2 +2D+1

where D is the differentiation operator. Find the matrix of this linear transformationrelative to the basis

{1,x,x2,x3

}. Find the matrix directly and then find the matrix

with respect to the differential operator D+1 and multiply this matrix by itself. Youshould get the same thing. Why?

20. Let L be the linear transformation taking polynomials of degree at most three topolynomials of degree at most three given by D2 + 5D+ 4 where D is the differen-tiation operator. Find the matrix of this linear transformation relative to the bases{

1,x,x2,x3}. Find the matrix directly and then find the matrices with respect to the

differential operators D+ 1,D+ 4 and multiply these two matrices. You should getthe same thing. Why?

21. Suppose A ∈ L (V,W ) where dim(V ) > dim(W ) . Show ker(A) ̸= {0}. That is,show there exist nonzero vectors v ∈V such that Av = 0.

22. A vector v is in the convex hull of a nonempty set if there are finitely many vectorsof S,{v1, · · · ,vm} and nonnegative scalars {t1, · · · , tm} such that

v =m

∑k=1

tkvk,m

∑k=1

tk = 1.

Such a linear combination is called a convex combination. Suppose now that S ⊆V,a vector space of dimension n. Show that if v = ∑

mk=1 tkvk is a vector in the convex

hull for m > n+1, then there exist other scalars{

t ′k}

such that

v =m−1

∑k=1

t ′kvk.

Thus every vector in the convex hull of S can be obtained as a convex combinationof at most n+1 points of S. This incredible result is in Rudin [37]. Hint: ConsiderL : Rm→V ×R defined by

L(a)≡

(m

∑k=1

akvk,m

∑k=1

ak

)

Explain why ker(L) ̸= {0} . Next, letting a ∈ ker(L) \ {0} and λ ∈ R, note thatλ a ∈ ker(L) . Thus for all λ ∈ R,

v =m

∑k=1

(tk +λak)vk.

Now vary λ till some tk +λak = 0 for some ak ̸= 0.

23. For those who know about compactness, use Problem 22 to show that if S ⊆ Rn andS is compact, then so is its convex hull.