Download Solution to 18.700 Problem Set 2 1. (3 points) Let V be the vector

Solution to 18.700 Problem Set 2
1. (3 points) Let V be the vector space of polynomials of degree at
most five with real coefficients. Define a linear map
T : V → R3 ,
T (p) = (p(1), p(2), p(3)).
That is, the coordinates of the vector T (p) are the values of p at 1, 2, and
3.
a) Find a basis of the null space of T .
The null space of T consists of those polynomials of degree at most five vanishing
at 1, 2, and 3. A polynomial vanishes at a if and only if it is divisible by x − a. (This
is an elementary algebra fact that you could just use without proving. The reason it
is true is that division by x − a allows you to write any polynomial p as
p(x) = (x − a)q(x) + r;
here the polynomial q is the “quotient” and the real number r is the “remainder.”
This equation implies that p(a) = (a − a)q(a) + r = r, so that p(a) = 0 if and only if
the remainder r is zero.)
Therefore p vanishes at all of the distinct values a1 , . . . , am if and only if it is
divisible by all of (x − a1 ), (x − a2 ),. . . ,(x − am ). Because of unique factorization for
polynomials, this happens if and only if p is divisible by the product
(x − a1 )(x − a2 ) · · · (x − am ).
(This is analogous to the fact that an integer is divisible by both 3 and 7 if and only if
it is divisible by 21. Again it’s an algebra fact that you may just use without proof.)
The conclusion is that the null space of T consists of all polynomials of degree at
most five that are divisible by
z3 (x) = (x − 1)(x − 2)(x − 3) = x3 − 6x2 + 11x − 6.
Multiplying polynomials adds their degrees; so the null space of T consists of
{(a0 + a1 x + a2 x2 )z3 (x) | ai ∈ R}
This formula shows that a basis of the null space consists of the three polynomials
(z3 (x), xz3 (x), x2 z3 (x)) = (x3 −6x2 +11x−6, x4 −6x3 +11x2 −6x, x5 −6x4 +11x3 −6x2 ).
b) Find a basis of the range of T .
Because V has dimension six and the null space has dimension three, the range
must have dimension 6 − 3 = 3. So the range is all of R3 , and we can take for a basis
((1, 0, 0), (0, 1, 0), (0, 0, 1))
1
. (You can also see explicitly that these three vectors are in the range of T by writing
down polynomials of degree two (which is less than five) mapping to these vectors:
for example
(x − 0)(x − 2)
T
= (0, 1, 0).)
(1 − 0)(1 − 2)
2. (3 points) Let V be the vector space of polynomials of degree at
most 999 with real coefficients. Define a linear map
T : V → R100 ,
T (p) = (p(1), p(2), . . . , p(100)).
a) Find the dimension of the null space of T .
Perhaps the easiest way to proceed is to solve (b) first; once you know the range
has dimension 100, then (since V has dimension 1000) the null space must have
dimension 1000 − 100 = 900. A direct solution is to argue as in 1(a) that the null
space consists of polynomials divisible by the degree 100 polynomial
z100 (x) = (x − 1)(x − 2) · · · (x − 100);
explicitly
null space of T = {q(x)z100 (x) | q(x) = a0 + a1 x + · · · + a899 x899 }.
This exhibits a basis of the null space consisting of the 900 polynomials
{xj z100 (x) | 0 ≤ j ≤ 899}.
b) Find the dimension of the range of T .
If you solved (a) directly, then you can say that the dimension of the range must
be
dim V − dim(null space of T ) = 1000 − 900 = 100.
If you want to solve this one independently of (a) (and so use it to solve (a) easily)
then you can for 1 ≤ j ≤ 100 write down a polynomial pj of degree 99 (which is less
than 999) and with the property that
T (pj ) = ej
This is
(the jth standard basis vector of R100 ).
pj (x) =
Q
Q
1≤i≤100,i6=j (x
1≤i≤100,i6=j (j
− i)
− i).
This shows that the range of T contains a basis of R100 , and so must have dimension
100.
3. (6 points) Let V be the vector space of polynomials of degree at
most 99 with real coefficients. Define a linear map
T : V → R1000 ,
T (p) = (p(1), p(2), . . . , p(1000)).
2
a) Find the dimension of the null space of T .
Any polynomial that vanishes at these 1000 real numbers must be divisible by
the degree 1000 polynomial z1000 . The only polynomial of degree at most 99 that is
divisible by one of degree 1000 is zero; so the null space is zero, and has dimension
zero.
b) Find the dimension of the range of T .
Since the null space has dimension zero, the dimension of the range is equal to
the dimension of V , which is 100.
c) (This one is hard.) Is the vector (−1, 1, −1, 1, −1, 1, . . . , −1, 1) in the range
of T ? That is, is there a polynomial of degree at most 99 whose values at
1, 2, . . . , 1000 alternate between −1 and 1?
The answer is no.
The straightforward way to do this is in three steps: first, write down one polynomial p with the values desired. For this, define (roughly as in 2(b))
Q
pj (x) = 1≤i≤1000,i6=j (x − i)
Q
1≤i≤1000,i6=j (j − i),
p(x) =
1000
X
(−1)k pk (x).
k=1
Since pj (j) = 1 and pj (i) = 0 for 1 ≤ i 6= j ≤ 1000, it’s clear that p takes the desired
values.
Second, write down all polynomials p0 taking the values desired. These are
p0 (x) = p(x) + q(x)z1000 (x);
here z1000 is the degree 1000 polynomial vanishing at 1, . . . , 1000, and q is anything.
Now the question is whether any of these p0 can have degree at most 99. Each pj
has degree exactly 999, so p has degree at most 999. The polynomial z1000 has degree
1000, so if q is nonzero and has degree m, then qz1000 has degree 1000 + m. Adding
a degree 1000 + m polynomial to a degree-at-most-999 polynomial gives a degree
1000 + m polynomial (no possibility of cancelling leading terms). So we conclude
if q 6= 0 has degree m, then p0 has degree 1000 + m.
So none of these has degree at most 99.
The last remaining possibility is q = 0; that is, p0 = p. So the question comes down
to this:
does p have degree at most 99?
The numbers were chosen so that you couldn’t reasonably compute p by hand. (You
could use a computer algebra system to compute p, but that’s not interesting.) So
how can you hope to answer this question? One way is to look at the coefficient of
x999 in pj (x): it’s

−1
!  1000
j−1
Y
Y

(j − i) 
(j − i) .
i=1
i=j+1
3
The j − 1 terms in the first product are all positive, and the 1000 − j terms in the
second product are all negative; so the coefficient has the same sign as (−1)1000−j =
(−1)j . Since the polynomial p is a sum of various (−1)j pj , all the terms being added
have a strictly positive coefficient of x999 . The conclusion is that p has degree exactly
999. Therefore the answer to the last question above is no.
This argument is hard to motivate, and looks like blind luck. I know a a few
shorter proofs. For one, Add 1 to p; the result is a nonzero polynomial vanishing at
the 500 odd numbers 1, 3, 5, . . . , 999. So p + 1 must be divisible by the product
(x − 1)(x − 3) · · · (x − 999),
(500 factors) and so must have degree at least 500 or just be 0. It can’t be 0, since
p(2) + 1 = 2. But p + 1 has degree at least 500 implies p also has degree at least 500.
For another, define
p1 (x) = p(x + 1) − p(x).
It’s clear from high school algebra that the degree of p1 is at most one less than the
degree of p: p1 has degree at most 98. Also
p1 (1) = 1 − (−1) = 2,
p1 (2) = −1 − 1 = −2, . . . p1 (999) = 2;
that is, the values of p1 at 1, 2, . . . , 999 alternate between 2 and −2.
Now repeat: define
p2 (x) = p1 (x + 1) − p1 (x).
The degree of p2 is at most one less than the degree of p1 : p2 has degree at most 97.
Also
p2 (1) = −2 − 2 = −4, p2 (2) = 2 − (−2) = 4, . . . p2 (998) = 4;
that is, the values of p2 at 1, 2, . . . , 998 alternate between −4 and −4.
Keep repeating. After 99 steps, you’ll get a polynomial p99 having degree at most
zero (that is, constant!) such that the values of p99 at 1, 2, . . . , 901 alternate between
299 and −299 . That’s a contradiction. (In fact this argument shows that p has to
have degree at least 999).
For a last and maybe shortest of all proof: the assumption on p guarantees that
p has a root between 1 and 2, another root between 2 and 3,. . . , and a root between
999 and 1000. So p has at least 999 roots. So p has degree at least 999.
4. (3 points) Suppose V and W are vector spaces over F ; that T ∈
L(V, W ); that (v1 , . . . , vn ) is any list of vectors in V ; and that (T v1 , . . . , T vn )
is linearly independent in W . Prove that (v1 , . . . , vn ) is linearly independent
in V .
This was included as a complement to exercise 5 on page 59 of Axler. For the
proof, suppose that (a1 , . . . , an ) are scalars, and that a1 v1 + · · · + an vn = 0 in V . We
must prove that all the ai are zero. Applying T to this equation, we get
a1 T v1 + · · · + an T vn = 0
in W . By the assumed linear independence of the T vi , this equation implies that all
the ai must be zero, as we wished to show.
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