Download APPM 2360 17 October, 2013 Worksheet #7 1. Consider the space

APPM 2360
17 October, 2013
Worksheet #7
1. Consider the space of all polynomials with real coefficients. Determine whether each of the
following is a subspace.
(a) All polynomials with integer coefficients.
(b) All polynomials with degree less than or equal to three.
(c) All polynomials with only even powers (i.e. P (t) = a0 + a1 t2 + a2 t4 + . . .)
√
(d) All polynomials with i = −1 as a root (i.e. f (i) = 0).
Solution:
(a) NO
√
√
√
√
Let u = x2 + x + 1 ∈ W and λ = 2 ∈ R then uλ = 2x2 + 2x + 2 ∈
/ W . Then W
fails to satisfy second condition of subspace definition. Therefore W is not subspace.
(b) YES
Let u = ax3 + bx2 + cx + d ∈ W and v = a1 x3 + b1 x2 + c1 x + d1 ∈ W then
i. u + v = (a + a1 )x3 + (b + b1 )x2 + (c + c1 )x + (d + d1 ) ∈ W
ii. ∀λ ∈ R we get λu = λax3 + λbx2 + λcx + λd ∈ W
Thus W is subspace of the space of all polynomials with real coefficients, since set W
satisfies both conditions of subspace definition.
(c) YES
Let P (t), Q(t) ∈ W such that P (t) = a0 +a1 t2 +a2 t4 +. . . and Q(t) = b0 +b1 t2 +b2 t4 +. . .
i. R(t) = P (t)+Q(t) = a0 +b0 +(a1 +b1 )t2 +(a2 +b2 )t4 +. . ., then R(t) = P (t)+Q(t) ∈
W
ii. ∀λ ∈ R we get λQ(t) = λb0 + λb1 t2 + λb2 t4 + . . ., then λu ∈ W
Since set W satisfies both conditions of subspace definition, W is subspace of the space
of all polynomials with real coefficients.
(d) YES
Let u = f (x) ∈ W and v = g(x) ∈ W then f (i) = g(i) = 0
i. u + v = f (x) + g(x) = h(x) since h(i) = f (i) + g(i) = 0, then u + v = h(x) ∈ W
ii. ∀λ ∈ R we get λu = λf (x), since λf (i) = λ0 = 0 then λu ∈ W
Since set W satisfies both conditions of subspace definition, W is subspace of the space
of all polynomials with real coefficients.
2. (a) Any subset of a vector space is also a vector space.
(b) A linearly independent set of vectors in Rn is a basis for Rn
(c) An n × n matrix is row equivalent to In×n if and only if the columns of A form a basis
for Rn
(d) The polynomials p1 (t) = t + 1 and p2 (t) = t − 1 form a basis for the vector space of all
polynomials of degree 1 or less, { P (t) = a + bt with a, b ∈ R }
(e) The set of all pairs of real numbers (x, y) such that x ≥ y is vector space.
Solution:
1
(a) FALSE
Consider V = R2 as our vector space, then S = {(1, 2), (1, 1)} is obviously subset of V .
However, S is not closed under addition of vectors:
(1, 2) + (1, 1) = (2, 3) ∈
/S
(b) FALSE
Consider R3 as vector space (n = 3). Then S = {(1, 0, 0), (0, 1, 0)} is linearly independent set of R3 . However, it is not a basis for R3 . That is because it has only 2 elements,
but dim(R3 ) = 3
(c) TRUE
An n × n matrix is row equivalent to In×n ⇐⇒ A is invertible ⇐⇒ rank(A) = n ⇐⇒
rank(A) = dim(col(A)) = n ⇐⇒ the columns of A form a basis for Rn
(d) TRUE
Let S = {p1 , p2 }. Then
0 = c1 p1 +c2 p2 = c1 (t+1)+c2 (t−1) = (c1 +c2 )t+(c1 −c2 ) ⇐⇒ c1 +c2 = 0 and c1 −c2 = 0 ⇐⇒ c1 = c2
Then S is linearly independent and set S has the same number of elements as the
dimension of P1 . Consequently, S is a basis for P1 .
(e) FALSE
Say considered set is S then (2, 1) ∈ S. Let c = −1 ∈ R, then c(2, 1) = (−2, −1) ∈
/ S.
Thus S is not a vector space, since it is not closed under scalar multiplication.
3. Decide whether the set S is a linearly independent subset of V, where V = P2 and S =
{t + 3, t2 − 1, 2t2 − t − 5}.
Solution:
c1 (t + 3) + c2 (t2 − 1) + c3 (2t2 − t − 5) = 0
If S is a linearly independent subset, then the only solution to this equation is c1 = c2 = c3 =
0. Rearranging, we have
(c2 + 2c3 )t2 + (c1 − c3 )t + (3c1 − c2 − 5c3 ) = 0
We now have a system of equations,
c2 + 2c3 = 0
c1 − c3 = 0
3c1 − c2 − 5c3 = 0
We can write this in matrix form Ax = b. The augmented matrix [ A | b ] is


0 1
2 0
 1 0 −1 0  .
3 −1 −5 0
Transforming to RREF gives us


1 0 −1 0
 0 1 2 0 .
0 0 0 0
2
Since the system is consistent and # pivots < # columns, there is infinitely many solutions.
Hence, the vectors are linearly dependent. We can also check this by noting that the determinant of A is zero.
4. Determine whether or not the set S
S = {(1, 0, 1), (1, 1, 0), (0, 1, 1)}.
Solution:
is a basis for the vector space V, where V = R3 and
1 1 0 0 1 1 = 2 6= 0.
1 0 1 The set S has the same number of elements as the dimension of R3 and S is linearly independent. Consequently, S is a basis for R3 .
5. Consider the following matrix and

1
A= 2
0
vector

2 0 0 0
4 1 1 1 ,
0 2 2 2


2
b= 3 
λ
(a) Col(A) is the span of the column vectors of matrix A. For what values of λ is b in
Col(A).
(b) Find the dimension of Col(A).
Solution:
(a) b ∈ Col(A) if and only if linear system Ax = b

1 2 0 0
 2 4 1 1
0 0 2 2
is consistent. Augmented matrix [A|b] is

0 2
1 3 
2 λ
reducing it to RREF using Gaussian elimination we get


1 2 0 0 0 2
 0 0 1 1 1 −1 
0 0 0 0 0 λ+2
Then linear system is consistent if and only if λ = −2. Thus only for λ = −2 we have
b ∈ Col(A)
(b) Since dim(Col(A)) = rank(A) and from part a of the problem we know that rank(A) =
2. Then dim(Col(A)) = 2
3