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Linear Algebra Take Home 1
Due: 6/29/2009
Homework by: Charles Ambat
1. The complex numbers
C = {x + iy | x, y ∈ R}
form a field.
(a) How is the ”sum” and ”product” of two complex numbers defined?
• Let c1 ∈ C|c1 =x1 +iy1 and c2 ∈ C|c2 =x2 +iy2 ,
– Sum: c1 + c2 = (x1 + x2 ) + i(y1 + y2 )
– Product: c1 c2 = (x1 + iy1 )(x2 + iy2 ) = (x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 )
(b) What is the multiplicative inverse of 3+i4?
• Given c ∈ C|c=3+i4 , the inverse of c is c−1 such that
cc−1 = 1 ⇒ (3 + i4)(x2 + iy2 ) = 1.
The product yields, (3x2 − 4y2 ) + i(3y2 + 4x2 ) = 1. This tells us that
3x2 − 4y2 = 1, 3y2 + 4x2 = 0.
Solving for x2 and y2 gives: x2 = (3 /25 ), y2 = (−4 /25 ).
Therefore, the multiplicative inverse of c = 3 + i4 is
3 − i4
.
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2. The collection P4 (R) of all polynomials with coefficients in R and degree less than or equal to
4 form a vector space over R. Does the zero polynomial together with all polynomials with
coefficients in R and degree equal to 4 form a vector space over R? Why or why not?
c−1 =
The zero polynomial together with all polynomials with coefficients in R and degree equal
to 4 does not form a vector space over R.
Proof:
Let the zero polynomial together with all polynomials with coefficients in R and degree equal
to 4 be
P40 (R) = {0, an xn + an−1 xn−1 + ... + a0 |n=4,ai ∈R(1≤i≤4) },
in order for P40 (R) to form a a vector space over R, the group (P40 (R), +) must be an Abelian
group closed under addition such that
∀f (x), g(x) ∈ P40 (R)
f (x) + g(x) ∈ P40 (R).
We see that P40 (R) is not a group if we take two degree 4 polynomials to be
f (x) = x4 + x3 , g(x) = −x4 .
Defining addition on P40 (R):
f (x) + g(x) = (x4 + x3 ) + (−x4 )
= (1 +R (−1))x4 + x3 = x3 .
By definition f (x) + g(x) = x3 is not closed under addition because x3 is not a polynomial
with degree equal to 4. Thus, (P40 (R), +) does not qualify as a group.
Therefore, (P40 (R), the zero polynomial together with all polynomials with coefficients in R
and degree equal to 4 does not form a vector space over R.
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3. Let W1 and W2 be subspaces of a vector space V over a field F . Prove that the set
W1 + W2 = {w1 + w2 ∈ V | w1 ∈ W1 and w2 ∈ W2 }
is also a subspace of V . Is W1 ⊂ W1 + W2 ? Why or why not?
Proof:
In order for W1 + W2 to be a subspace of V , W1 + W2 must also be a vector space.
By THM 1, W1 + W2 must satisfy the following: (1) must have a zero vector in V , (2) closure
under ‘+’, (3) closure under ‘•’
(1) 0 ∈ W1 + W2
• Since W1 and W2 are already taken to be subspaces of V , THM 1 makes the following
true.
0 ∈ W1 , 0 ∈ W2
Now, consider 0 = 0 + 0.
0 = 0∈W1 + 0∈W2
We now see that the zero vector exists and belongs to the set, therefore proving that
0 ∈ W1 + W2 .
(2) ∀x, y ∈ W1 + W2
x + y ∈ W1 + W2 (closure under ‘+’)
• Let x = w1 + w2 and y = w10 + w20 .
Define addition of x + y:
x + y = (w1 + w2 ) + (w10 + w20 )
= (w1 + w10 ) + (w2 + w20 )
We know w1 , w10 ∈ W1 and w2 , w20 ∈ W2 , which implies
x + y = (w1 + w10 )|∈W1 + (w2 + w20 )|∈W2 .
Therefore proving that x + y ∈ W1 + W2 .
(3) ∀a ∈ F and ∀w ∈ W1 + W2
aw ∈ W1 + W2 (closure under ‘ ’)
• Let w = w1 + w2 .
Define scalar action of aw:
aw = a(w1 + w2 )
= aw1 + aw2
Since W1 and W2 are already taken to be subspaces of V , THM 1 makes the following
true.
aw1 ∈ W1 , aw2 ∈ W2 ⇒ aw ∈ W1 + W2
Therefore proving that aw ∈ W1 + W2 .
In conclusion, W1 + W2 proves to be a subspace of V by the definition of Theorem 1.
W1 is a subset of W1 + W2 .
Proof:
We want to show ∃a ∈ W1 that is also in W1 +W2 so that we may conclude that W1 ⊂ W1 +W2 .
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Let a ∈ W1 , and consider the following to be true
a ∈ W1 + W2 ,
Since W2 is already given as a subspace, we can use 0 element in W2 to prove the claim.
From this we can get
⇒ a ∈ W1 + 0,
which is already a truth in itself. This further implies that a ∈ W1 +W2 . Because a ∈ W1 +W2 ,
this proves that W1 is a subset of W1 + W2 .
4. If f (x), g(x) ∈ P (R) are defined by
f (x) = x3 + 2x2 + πx + 2
and
g(x) = −πx + 14,
compute
(a) 2(f+g)(x)
= 2(f (x) + g(x))
= 2(x3 + 2x2 + (π + (−π))x + (2 + 14))
= (2 · 1)x3 + (2 · 2)x2 + (2 · 16)
= 2x3 + 4x2 + 32.
(b) 6(f-g)(x)
= 6(f (x) − g(x))
= 6(x3 + 2x2 + (π +− (−π))x + (2 + (−14))
= (6 · 1)x3 + (6 · 2)x2 + (6 · 2π)x + (6 · (−12))
= 6x3 + 12x2 + 12πx − 72.
5. Let V be a vector space over a field F with basis {u, v}. If a ∈ F and a 6= 0, is {u + v, av} a
basis for V ? Why or why not?
{u + v, av} is a basis for V.
Let S = {u + v, av}, B = {u, v}.
In order for S to be a basis for V , by definition for a basis, the following must be true.
(a) SPAN(S) = V
(b) S is linearly independent
Proof for (a): We need to show for any vector w ∈ V , w ∈ SPAN(S), that is, w = a1 (u + v) +
a2 (av) for some scalars a1 , a2 ∈ F . Let w ∈ V . Since SPAN(B) = V , w = b1 u + b2 v for some
b1 , b2 ∈ F . Since a 6= 0,
w
=
=
b1 u + b2 v
b1 (u + v) + [(b2 − b1 )a−1 ](av).
Therefore w ∈ SPAN(S) and we’re done with (a).
Proof for (b): To show S is linearly independent, we must take a linear combination of S and
set it to equal to zero, then prove that the coefficients must be zero. Now note,
0
=
a1 (u + v) + a2 (av)
=
a1 u + (a1 + aa2 )v
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and since B is linearly independent, this means a1 = (a1 + aa2 ) = 0 or a1 = aa2 = 0 but since
a 6= 0 this is equivalent to a1 = a2 = 0 and so S linearly independent. Therefore, the set S is
linearly independent.
In conclusion, since S is linearly independent and SPAN(S) = V , the set {u + v, av} is a
basis for V .
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