Grad at 06:55, 16 November 2015

2015-11-16T06:55:14Z

Grad: Created page with "'''1.''' Find a subset $C \subset \mathbb{F}^2$ that is closed under scalar multiplication but not under addition of vectors.

''Solution'' There are m..."

2015-11-09T21:00:58Z

Created page with "'''1.''' Find a subset <math>C \subset \mathbb{F}^2</math> that is closed under scalar multiplication but not under addition of vectors. ''Solution'' There are m..."

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'''1.''' Find a subset <math>C \subset \mathbb{F}^2</math> that is closed under scalar multiplication but not under addition of vectors. 
 

''Solution''
There are many possible answers that work. Here is one of them. 
<math>\mathbb{F}^2 = \mathbb{R}^2</math> with <math>C = \{(x,y): x^2 = y^2\}</math>. Then if <math>(x,y)\in C</math> and <math>\alpha \in \mathbb{R}</math> we have 
<math>\alpha (x,y) = (\alpha x, \alpha y) \in C</math> because <math>x^2 = y^2 \Rightarrow (\alpha x)^2 = \alpha^2 x^2 = \alpha^2 y^2 = (\alpha y)^2</math> so that <math>C</math> is closed under scalar multiplication. However, <math>(1,1) \in C</math> and <math>(1,-1) \in C</math>, but <math>(1,1)+(1,-1) = (2,0) \notin C</math> so that <math>C</math> is not closed under addition of vectors.

'''2.''' Find a subset <math>A \subset \mathbb{C}^2</math> that is closed under vector addition but not under multiplication by complex number. 

''Solution''
Many possible answers again. Here are a few: 
<math>A = \mathbb{N}^2</math>, <math>A = \mathbb{Z}^2</math>, <math>A = \mathbb{Q}^2</math>, <math>A=\mathbb{R}^2</math>, all are closed under addition, but if you multiply <math>(1,1) \in A</math> for all of these <math>A</math> by <math>i</math> then you get <math>(i,i) \notin A</math>.

'''3.''' Find a subset <math>Q \subset \mathbb{R}</math> that is closed under addition but not scalar multiplication by real scalars. 
 

''Solution''
Here them using <math>Q</math> can be a hint. If you let <math>Q = \mathbb{Q}</math>, the rational numbers, then they will be closed under addition, but not scalar multiplication. That is because <math>1 \in Q</math> and <math>\sqrt{2} \in \mathbb{R}</math>, but <math>\sqrt{2} = \sqrt{2} \cdot 1 \notin Q</math>. 
Another possible answer is <math>Q = \{ x \in \mathbb{R}: x > 0\}</math>. Then this will be closed under addition since the sum of two positive numbers is still positive, but <math>1 \in Q</math> and <math>-1 \in \mathbb{R}</math> and <math>-1 = -1 \cdot 1 \notin Q</math>.

'''6.''' Let <math>V</math> be a complex vector space i.e., a vector space where the scalars are <math>\mathbb{C}</math>. Define <math>V^*</math> as the complex vector space whose additive structure is that of <math>V</math> but where complex scalar multiplication is given by <math>\lambda * x = \bar{\lambda} x</math>. Show that <math>V^*</math> is a complex vector space. 
 

''Solution''
To begin we don’t need to show any of the first four axioms are true as they only involve addition of vectors and since <math>V^*</math> has the same additive structure as <math>V</math> and <math>V</math> is a vector space, the first four axioms will still be true. For the remaining four properties we simply check that they will hold. 
5) <math>1 * x = \bar{1} x = 1x = x</math> 
6) <math>\alpha * (\beta * x) = \alpha * (\bar{\beta} x) = \bar{\alpha} \bar{ \beta} x = \overline{ \alpha \beta} x = (\alpha \beta) * x</math> 
7) <math>\alpha * (x + y) = \bar{\alpha} (x+y) = \bar{\alpha} x + \bar{\alpha} y = \alpha * x + \alpha * y</math> 
8) <math>(\alpha + \beta) * x = \overline{\alpha + \beta} x + (\bar{\alpha} + \bar{\beta}) x = \bar{\alpha}x + \bar{\beta} x = \alpha * x + \beta * x</math> 
Therefore <math>V^*</math> is a complex vector space.

'''7.''' Let <math>P_n</math> be the set of polynomials in <math>\mathbb{F}[t]</math> of degree <math>\leq n</math>. 
(a) Show that <math>P_n</math> is a vector space. 
 
''Solution''
Suppose <math>x,y \in P_n</math> and <math>\alpha \in \mathbb{R}</math>. Then <math>x = a_n t^n + \cdots + a_0</math> and <math>y = b_n t^n + \cdots + b_0</math>. So that <math>x+y = (a_n+b_n)t^n + (a_0 + b_0) \in P_n</math>. Also <math>\alpha x = (\alpha a_n)t^n + \cdots \alpha a_0 \in P_n</math> so that <math>P_n</math> is closed under addition and scalar multiplication. Also <math>x + y = (a_n+b_n)t^n + (a_0 + b_0) = (b_n+a_n)t^n + (b_0 + a_0) = y + x</math> so addition is commutative. Similarly for associative. The zero vector in <math>P_n</math> is the zero polynomial with all coefficients equal to 0. The additive inverse of <math>x</math> is <math>(-x) = (-a_n)t^n + (-a_0)</math>. Also, <math>(1) x = x</math>, <math>\alpha(\beta x) = (\alpha \beta) x</math>, <math>(\alpha + \beta) x = \alpha x + \beta x</math> and <math>\alpha (x + y) = \alpha x + \alpha y</math> just by writing each of them out. Therefore <math>P_n</math> is a vector space. 
 
(b) Show that the space of polynomials of degree <math>n \geq 1</math> is <math>P_n - P_{n-1}</math> and does not form a subspace. 
 
''Solution''
First off, the space is equal to <math>P_n - P_{n-1}</math> because the polynomials that have exactly degree <math>n</math> are in <math>P_n</math> but not in <math>P_{n-1}</math> and there are no other polynomials in <math>P_n</math> that aren’t also in <math>P_{n-1}</math>. Now this set does not form a subspace because it is not closed under addition. <math>p(t) = t^n</math> and <math>q(t) = -t^n + 1</math> are both polynomials of degree exactly <math>n</math>. However, <math>p+q = 1</math> is a polynomial of degree 0 not <math>n</math>. 
 
(c) If <math>f(t): \mathbb{F} \to \mathbb{F}</math>, show that <math>V = \{ p(t)f(t): p \in P_n\}</math> is a subspace of <math>Func(\mathbb{F}, \mathbb{F})</math>. 
 
''Solution''
To show it is a subspace, we only need to check that it is closed under addition and scalar multiplication. Let <math>x,y \in V</math> and <math>\alpha \in \mathbb{F}</math>. Then <math>x = p(t)f(t)</math> for some polynomial <math>p \in P_n</math> and <math>y = q(t)f(t)</math> for some polynomial <math>q \in P_n</math>. Then <math>x + y = p(t)f(t)+q(t)f(t) = (p(t)+q(t))f(t)</math>. But since <math>p(t)+q(t)</math> is just another polynomial in <math>P_n</math>, then <math>x+y</math> is exactly of the form polynomial times <math>f(t)</math>. Thus <math>x+y \in V</math>. Also <math>\alpha x = \alpha (p(t)f(t)) = (\alpha p(t)) f(t)</math> which is again of the form polynomial times <math>f(t)</math> so <math>\alpha x \in V</math>. Therefore <math>V</math> is a subspace. 8) <math>(\alpha + \beta) * x = \overline{\alpha + \beta} x + (\bar{\alpha} + \bar{\beta}) x = \bar{\alpha}x + \bar{\beta} x = \alpha * x + \beta * x</math> 
Therefore <math>V^*</math> is a complex vector space.

'''8.''' Let <math>V = \mathbb{C}^\times = \mathbb{C} - \{0\}</math>. Define addition on <math>V</math> by <math>x \boxplus y = xy</math>. Define scalar multiplication by <math>\alpha \boxdot x = e^{\alpha}x</math>. 
(a) Show that if we use <math>0_V = 1</math> and <math>-x = x^{-1}</math>, then the first four axioms for a vector space are satisfied. 
 
''Solution''
1) <math>x\boxplus y = xy = yx = y \boxplus x</math> so addition is commutative. 
2) <math>(x \boxplus y) \boxplus z = (xy) \boxplus z = (xy)z = xyz = x(yz) = x(y \boxplus z) = x\boxplus (y \boxplus z)</math> and addition is associative. 
3) <math>x \boxplus 0_V = x \boxplus 1 = 1x = x</math> 
4) <math>x \boxplus -x = x \boxplus x^{-1} = xx^{-1} = 1 = 0_V</math> 
 
(b) Which of the scalar multiplication properties do not hold? 
 
''Solution''
The only property that won’t hold is associativity of scalar multiplication. 
<math>\alpha \boxdot (\beta \boxdot x) = \alpha (e^\beta x) = e^\alpha e^\beta x</math> which is not the same as <math>(\alpha \beta) \boxdot x = e^{\alpha \beta} x</math>.

Section 1.4 Homework - Revision history

Grad at 06:55, 16 November 2015

Grad: Created page with "'''1.''' Find a subset C \subset \mathbb{F}^2 that is closed under scalar multiplication but not under addition of vectors. ''Solution'' There are m..."

Grad: Created page with "'''1.''' Find a subset $C \subset \mathbb{F}^2$ that is closed under scalar multiplication but not under addition of vectors.

''Solution'' There are m..."