Download Recitation Notes Spring 16, 21-241: Matrices and Linear Transformations January 26, 2016

Recitation Notes
Spring 16, 21-241: Matrices and Linear Transformations
January 26, 2016
Abstract
[TOPICS COVERED] Scope of qualifiers on statements. Negating statements. Axioms of
vector spaces. Examples of vector spaces. Functions in polynomial spaces.
1
Administrative Matters
• Homework deadline: Wednesdays 3:20pm.
• Weekly online assignments: Monday 10:00pm
2
Definitions
1. Axioms (hence can you prove axioms?)
2. Vector spaces
3
Problems
1. Negate the following statement:
∀{an ∈ R : i ∈ N},
∀ ∈ R, > 0. ∃n0 ∈ N. ∃` ∈ R. ∀n ∈ N, n ≥ n0 . |an − `| < =⇒ ∀ ∈ R, > 0. ∃n0 ∈ N. ∀n ∈ N, n ≥ n0 . ∀m ∈ N, m ≥ n0 . |an − am | < .
Solution.
∃∀{an ∈ R : i ∈ N},
(∀ ∈ R, > 0. ∃n0 ∈ N. ∃` ∈ R. ∀n ∈ N, n ≥ n0 . |an − `| < )
∧ (∃ ∈ R, > 0. ∀n0 ∈ N. ∃n ∈ N, n ≥ n0 . ∃m ∈ N, m ≥ n0 . |an − am | ≥ )
Note: Do not be intimidated by complicated statements. Always identify qualifiers and
sub-statements.
2. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 1.1.7,9. Compute the
following:
1
2
(a)
+
3
3
−2
(b)
−
−2
3
3
0
Solution.
(a) Addition is done component-wise.
3
2
3+2
5
+
=
=
0
3
0+3
3
(b) Perform operation component-wise.
3
−2
3 − (−2)
5
−
=
=
−2
3
−2 − 3
−5
How are these expressed geometrically?
3. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 1.1.15. Simplify the given
expression:
2(x − 3y) + 3(2y + x)
Solution.
2(x − 3y) + 3(2y + x)
= 2x − 6y + 6y + 3x
= 5x + 0y
= 5x
4. Prove the following properties for vectors in Rn :
(a) x + y = y + x
(b) (x + y) + z = x + (y + z)






x1
y1
z1






Solution. Let x =  ... , y =  ... , and z =  ... .
xn
yn
zn
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(a)
x+y

 

x1
y1

 

=  ...  +  ... 
xn
yn


x 1 + y1


..
= 

.
by definition of vector addition in Rn
x n + yn


y1 + x 1


..
= 

.
yn + x n

 

y1
x1

 

=  ...  +  ... 
yn
xn
by commutativity of real addition
by definition of vector addition in Rn
= y+x
(b)
(x + y) + z

 
 
x1
y1

 
 
=  . . .  +  . . .  + 
xn
yn

 

x 1 + y1
z1

  .. 
..
= 
+ . 
.
x n + yn
zn


(x1 + y1 ) + z1


..
= 

.
(xn + yn ) + zn


x1 + (y1 + z1 )


..
= 

.
xn + (yn + zn )

 

y1 + z 1
x1

 

..
=  ...  + 

.
yn + z n

 
x1
y1

 
 
=  . . .  +  . . .  + 
xn
yn

z1
.. 
. 
zn
by definition of vector addition in Rn
by definition of vector addition in Rn
by associativity of real addition
by definition of vector addition in Rn
xn



z1
.. 
. 
zn
= x + (y + z)
Page 3
by definition of vector addition in Rn
5. David Poole Linear Algebra: a modern introduction (4th Ed.) Ex 6.1.5,7. Are the following
vector spaces?
(a) R2 ,with the usual addition but scalar multiplication defined by
x
cx
c
=
y
y
(b) The set of positive real numbers, with addition ⊕ defined by x ⊕ y = xy and scalar
multiplication defined by c x = xc .
Solution.
(a) No. Does not fulfill (c + d)x = cx + dx.
1
1
2
(1 + 1)
=2
=
1
1
1
but
1
1
1
+1
1
1
=
1
1
+
1
1
=
2
2
.
(b) Yes. It satisfies all axioms of a vector space.
(a) x ⊕ y = xy ∈ R.
(b) x ⊕ y = xy = yx = y ⊕ x by commutativity of real multiplication.
(c) (x ⊕ y) ⊕ z = xy ⊕ z = (xy)z = x(yz) = x ⊕ yz = x ⊕ (y ⊕ z) by associativity of real
multiplication.
(d) 1 satisfies the condition. x ⊕ 1 = x 1 = x since 1 is the multiplicative identity of the
reals.
1
x ∈
R+ .
(e) Since x ∈ R+ ,
(f) c x = xc ∈
R+ is well defined. Then x ⊕
1
x
=x
1
x
= 1.
(g) c (x ⊕ y) = c xy = (xy)c = xc y c = (c x)(c y) = (c x) ⊕ (c y).
(h) (c + d) x = xc+d = xc xd = (x c)(x d) = (x c) ⊕ (x d).
(i) c (d x) = c (xd ) = (xd )c = x(dc) = (dc) x = (cd) x by commutativity of real
multiplication.
(j) 1 x = x1 = x.
4
Additional Notes
1. Generally there are two kinds of theorems: Implications and Equivalent statements.
(a) Example of implication:
Let x ∈ Z. If x is a positive integer, then 2x is a positive integer.
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(b) Example of equivalent statements:
Let x ∈ Z. The following are equivalent:
x is even
3x is even
−x is even
2. You cannot always commute qualifiers.
Take for example the axioms of vector spaces:
∀x ∈ V, ∃y ∈ V such that x + y = 0
(note that this y is commonly denoted −x) and
∃x ∈ V such that ∀y ∈ V, y + x = y
(note that this x is commonly denoted 0) The first statement means that every element in
the vector space has an inverse. The second statement means that there exists a zero vector.
3. In your work, it may be advantageous to write vectors and scalars differently in order to avoid
confusion.
4. Note that it is not "immediately obvious" that
(−1)x = −x.
This requires proof.
5. Proving that a set is a vector space is usually tedious. It is inadvisable to embark on such
a proof without some degree of certainty that the set (with operations defined) is indeed a
vector space. The following serve as simple checks.
(a) What is the zero vector?
(b) Given an element, does the inverse exist?
(c) Is the set closed under vector addition?
(d) Is the set closed under scalar multiplication?
5
Exercises
1. Prove the following:
(a) The zero vector is unique
(b) The inverse is unique
(c) 0 · x = 0
(d) (−1) · x = −x
2. Prove or disprove that the power set of Z with symmetric set difference as vector addition and
{0, 1} as scalars is a vector space.
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