Download Solutions

Linear Algebra — Homework 1
Solutions
1. (a) [5, ∞) is closed under addition since R is closed under addition (so the sum is a real
number), and if a, b ≥ 5 then a + b ≥ 10, and so greater than 5. It is also closed under
multiplication since product of two positive real numbers is a positive real number, and
multiplying a real number by any number greater than 1 makes it bigger.
(b) (−∞, 0] is closed under addition since the sum of two negative numbers is another
negative negative number (and 0 leaves all numbers unchanged under addition). This
set is not closed under multiplication since, e.g., (−1)(−1) = 1.
(c) {1, 2, 4, 8, 16, . . . } is closed under multiplication since we can choose 2n and 2m as representative elements (where n and m are natural numbers). Then (2n )(2m ) = 2n+m , which
again is in the set. This set is not closed under addition since, e.g., 1 + 2 = 3. 1 and 2
are in the set but 3 is not.
2.
• (~v ⊕ w)
~ ⊕ ~x = ~v ⊕ (w
~ ⊕ ~x).
• There is an additive identity vector ~0, such that ~v ⊕ ~0 = ~v for each vector ~v .
• Each vector ~v has an additive inverse, −~v , such that ~v ⊕ (−~v ) = ~0.
• ~v ⊕ w
~ =w
~ ⊕ ~v .
• r (~v ⊕ w)
~ = (r ~v ) ⊕ (r w).
~
• (r + s) ~v = (r ~v ) ⊕ (s ~v ). (Note the different plus signs here.)
• (rs) ~v = r (s ~v )). (Note the different kinds of multiplication.)
• 1 ~v = ~v .
3. In the expression (−r)~v , the minus sign simply negates the real number r. In the expression
−(r~v ), the minus sign is the symbol for the additive inverse in a vector space. −r should be
read “negative r,” but −(r~v ) should be read “the additive inverse of r~v .”
4. (a) The set of matrices with trace 0 is closed under addition since
a+x b+y
a b
x y
+
=
,
c d
z w
c+z d+w
and if a + d = 0 and x + w = 0 then a + x + d + w = (a + d) + (x + w) = 0 + 0 = 0. The
set of trace 0 matrices is closed under multiplication since
ra rb
a b
r
=
,
c d
rc rd
and if a + d = 0 then ra + rd = r(a + d) = r0 = 0.
0 0
There is an additive identity since the zero matrix
has trace 0. Every trace 0
0 0
matrix has an inverse with trace 0 since
a b
−a −b
0 0
+
=
,
c d
−c −d
0 0
and if a + d = 0 then (−a) + (−d) = −(a + d) = −0 = 0. The other axioms follow as for
ordinary 2 × 2 matrices.
(b) The set of matrices with trace 1 is not a vector space since (for example) it is not closed
under addition:
1 0
0 0
1 0
+
=
,
0 0
0 1
0 1
which does not have trace 1. Also acceptable: you could show this set is not closed
under scalar multiplication, does not contain an additive identity, or does not contain
additive inverses for all its elements.
5. Z2 is not a vector space since it is not closed under scalar multiplication. For example,
1
(3, 4) = ( 23 , 2) ∈
/ Z2 .
2