Download Math 5A: Homework #10 Solution

Math 5A: Homework #10 Solution
12. This is the set of vectors (x, y, z) such that x > 0, y > 0, z > 0. However, this does not form a vector space. It violates closure under scalar
multiplication (C2), since (1, 1, 1) is in the first octant, but −1(1, 1, 1) =
(−1, −1, −1) is not.
14. The set of all polynomials of degree 2 does form a vector space. The sum
of any two degree 2 polynomials is another polynomial of degree less than
or equal to 2. And multiplying a degree 2 polynomial by a constant is
another degree 2 polynomial. The other properties are easy to check as
well.
a 0
16. The set of all diagonal 2x2 matrices is a vector space. Since
+
0 b
c 0
a+c
0
=
, it is clear that the sum of any 2 diagonal
0 d
0
b+d
a 0
ac 0
matrices is a diagonal matrix. Also c
=
, so a
0 b
0 bc
diagonal matrix times a scalar is again a diagonal matrix. The other
properties follow from matrix addition rules.
1 0
18. The set of all invertible 2x2 matrices is not a vector space.
is
0 1 −1 0
1 0
−1 0
invertible and
is invertible, but
+
=
0 −1
0 1
0 −1
0 0
is not invertible.
0 0
52. The first quadrant is closed under addition but not scalar multiplication.
Any vector in the first quadrant can be expressed (x, y) where x > 0
and y > 0. Adding any two vectors of this form, we get (x, y) + (z, t) =
(x + z, y + t) and x + z > 0 since x, z > 0 and y + t > 0, since y, t > 0. So
the first quadrant is closed under addition. However −1(1, 1) = (−1, −1)
is not in the first quadrant, so the first quadrant is not closed under scalar
multiplication.
54. Consider the set of vectors of the form (x, 0) and (0, y) where x > 0 and
y > 0. Since (1, 0) and (0, 1) are in this set, but (1, 0) + (0, 1) = (1, 1) is
not, this set is not closed under addition. And since −1(1, 0) = (−1, 0) is
not in this set, it is not closed under scalar multiplication either.
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