If V is a vector space and W is a subset of V then W is a subspace of V if it is also a vector space under the rules of addition and scalar multiple defined on V .
1. Determine which of the following are subspaces of the vector space V
3 over the real numbers .
a) The subset W of all vectors of the for a 0 0 , or in math notation W a 0 0 a b) the subset W of all vectors of the for a 1 1 , or in math notation W a 1 1 a
Solution:
We use the following theorem
————————————————————————————————-
W is a subspace of V if and only if the two axioms of Closure are satisfied, given v u W and k F then: (1) v u W and (5) ku W .
————————————————————————————————–
(a) We use the standard operations of addition and scalar multiple. We test the subset described in
(a). given the vectors of W then a 0 0 b 0 0 a b 0 0 a b 0 0 W which satifies the first closure axiom. Now if k is a scalar of F k a 0 0 ka 0 0 ka 0 0 then which proves the scalar multiple axiom is also satisfied and therefore the subset W is a sbspace.
(b) In a similar fashion, the addition implies a 1 1 b 1 1 a b 2 2 a b 2 2 W therefore W in this case cannot be a subspace.
2. Determine if the following are subspaces of V M
22
All 2x2 Real Matrices over the real numbers
.
a) All 2x2 matrices with integer entries b) All matrices a b c d where a b c d 0.
Solution:
Again we use the theorem for subspaces. We only need to prove that the closure axioms are valid in each case.
1

(a) The set of integer numbers
1 2
, are denoted with the letter . The subset described in (a) would be
W a b c d a b c d now the closure rule under standard addition would be a b c d
A B
C D a A b B c C d D but because the sum of the integers a A, b B, etc. gives also an integer then we conclude that a A b B c C d D
W and the closure under addition is satified. Let us now check the closure under multiplication by a scalar. We always use the same field F that we use in the original vector space, in this case F .
therefore a b ka kb k c d kc kd but now we notice that the components of the new matrix are not integers anymore since ka real integer real sowe conclude that ka kb kc kd
W a b and we say that W is not a subspace of V a b c d c d
(b) We now test the subset (b), in this case the subset is defined by
V a b c d a b c d 0 ; a b c d
First we check the addition a c
1
1 b d
1
1 a
2 c
2 b
2 d
2 let us check if the result is also in the subset W : a c
1
1 a
2 c
2 b
1 d
1 b
2 d
2 a
1 a
2 b
1 b
2 c
1 c
2 d
1 d
2 where in the second row we use the knowledge that therefore a
1 b
1 c
1 d
1
0 and a
2 b
2 c
2 d
2 a
1 a
2 b
1 b
2 c
1 c
2 d
1 d
2 a
1 b
1 c
1 d
1 a
2 b
2 c
2 d
2
0 0 0 a
1 b
1 and a
2 b
2 where in W and c
1 d
1 c
2 d
2
0 were automaticaly satisfied. In conlcusion
W
1 that means the set of numbers
3 2 1 0 1 2 3
2
Remember, the natural numbers are denoted by
1 2 3
2

The second test: closure under multiplication by a real k a b c d ka kb kc kd to verify that this matrix belongs to W we analyse the sum ka kb kc kd k a b c d
k0 0 a b c d where in the second step we use the fact that our matrix a b c d 0. That is ka kb kc kd and therefore we can now say that W is a subspace of V .
W
W so it is satisfied that
3