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Homework week 13

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

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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 .

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(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

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