1/ Determine whether the given set is a vector space.
If it isn't, state the axioms that fail.(click here for list of axioms)

a) The set of diagonal n x n matrices under the standard definitions
of matrix addition and scalar multiplication.

b) All vectors in R² in the first and third quadrants.

c) The set of all vectors in R³ in which x = y = z.

d) The set of 2 x 2 matrices of the form under the
standard definitions of addition and scalar multiplication.

e) Vectors in R² with the standard definition of scalar multiplication and
addition defined by

f) All vectors in R³ such that u ₁ + 2u ₂ – 3u ₃ = 0.

(18)

2/ Given that x and y are vectors in a vector space V, prove that
there exists a unique vector z in V, such that x + z = y.

(5)

3/ Determine whether subset H is a subspace of vector space V.

a) H = ; V = R³

b) H = ; V = R³

c) H = ; V = R²

d) H = all 2 x 2 matrices with a₂₁ , a₂₂ = 0 , V = M_{2 x 2}

e) H = all points on the line

(10)

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4)

(4)

b)

(5)

c) Do the vectors

span R³ ? Why or why not?

(3)

TOTAL (45)

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Solutions

1/

a) it's a vector space	b) no, axiom 1 fails	c) it's a vector space
d) it's a vector space	e) no, axioms 7 & 8 fail	f) it's a vector space

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2/ Proof:

assume x + z = y

since x ` V, then –x ` V (axiom 5)

now, – x + (x + z) = – x + y

so, (– x + x) + z = – x + y

which means that 0 + z = –x + y

so, since z is the sum of 2 vectors in V, z is a vector in V.

now we prove it is unique.

Assume w ` V such that x + w = y

then – x + (x + w) = – x + y

and w = –x + y = z

Therefore z is unique.

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3)

a) yes

b) no, ku " H

c) yes

d) yes

e) no, not through origin.

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4)

a) 2u – v = w

b) 3B + 2C = A

c) yes since det A = –1 if A = matrix of vectors

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vector space axioms

Note: ( means for all, means there exists)

1. For any u and v in V, u + v is in V.

2. u + v = v + u for all elements of V.

3. u + (v + w) = (u + v) + w, u, v and w in V.

4. a vector 0 in V, such that u + 0 = 0 + u = u, u in V.

5. a vector (– v), such that v + (– v) = 0, v in V.

6. kv is in V, v in V, and k in R.

7. k(u + v) = ku + kv, u and v in V, and k in R.

8. (k + l)u = ku + lu, u in V and k and l in R.

9. (kl)u = k(lu), u in V and k and l in R.

10. 1u = u, u in V.

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