Let V be the set of ordered pairs [a,b] of real
numbers. Decide with clear reason, whether or
not V is a vector space over a field of real
numbers with addition in V and scaler multiplication on V defined by:
a) [a,b]+[c,d] =[ac, bd]
b) k[a,b] = [ka, kb]
c) [a,b]+[c,d] = [(a+c), (b+d)]
c) k[a,b] = [a,b]
step plz i beg idealess
Scalar multiplication properties makes d wrong, and b correct.
Vector addition rules make a) very wrong, and c correct.
I look at the words "clear reason", and wonder. There properties of vectors are pretty basic, http://www.math.ubc.ca/~feldman/m226/vectorppties.pdf
Vector addition and subtraction
Combined vector operations
To determine if the set of ordered pairs [a, b] of real numbers, denoted as V, is a vector space over a field of real numbers, we need to check if it satisfies the properties of vector addition and scalar multiplication.
Here are the steps to determine if V is a vector space:
1. Check closure under vector addition: For any [a,b] and [c,d] in V, compute [a, b] + [c, d] using the given addition operation.
a) [a, b] + [c, d] = [ac, bd]
b) [a, b] + [c, d] = [(a + c), (b + d)]
Verify if both results are also in V, i.e., if ac, bd, (a + c), and (b + d) are all real numbers. If they are, then V satisfies closure under vector addition.
2. Check existence of additive identity: There must be an identity element, denoted as 0, such that for any [a, b] in V, [a, b] + 0 = [a, b].
Substitute [a, b] and 0 into the addition operation obtained in step 1 to verify if the equation holds true. If it does, then V has an additive identity.
3. Check existence of additive inverse: For any [a, b] in V, there must exist another element, denoted as -[a, b], such that [a, b] + (-[a, b]) = 0.
Substitute [a, b] and -[a, b] into the addition operation obtained in step 1 and verify if the equation yields the additive identity (0). If it does, then V has additive inverses.
4. Check closure under scalar multiplication: For any scalar k and [a, b] in V, compute k[a, b] using the given scalar multiplication operation.
a) k[a, b] = [ka, kb]
b) k[a, b] = [a, b]
Verify if both results are in V, i.e., if ka, kb, a, and b are all real numbers. If they are, then V satisfies closure under scalar multiplication.
5. Check compatibility with field operations: For any scalars k, l and [a, b] in V, test if the following properties hold:
- Distributive property: k([a, b] + [c, d]) = k[a, b] + k[c, d]
- Distributive property: (k + l)[a, b] = k[a, b] + l[a, b]
- Associative property: (kl)[a, b] = k(l[a, b])
- Multiplicative identity: 1[a, b] = [a, b]
Substitute values into the corresponding operations from earlier steps and verify if the properties hold true. If they do, then V satisfies compatibility with field operations.
If V satisfies all of these properties, then it can be considered a vector space over a field of real numbers. If any one of the properties fails, then V is not a vector space.