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In mathematics, a vector space (also called a linear space) is a collection of objects, called vectors, that can be added together and multiplied by scalars (usually real or complex numbers) in a consistent and meaningful way. More formally, a vector space is a set $V$ together with two operations: vector addition and scalar multiplication, which satisfy the following axioms:
- Closure under vector addition: For any vectors $\mathbf{u},\mathbf{v} \in V$, their sum $\mathbf{u}+\mathbf{v}$ is also in $V$.
- Associativity of vector addition: For any vectors $\mathbf{u},\mathbf{v},\mathbf{w} \in V$, we have $(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})$.
- Commutativity of vector addition: For any vectors $\mathbf{u},\mathbf{v} \in V$, we have $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$.
- Existence of a zero vector: There exists a vector $\mathbf{0} \in V$ such that $\mathbf{v}+\mathbf{0}=\mathbf{v}$ for any $\mathbf{v} \in V$.
- Existence of additive inverses: For any vector $\mathbf{v} \in V$, there exists a vector $-\mathbf{v} \in V$ such that $\mathbf{v}+(-\mathbf{v})=\mathbf{0}$.
- Closure under scalar multiplication: For any scalar $a$ and any vector $\mathbf{v} \in V$, their product $a\mathbf{v}$ is also in $V$.
- Distributivity of scalar multiplication over vector addition: For any scalars $a$ and $b$ and any vector $\mathbf{v} \in V$, we have $a(b\mathbf{v})=(ab)\mathbf{v}$.
- Distributivity of scalar multiplication over scalar addition: For any scalars $a$ and $b$ and any vector $\mathbf{v} \in V$, we have $(a+b)\mathbf{v}=a\mathbf{v}+b\mathbf{v}$.
- Multiplicative identity: For any vector $\mathbf{v} \in V$, we have $1\mathbf{v}=\mathbf{v}$.
These axioms capture the basic properties of linear algebra and allow us to define and study a wide range of mathematical objects, including vectors, matrices, linear transformations, and more. Examples of vector spaces include the space of real-valued $n$-dimensional vectors $\mathbb{R}^n$, the space of polynomials of degree at most $n$ with coefficients in $\mathbb{R}$, and the space of continuous functions on a closed interval.
If V is the set of ordered pairs (x, y) of real numbers, show that it is a vector space with addition(x, y) + (x1, y1) = (x + x1, y + y1 + 1) and scalar multiplication a(x, y) = (ax, ay + a − 1). What is the zero vector in V?
To show that V is a vector space with the given operations of addition and scalar multiplication, we need to verify the eight axioms of a vector space. These are:
- Closure under addition: For any (x, y), (x1, y1) in V, (x, y) + (x1, y1) = (x + x1, y + y1 + 1) is also in V.
- Commutativity of addition: For any (x, y), (x1, y1) in V, (x, y) + (x1, y1) = (x1, y1) + (x, y).
- Associativity of addition: For any (x, y), (x1, y1), and (x2, y2) in V, (x, y) + [(x1, y1) + (x2, y2)] = [(x, y) + (x1, y1)] + (x2, y2).
- Existence of zero vector: There exists a vector 0 in V such that for any (x, y) in V, (x, y) + 0 = (x, y).
- Existence of additive inverse: For any (x, y) in V, there exists a vector (-x, -y-1) in V such that (x, y) + (-x, -y-1) = 0.
- Closure under scalar multiplication: For any scalar a and any (x, y) in V, a(x, y) = (ax, ay + a − 1) is also in V.
- Distributivity of scalar multiplication over vector addition: For any scalar a and any (x, y), (x1, y1) in V, a[(x, y) + (x1, y1)] = a(x, y) + a(x1, y1).
- Distributivity of scalar multiplication over scalar addition: For any scalars a and b, and any (x, y) in V, (a + b)(x, y) = a(x, y) + b(x, y).
Now we will prove each axiom:
- (x, y) + (x1, y1) = (x + x1, y + y1 + 1) is in V because it is an ordered pair of real numbers.
- (x, y) + (x1, y1) = (x + x1, y + y1 + 1) = (x1 + x, y1 + y + 1) = (x1, y1) + (x, y).
- (x, y) + [(x1, y1) + (x2, y2)] = (x, y) + (x1 + x2, y1 + y2 + 1) = (x + x1 + x2, y + y1 + y2 + 2) = [(x + x1, y + y1 + 1) + (x2, y2)] = [(x, y) + (x1, y1)] + (x2, y2).
- The zero vector in V is (0, -1), since for any (x, y) in V, (x, y) + (0, -1) = (x + 0, y + -1 + 1) = (x, y).
- For any (x, y) in V, the additive inverse is (-x, -y-1
The set V of 2×2 matrices with zero determinant;usual matrix operations.
The set V of 2×2 matrices with zero determinant is a vector space with the usual matrix operations of addition and scalar multiplication. To show this, we need to verify the eight axioms of a vector space:
- Closure under addition: For any matrices A and B in V, A + B is also in V, since the determinant of A + B is the sum of the determinants of A and B, both of which are zero.
- Commutativity of addition: For any matrices A and B in V, A + B = B + A, since matrix addition is commutative.
- Associativity of addition: For any matrices A, B, and C in V, (A + B) + C = A + (B + C), since matrix addition is associative.
- Existence of zero vector: The zero matrix is in V, since its determinant is zero.
- Existence of additive inverse: For any matrix A in V, the additive inverse -A is also in V, since the determinant of -A is the negative of the determinant of A, which is zero.
- Closure under scalar multiplication: For any scalar k and any matrix A in V, kA is also in V, since the determinant of kA is k times the determinant of A, which is zero.
- Distributivity of scalar multiplication over vector addition: For any scalar k and any matrices A and B in V, k(A + B) = kA + kB, since scalar multiplication distributes over matrix addition.
- Distributivity of scalar multiplication over scalar addition: For any scalars k and l and any matrix A in V, (k + l)A = kA + lA, since scalar multiplication distributes over scalar addition.
Therefore, the set V of 2×2 matrices with zero determinant is a vector space with the usual matrix operations. The zero vector in V is the zero matrix.
By calculating (1 + 1)(v + w) in two ways (using axioms S2 and S3), show that axiom A2 follows from the other axioms.
Let V be a vector space and let v and w be vectors in V. We want to show that axiom A2, which states that (a + b)v = av + bv for any scalars a and b and any vector v in V, follows from the other axioms.
Using axiom S2, we have:
(1 + 1)(v + w) = 2(v + w) (distributing scalar multiplication) = v + w + v + w (using axiom S3 twice) = (1v + 1w) + (1v + 1w) (distributing scalar multiplication) = (1 + 1)v + (1 + 1)w (distributing scalar multiplication) = 2v + 2w (simplifying scalar multiplication)
Using axiom S3, we have:
(1 + 1)(v + w) = 1(v + w) + 1(v + w) (distributing scalar multiplication) = v + w + v + w (using axiom S2 twice) = (1v + 1v) + (1w + 1w) (distributing scalar multiplication) = (1 + 1)v + (1 + 1)w (simplifying scalar multiplication) = 2v + 2w
Since we have shown that (1 + 1)(v + w) equals 2v + 2w using both axioms S2 and S3, we can equate the two expressions:
2v + 2w = (1 + 1)v + (1 + 1)w
Simplifying both sides, we get:
2(v + w) = 2v + 2w
Canceling the common factor of 2, we get:
v + w = v + w
This is the same as the left-hand side of axiom A2. Therefore, axiom A2 follows from the other axioms.
