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Chapter 6: Problem 10

Suppose \(V\) and \(W\) are vector spaces. Define \(V \times W\) by $$V \times W=\\{(f, g): f \in V \text { and } g \in W\\}$$ Define addition and scalar multiplication on \(V \times W\) by $$ \left(f_{1}, g_{1}\right)+\left(f_{2}, g_{2}\right)=\left(f_{1}+f_{2}, g_{1}+g_{2}\right) \quad \text { and } \quad \alpha(f, g)=(\alpha f, \alpha g)$$ Prove that \(V \times W\) is a vector space with these operations.

Short Answer

Expert verified
In order to prove that $V \times W$ is a vector space, we need to verify that it satisfies the vector space axioms under the given operations of addition and scalar multiplication. By going step by step, we have shown that: 1. Associativity of addition holds true for $V \times W$. 2. Commutativity of addition holds true for $V \times W$. 3. The identity element of addition exists in $V \times W$. 4. The inverse element of addition exists in $V \times W$. 5. Compatibility of scalar multiplication holds true for $V \times W$. 6. The identity element of scalar multiplication exists in $V \times W$. 7. The distributive property for scalar multiplication holds true for $V \times W$. 8. The distributive property for scalar addition holds true for $V \times W$. Since all the vector space axioms are satisfied, we can conclude that $V \times W$ is indeed a vector space with the given operations of addition and scalar multiplication.

Step by step solution

01

1. Associativity of addition

First, we need to prove that for any elements (f1, g1), (f2, g2), and (f3, g3) in V x W, the following holds: \[((f_{1}, g_{1}) + (f_{2}, g_{2})) + (f_{3}, g_{3}) = (f_{1}, g_{1}) + ((f_{2}, g_{2}) + (f_{3}, g_{3}))\] Since V and W are vector spaces, f1 + f2 + f3 = f1 + (f2 + f3) and similarly for g. Calculating the left side, we get: \[((f_{1}+f_{2}, g_{1}+g_{2}) + (f_{3}, g_{3}) = (f_{1}+f_{2}+f_{3}, g_{1}+g_{2}+g_{3})\] Calculating the right side, we get: \[(f_{1}, g_{1}) + (f_{2}+f_{3}, g_{2}+g_{3}) = (f_{1}+(f_{2}+f_{3}), g_{1}+(g_{2}+g_{3})) = (f_{1}+f_{2}+f_{3}, g_{1}+g_{2}+g_{3})\] Since the left and right sides are equal, associativity of addition holds true for V x W.
02

2. Commutativity of addition

Next, we need to prove that for any elements (f1, g1) and (f2, g2) in V x W, the following holds: \[(f_{1}, g_{1}) + (f_{2}, g_{2}) = (f_{2}, g_{2}) + (f_{1}, g_{1})\] Since V and W are vector spaces, f1 + f2 = f2 + f1 and similarly for g. Calculating both sides, we get: \[(f_{1}+f_{2}, g_{1}+g_{2}) = (f_{2}+f_{1}, g_{2}+g_{1})\] Since they are equal, commutativity of addition holds true for V x W.
03

3. Identity element of addition

Next, we need to show that there exists an element (0, 0) in V x W such that for any element (f, g) in V x W, the following holds: \[(f, g) + (0, 0) = (f, g)\] Since V and W are vector spaces, f + 0 = f and g + 0 = g. Thus, we can conclude that for any element (f, g) in V x W, \[(f, g) + (0, 0) = (f+0, g+0) = (f, g)\] This shows that the identity element of addition exists in V x W.
04

4. Inverse element of addition

Next, we need to show that for any element (f, g) in V x W, there exists an element (-f, -g) in V x W such that the following holds: \[(f, g) + (-f, -g) = (0, 0)\] Since V and W are vector spaces, f + (-f) = 0 and g + (-g) = 0. Thus, we can conclude that for any element (f, g) in V x W, \[(f, g) + (-f, -g) = (f+(-f), g+(-g)) = (0, 0)\] This shows that the inverse element of addition exists in V x W.
05

5. Compatibility of scalar multiplication

Next, we need to prove that for any scalars 伪 and 尾, and any element (f, g) in V x W, the following holds: \[\alpha(\beta(f, g)) = (\alpha\beta)(f, g)\] Since V and W are vector spaces, 伪(尾f) = (伪尾)f and 伪(尾g) = (伪尾)g. Calculating both sides, we get: \[\alpha(\beta f, \beta g) = (\alpha (\beta f), \alpha (\beta g)) = ((\alpha \beta) f, (\alpha \beta) g)\] Since they are equal, compatibility of scalar multiplication holds true for V x W.
06

6. Identity element of scalar multiplication

Next, we need to show that for any element (f, g) in V x W, there exists a scalar 1 such that the following holds: \[1(f, g) = (f, g)\] Since V and W are vector spaces, 1f = f and 1g = g. Thus, we can conclude that for any element (f, g) in V x W, \[1(f, g) = (1f, 1g) = (f, g)\] This shows that the identity element of scalar multiplication exists in V x W.
07

7. Distributive property for scalar multiplication

Next, we need to prove that for any scalars 伪, and any elements (f1, g1), (f2, g2) in V x W, the following holds: \[\alpha((f_{1}, g_{1}) + (f_{2}, g_{2})) = \alpha(f_{1}, g_{1}) + \alpha(f_{2}, g_{2})\] Since V and W are vector spaces, 伪(f1 + f2) = 伪f1 + 伪f2 and 伪(g1 + g2) = 伪g1 + 伪g2. Calculating both sides, we get: \[\alpha(f_{1}+f_{2}, g_{1}+g_{2}) = (\alpha(f_{1}+f_{2}), \alpha(g_{1}+g_{2})) = (\alpha f_{1}+\alpha f_{2}, \alpha g_{1}+\alpha g_{2})\] \[\alpha(f_{1}, g_{1}) + \alpha(f_{2}, g_{2}) = (\alpha f_{1}, \alpha g_{1}) + (\alpha f_{2}, \alpha g_{2}) = (\alpha f_{1}+\alpha f_{2}, \alpha g_{1}+\alpha g_{2})\] Since they are equal, the distributive property for scalar multiplication holds true for V x W.
08

8. Distributive property for scalar addition

Finally, we need to prove that for any scalars 伪 and 尾, and any element (f, g) in V x W, the following holds: \[(\alpha + \beta)(f, g) = \alpha(f, g) + \beta(f, g)\] Since V and W are vector spaces, (伪 + 尾)f = 伪f + 尾f and (伪 + 尾)g = 伪g + 尾g. Calculating both sides, we get: \[(\alpha + \beta)(f, g) = ((\alpha + \beta)f, (\alpha + \beta)g) = (\alpha f+\beta f, \alpha g+\beta g)\] \[\alpha(f, g) + \beta(f, g) = (\alpha f, \alpha g) + (\beta f, \beta g) = (\alpha f+\beta f, \alpha g+\beta g)\] Since they are equal, the distributive property for scalar addition holds true for V x W. We have shown that all of the vector space axioms are satisfied for V x W and its operation of addition and scalar multiplication. Therefore, V x W is indeed a vector space as the exercise asked to prove.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Associativity of Addition
Understanding the associativity of addition is like realizing that no matter how you group friends for a group photo, everyone is still there. In the realm of vector spaces, this means you can add vectors in any order without changing the result. For instance, suppose we have three vectors called 'A', 'B', and 'C'. If we add them like this \(A+B)+C\), or as \(A+(B+C)\), it's all the same by the end. They group nicely, just like friends do, ensuring that the sum doesn't depend on how they are partnered up.
Commutativity of Addition
Commutativity of addition is like deciding whether socks or pants go on first; the result is you're dressed. In vector spaces, this principle affirms that the order of addition doesn't matter. Take two vectors, 'X' and 'Y'鈥攁dding them as \(X+Y\) or \(Y+X\) will lead to the same stylish outcome. This fundamental property eases up computations by allowing flexibility in the order of operations.
Identity Element of Addition
Imagine going out just to take a walk around the block and ending up right back at home鈥攖his is similar to the identity element of addition in vector spaces. It's the zero vector, often noted as '0', that, when added to any vector 'V', leaves 'V' unchanged: \((V+0) = V\). The zero vector is the home base; adding it to any vector brings you back to the original vector.
Inverse Element of Addition
Think of the inverse element of addition as a reset button. Every vector 'R' in a vector space has a unique inverse \(-R\) that when added together, they cancel each other out: \(R+(-R)=0\). It's like having a debt and exactly enough money to pay it off, landing at a net worth of zero.
Compatibility of Scalar Multiplication
Imagine you have a photo and you want to zoom in by doubling its size twice. You could double the size and then double again, or you could just quadruple it in one go. In vector spaces, this is known as the compatibility of scalar multiplication. Mathematically, it says \(a(bV) = (ab)V\) for any scalar 'a' and 'b', and vector 'V'. No matter if you scale in steps or all at once, the vector's end size is the same.
Identity Element of Scalar Multiplication
If we think about the number 1 as a 'do nothing' multiplier in a recipe, that's our identity element of scalar multiplication. In the kitchen of vector spaces, multiplying a vector 'U' by 1 does not change its flavor: \((1U = U\)). It鈥檚 the 鈥榢eep it as it is鈥� of the scalar world, confirming that multiplying a vector by 1 gives you back the same vector.
Distributive Property for Scalar Multiplication
Baking two batches of cookies separately or combined should give the same total amount of cookies. This is like the distributive property for scalar multiplication with vectors. Mathematically, this baking wisdom translates to taking a scalar 's' and vectors 'D' and 'E', then the distributive law says \((s(D+E) = sD+sE\)). So whether you distribute the scalar before or after adding the vectors, you get the same scrumptious result.
Distributive Property for Scalar Addition
When you're at a buffet and you take a plate of food, adding a little bit of two different items doesn't change if you get them together or separately. Similarly, in vector math, for scalars 'm' and 'n' and a vector 'F', the distributive property of scalar addition assures that \((m+n)F = mF+nF\). Divide your attention between dishes or take a bit of everything at once; your plate holds the same amount of food either way.

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Most popular questions from this chapter

Show that there exists a linear functional \(\varphi: \ell^{\infty} \rightarrow \mathbf{F}\) such that $$\left|\varphi\left(a_{1}, a_{2}, \ldots\right)\right| \leq\left\|\left(a_{1}, a_{2}, \ldots\right)\right\|_{\infty}$$ for all \(\left(a_{1}, a_{2}, \ldots\right) \in \ell^{\infty}\) and $$\varphi\left(a_{1}, a_{2}, \ldots\right)=\lim _{k \rightarrow \infty} a_{k}$$ for all \(\left(a_{1}, a_{2}, \ldots\right) \in \ell^{\infty}\) such that the limit above on the right exists.

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