Let ℱ(ℝ^2, ℝ^2) denote the vector space consisting of all functions 𝐟: ℝ^2 →ℝ^2. (a) Which of th

Let ℱ(ℝ^2, ℝ^2) denote the vector space consisting of all...

Let $\mathcal{F}\left(\mathbb{R}^2, \mathbb{R}^2\right)$ denote the vector space consisting of all functions $\mathbf{f}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$. (a) Which of the following functions $\mathbf{f}(x, y)$ are elements? (iii) $\left(\begin{array}{c}e^x \\ \cos y\end{array}\right)$, (iv) $\left(\begin{array}{l}1 \\ 3\end{array}\right)$ (v) $\left(\begin{array}{cc}x & y \\ -y & x\end{array}\right)$, (vi) $\left(\begin{array}{c}x \\ y \\ x+y\end{array}\right)$. (i) $x^2+y^2,($ ii $)\left(\begin{array}{c}x-y \\ x y\end{array}\right)$, (b) Sum all of the elements of $\mathcal{F}\left(\mathbb{R}^2, \mathbb{R}^2\right)$ you identified in part (a). Then multiply your sum by the scalar -5 . (c) Carefully describe the zero element of the vector space $\mathcal{F}\left(\mathbb{R}^2, \mathbb{R}^2\right)$.

Let $\mathcal{F}\left(\mathbb{R}^2, \mathbb{R}^2\right)$ denote the vector space consisting of all functions $\mathbf{f}: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.
(a) Which of the following functions $\mathbf{f}(x, y)$ are elements?
(iii) $\left(\begin{array}{c}e^x \\ \cos y\end{array}\right)$,
(iv) $\left(\begin{array}{l}1 \\ 3\end{array}\right)$
(v) $\left(\begin{array}{cc}x & y \\ -y & x\end{array}\right)$,
(vi) $\left(\begin{array}{c}x \\ y \\ x+y\end{array}\right)$.
(i) $x^2+y^2,($ ii $)\left(\begin{array}{c}x-y \\ x y\end{array}\right)$,
(b) Sum all of the elements
of $\mathcal{F}\left(\mathbb{R}^2, \mathbb{R}^2\right)$ you identified in part (a). Then multiply your sum by the scalar -5 .
(c) Carefully describe the zero element of the vector space $\mathcal{F}\left(\mathbb{R}^2, \mathbb{R}^2\right)$.

Key Concepts

Vector Space of Functions

This concept refers to a collection of functions that share common domain and codomain—in this case, functions mapping from ?² to ?²—which satisfy the axioms of vector spaces such as closure under addition and scalar multiplication, existence of a zero element, and existence of additive inverses. The vector space structure allows one to combine functions and scale them, just as one does with finite-dimensional vectors.

Well-Defined Mappings

For any function to be an element of the vector space, it must be a well-defined rule that assigns to each element in the domain (?²) exactly one element in the codomain (?²). This means that the output must always be a vector in ?² irrespective of the input, and the function must be defined on the entire domain.

Function Output Consistency

In the context of functions from ?² to ?², each function must output a 2-dimensional vector. This concept emphasizes that for a mapping to be an element of the vector space, its output must consist of exactly two components, ensuring consistency across the domain.

Constant Functions

Constant functions are functions that return the same vector for every input in ?². In this vector space, a constant function is a valid element provided that the constant vector itself is in ?². This showcases an important class of functions within any function space.

Matrix Representation vs. Function

It is important to distinguish between an entity that represents a matrix and one that represents a function. While a function must assign a vector to every point in ?², a static matrix does not, unless it is interpreted as a constant function. Understanding this difference is crucial in identifying elements of the vector space.

Addition and Scalar Multiplication in Function Spaces

These operations are defined pointwise. That is, the sum of two functions f and g is a function where (f+g)(x,y) = f(x,y) + g(x,y), and scalar multiplication is defined by (c*f)(x,y) = c * f(x,y). These properties are essential to the structure of any vector space, ensuring that functions within the space can interact linearly.

Zero Element of a Function Space

In any vector space, the zero element acts as the additive identity. For a function space from ?² to ?², the zero element is the function that maps every point in the domain to the zero vector in ?². This function is critical as it satisfies f(x,y) + 0 = f(x,y) for all functions f in the space.

Transcript

00:01 We'd like to show that r -squared, with the usual definitions for addition and scalar multiplication, is a vector space.

00:10 I'll split this into two parts, i'm going to call them the group axioms and the field action axioms.

00:16 Not things that you need to worry about if it's not terminology you're familiar with, it's just a way for me to organize what the axioms are.

00:22 So first of all, we need there to be some additive identity.

00:26 Well, 0 ,0 plus x ,y is equal to 0 plus x, 0 plus y, by definition of our vector addition, which is equal to x ,y.

00:40 So we have an additive identity.

00:42 That's good, that's nice.

00:45 We also need inverses.

00:49 And so x ,y, for an arbitrary vector in r -squared, plus negative x, negative y, which is also in r -squared, is equal to the identity 0 ,0.

01:00 That's good.

01:02 We need the addition to be associative, so let's do that real quick.

01:08 A ,b plus c ,d plus e ,f.

01:15 If we resolve this guy first, we get a ,b plus c plus e over d plus f.

01:24 And then that ends up being a plus c plus e, b plus d plus f.

01:35 Or, if we resolve this guy first, then it becomes a plus c over b plus d plus e ,f.

01:51 And that is a plus c, parenthetically, plus e, b plus d plus f.

01:59 And then these are equal by the ordinary associativity of real number addition.

02:05 This operation is associative.

02:08 And finally, we have x ,y plus a ,b is equal to, by definition, x plus a over y plus b.

02:19 Which, because real addition is commutative, gives us a plus x over b plus y, which is, of course, a ,b plus x ,y.

02:31 Which is great, that means that our vector addition is commutative.

02:37 So, the ordinary addition of vectors gives us everything we need for addition to work.

02:43 Now all we need is to do a scalar multiplication.

02:48 First of all, it needs to be associative.

02:52 So we've got alpha times beta times x ,y is what we're doing here...

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