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So the question asks: Let $R^R$ be the vector space of all functions f : R −→ R, and let $R^N$ be the vector space of all infinite sequences, or tuples, (x1, x2, . . . , xn, . . .) of real numbers.

(a) Show that both $R^R$ and $R^N$ are infinite-dimensional.

(b) Prove that the function T : $R^R$ −→ $R^N$ given by T(f) = (f(1), f(2), . . .) is a linear transformation.Why doesn’t it make sense to talk about the matrix of the linear transformation T?

So so far I have:

For (a) ,

Since the functions from R to R has polynomial, trigonometric, exponential and etc.

By Taylor series expansion, every function has a polynomial representation of either finite degree or infinite degree.

So, the set of all functions over real numbers can be written as

$R^R$ = {$a_0+ a_1x + a_2x_2+…+ a_nx_n + … |a_i∈R$}

So each function can be written as n-tuples where n is finite or infinite as

$a_0+ a_1x + a_2x_2+…+ a_nx_nn+ … =(a_1, a_2, …, a_n, …)$

So $R^R$ = {$(a_1, a_2, …, a_n, …)|a_i∈R$}

This forms a vector space under addition of n- tuples and scalar multiplication.

Therefore, the function from R to R forms an infinite dimensional vector space $R^R$.

The space of all infinite sequences of real numbers can be written as

$R^N=$ {$(a_1, a_2, …, a_n, …)|a_i∈R$}

Each sequence can be written as a linear combination of

S = {(1,0,0,…,0,…),(0,1,0,…,0,…),(0,0,1,0,…,0,…),…}

So there are infinite vectors in S which span $R^N$

Therefore, $R^N$ is an infinite dimensional vector space.

For (b),

I know to prove a transformation is linear, it has to satisfies:

  1. $T(v_1+v_2)=T(v_1)+T(v_2) $for any vectors $v_1$ and $v_2$ in V, and

  2. $T(kv)=kT(v)$ for any scalar k.

So I need to prove that

  1. T(f1) +T(f2) = T(f1+f2)

  2. T(k*f1) = k T(f1)

But how am I supposed to do this? And the question asks that why it makes no sense to talk about the matrix of T, is it because the transformation is from an infinite-dimensional vector space to the another infinite-dimensional vector space, and the matrix is abstract which has infinite basis?

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    $\begingroup$ "By Taylor series expansion, every function has a polynomial representation of either finite degree or infinite degree". This is not true. $\endgroup$ – user258700 Feb 21 '16 at 5:43
  • $\begingroup$ You're argument for infinite dimension is wrong. Also, linear combinations have only finitely many terms. You can't do series in plain linear spaces without topology. $\endgroup$ – user251257 Feb 21 '16 at 6:31
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First, not all functions from the reals to the reals are "polynomial, triginometric, exponential". These form in fact a very tiny subset of all functions. A function is not a formula. It's just that for every $x \in \mathbb{R}$ there exists a unique value $f(x) \in \mathbb{R}$. The unicity is what makes it a function. The asignment of $f(x)$ to $x$ is completely arbitrary, e.g. I could take the 7-th digit in the decimal (infinite) representation of $x$ for all $x > \sqrt{17}$, the 5-th digit of that representation for all $x < \sqrt{17}$ and $f(\sqrt{17}) = \pi$. And this is even relatively nice, because I can write a program for it, but this need not be the case in general. A truly "random" function would just be an infinite list of values, one for each real, assigned all independently of each other. So there is no hope for Taylor expansions etc. Realise that $\mathbb{R}^\mathbb{R}$ is truly a huge set.

The operations on $\mathbb{R}^\mathbb{R}$ are pointwise, so if we have two such functions $f,g$ then we define $f+g$ as a new function, by telling what its value on an arbitrary $x \in \mathbb{R}$ is: $(f+g)(x) = f(x) + g(x)$, i.e. we just add the values of $f$ and $g$ at $x$. Similarly for a scalar $c \in \mathbb{R}$, we define $(c\cdot f)(x) = cf(x)$ for all $x \in \mathbb{R}$, where the latter is just standard multiplication in $\mathbb{R}$. The $0$ is just the function where all values are equal to $0$.

Both have infinite sets of linearly independent elements (or vectors, as elements of a vector space are called, even though they are not "vectors" in the old fashioned sense, like the functions in $\mathbb{R}^\mathbb{R}$): take the functions $f_p$, defined for a fixed $p \in \mathbb{R}$: $f_p(x) = 1$ if $x = p$, and $f_p(x) = 0$ if $x \neq p$. So all $0$ except for a spike at $p$.

Why are the $f_p$ linearly independent? By definition, we need to consider a finite linear combination of distinct $f_{p_i}$: $c_{p_1} \cdot f_{p_1} + \ldots + c_{p_n} \cdot f_{p_n} = 0$ (equality as functions, which means just that they have the same values for all $x$), and we need to show that then all $c_{p_i}$ are $0 \in \mathbb{R}$. Because the equality holds for all $x$, we can use $x = p_1$ in particular. Then $$(c_{p_1} \cdot f_{p_1} + \ldots + c_{p_n} \cdot f_{p_n})(p_1) = c_{p_1}f_{p_1}(p_1) + \ldots + c_{p_n} f_{p_n}(p_1) = 0$$ As $f_{p_2}(p_1) = 0$, as $p_2 \neq p_1$, and so on, but $f_{p_1}(p_1) = 1$ this comes down to $c_{p_1} = 0$. The same idea works for all other coefficients as well. So the set of $f_p$ is linearly independent.

The same idea works in $\mathbb{R}^\mathbb{N}$, the set of sequences, which look more like normal vectors, but of infinite length. We just see this as the set of functions from $\mathbb{N}$ to $\mathbb{R}$ and the same operations and independent functions apply. So both spaces are infinite-dimensional.

If we see $\mathbb{R}^\mathbb{N}$ as a set of functions (as we should) then the function $T$ is just the restriction of $f$ to $\mathbb{N}$. To see that this is linear, take $f,g \in \mathbb{R}^\mathbb{R}$. Then $T(f+g)$ is defined for all $n$ as $(f+g)(n)$, which is in turn defined as $f(n) + g(n)$, and this equals $T(f)(n) + T(g)(n)$, as $T$ does "nothing", it's just the restriction of a function to a smaller domain. The latter sum is just by definition $(T(f) + T(g))(n)$, and as this holds for all $n$ we have equality of functions (or sequences, because we index by $\mathbb{N}$) and $T(f+g) = T(f) + T(g)$. The same can be done for scalars as well.

A matrix for a linear map $T$ is formed by choosing bases for both spaces and computing the base expansion for every $T(b)$ for all basis elements in the domain (as the columns). But here a base for $\mathbb{R}^\mathbb{R}$ is uncountable, so we cannot write it down as a matrix: these can be at most countably infinite in both dimensions.

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Since by (usual, standard) definition

$$\;(f_1+f_2)(x):=f_1(x)+f_2(x)\;\;,\;\;\;(kf)(x):=k\cdot f(x)\;,\;\;k,\,x\in\Bbb R$$

we then get

$$T(f_1+f_2)=\left(f_1(1)+f_2(1),\,f_1(2)+f_2(2),\,\ldots\right)=$$

$$=(f_1(1),f_1(2),\ldots)+(f_2(1),\,f_2(2),\,\ldots)=T(f_1)+T(f_2)$$

andalso

$$T(Kf):=\left(kf(1),\,kf(2),\,\ldots\right)=k\cdot(f(1),f(2),\ldots)=kT(f)$$

and thus $\;T\;$ is a linear map.

As you said: since both linear spaces are infinite-dimensional it'd be impossible to write down its matrix with respect to any basis of both spaces (though I've read sometimes about matrices of infinite degree but this seems to be more a way to try to imagine this abstract).

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  • $\begingroup$ Why is it down voted? $\endgroup$ – user251257 Feb 21 '16 at 6:26
  • $\begingroup$ @user251257 I have no idea. I think the above is completely accurate, but apparently someone didn't like something and he (or she) didn't write anything. $\endgroup$ – DonAntonio Feb 21 '16 at 6:32

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