# Is it true that finite-dimensional vector spaces can have infinite sequences as elements?

I recently encountered the following definition of a finite-dimensional vector space in Axler's Linear Algebra Done Right: A vector space is called finite-dimensional if some list of vectors in it spans the space.

This I was then thinking about $$V = \{(x, 0, 0, \cdots): x \in \mathbb{R} \}$$ i.e., the set of all infinite sequences whose first value is some real number and for which all other values are 0.

The book defines $${\mathbb{R}}^{\infty} = \{(x_1, x_2, \cdots): x_j \in \mathbb{R} \text{ for } j = 1, 2, \cdots \}.$$ He defines addition and scalar multiplication with $${\mathbb{R}}^{\infty}$$ as you'd expect: $$(x_1, x_2, \cdots) + (y_1, y_2, \cdots) = (x_1+y_1, x_2+y_2, \cdots)$$ and $$\lambda(x_1, x_2, \cdots) = (\lambda x_1, \lambda x_2, \cdots)$$.

Clearly $$V$$ is a subspace of $$\mathbb{R}^{\infty}$$. Now since any $$v = (x, 0, 0 \cdots) \in V$$ can be written $$v = x(1, 0, 0, \cdots)$$ where $$x \in \mathbb{R}$$ and $$(1, 0, 0, \cdots) \in V$$, $$(1, 0, 0, \cdots)$$ alone spans $$V$$. Since this list of just one vector in $$V$$ spans $$V$$, I am inclined to think that $$V$$ is finite-dimensional.

This does not mesh with any pre-existing intuition I had about what finite-dimensional vector spaces were, nor can I find any examples like this of a finite-dimensional vector space online, which leads me to think there is an error in my reasoning. Is this correct? If not, can someone point me to the error in my reasoning?

• Yes ,$V$ is finite-dimensional. Why does this bother you ? Jul 11 at 17:15
• Just to make sure: A finite-dimensional space is spanned by a finite list of vectors. But of course a finite-dimensional space can have infinitely many elements. Jul 11 at 17:17
• $\{(x,y,z) : x+y+z=0\}$ is two dimensional, but it consists of 3-tuples of numbers. Does that bother you? Jul 11 at 17:37
• Am I right that this bothers you because it's hard to imagine this subspace geometrically as living inside of an infinite dimensional space? This failure of the "arrows with magnitude and direction" picture is exactly why we find the abstract notion of vector space useful. It keeps the features of "arrows with magnitude and direction" that are useful without being limited by our geometric imagination in terms of what spaces we're able to consider. Jul 11 at 21:08
• Here's another perspective: a line is one-dimensional. It is still one-dimensional if it's in a plane. Or in 3D space. And this extends to any dimensional space. Jul 12 at 5:01

Vector spaces can be thought of abstractly as containing objects that can be added together and also can be multiplied by scalars and still stay within $$V$$ -- it doesn't matter what the vectors actually are, at least from a theoretical viewpoint. In applications, we care very much about what they are because we are not trying to prove theorems but usually get a concrete answer.

There are many examples of vectors that look nothing like lists of numbers at all. The one that helped seal this abstract viewpoint for me was function spaces (spaces where each element is a function!):

Here's how this works using a simple example of the vector space of polynomials with degree $$\leq 2$$ over $$\mathbb{R}$$ with real coefficients: $$V:=\mathcal{P}_2(\mathbb{R})[\mathbb{R}]$$ (i.e, each element $$v \in V$$ is a function $$f:\mathbb{R} \to \mathbb{R}$$ of the form $$ax^2 + bx + c \textrm{ with } a,b,c\in\mathbb{R}$$).

How big is each element?

Each element is a continuous function $$f$$ over the real numbers, so technically it represents an uncountably infinite number of pairs $$f = \{(x,f(x))\quad x \in \mathbb{R} \}$$

What is the dimension of $$V$$?

$$\dim V = 3$$ because the vectors $$\{1\}, \{x\}, \{x^2\},$$ span $$V$$

So, as you can see, the nature of the vectors in a vector space has nothing to do with the dimension of the vector space. The dimension is the smallest number of vectors $$v \in V$$ needed to span $$V$$.

Note also that $$V$$ is a finite dimensional subspace of the uncountably infinite dimensional space of continuous functions over the reals: $$C(\mathbb{R})$$.

So you can have a finite-dimensional vector space as a subspace of an infinite dimensional vector space which itself may contain objects that represent infinite lists. The two concepts are separate.

You can see this with simpler objects too:

Let $$\mathbf{b} = [1,1,1,1,1,1]$$ be a basis for $$V(\mathbb{R})$$, then $$\dim V = 1$$ even though $$\bf b \in \mathbb{R^6}$$

What is this subspace of $$\mathbb{R^6}$$? Its the line $$L = a\mathbf{b}\; a \in \mathbb{R}$$ which passes through the origin and is parallel to $$\bf b$$.

• The example with function spaces really does help; I hadn't really the fact that when the elements are continuous functions, each is technically representing an uncountably infinite number of points. Jul 13 at 16:12

Perhaps you're confused about the number of entries of an array/sequence, versus the dimension of a vector space. For example, consider the following two arrays \begin{align} v = (1, 1, 1),\\ w = (1, -1, 1). \end{align} Together, they span a 2 dimensional vector space $$V_{vw}$$, where every element is of the form $$(a, b, a)$$. Since $$v$$ and $$w$$ both have 3 entries, we can also think of them as vectors in the 3 dimensional vector space $$\mathbb{R}^3$$. Attached plots of $$V_{vw}$$ in $$\mathbb{R}^3$$ below to help you understand it.
As others have already mentioned, yes this vector space is finite dimensional, specifically 1-dimensional, if we're talking about vector spaces over $$\mathbb{R}$$. You already showed it pretty well:
Each $$v\in V$$ has a unique representation as $$x\cdot (1,0,0,...)$$ for one and only one $$x\in\mathbb{R}$$. So $$\{(1,0,0,...)\}$$ is a base of $$V$$.
I think your confusing cardinality and the number of dimensions a little bit. There are lots of finite-dimensional Vector Spaces with infinitely many elements. In fact, most vector spaces you would "commonly" think about, like $$\mathbb{R}^2$$ or $$\mathbb{R}^3$$ also have infinitely many elements.