TITLE: Let A be a m by n matrix, then the row space of matrix A is a subspace of the n-space and the column space a subspace of the m-space.
QUESTION [0 upvotes]: My book states the following:

The proof given:

My doubt arises from this part of the proof:

If I'm not mistaken, $\Bbb{R^{n}}$ is a vector space over the field of real numbers, which is $n$-dimensional. That is, the number of basis vectors is $n$. Thus, we need $n $ linearly independent vectors to form a basis for $\Bbb{R^{n}}$. How does this relate to the number of entries in a row vector? I guess, if we need to span vectors of the form $\begin{bmatrix}a_{1}&a_{2}&...&a_{n}\end{bmatrix}^{t}$ we need $n$ linearly independent vectors of the same form and, because we working on the $n$-space, this space includes those types of vectors.
In other words, my question is: why is having a row vector such as $r_{1}$ with $n$ real entries implies it's part of $\Bbb{R^{n}}$?

Note:  □ represents $\Bbb{R}$. For some reason, the character didn't load.

REPLY [0 votes]: $\Bbb R^n$ is a specific vector space over the real numbers, namely: As as set, it consist of all $\pmatrix{x_1\\x_2\\\vdots\\x_n}$ with $x_1,\ldots,x_n\in\Bbb R$. (And addition and scalar multiplication are defined component-wise). As the row vetors are defined to be of this form, it is clear that they are elements of $\Bbb R^n$.
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