Is a 'column vector' actually a vector or a matrix?

Question

I've read this post and posted my own question, but I think I will write a more direct question related to this topic, I understand how row and column vectors can be used to represent vectors as answered in the question, but is it actually a vector, or a way of us trying to give the idea of components by putting them in a matrix? For example, Euclidian vectors have no concept of 'transpose', if the components are equal, its the same vector, yet row and column vectors are transpose of each other?

A question I have is say we have:

$y = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$

Can we have also $y = [x_{1},x_{2},x_{3}]$?

as then we have \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = $[x_{1},x_{2},x_{3}]$

and this gives $y=y^T$ which would be an incorrect result.

Answer 1

A vector is an element of a vector space. A vector space is a set that satisfies a few specific requirements.

The set of $m \times n$ matrices with entries from a field $\mathbb{F}$ is always a vector space over $\mathbb{F}$, so all $m \times n$ matrices are vectors, including $m \times 1$ and $1 \times n$ row and column vectors.

As it so happens, there are also relationships between vector spaces - for example, the transpose of an $m \times n$ matrix is an $n \times m$ matrix, so we can say that $^T: \mathbb{F}^{m \times n} \rightarrow \mathbb{F}^{n \times m}$ is a relation between those two vector spaces. There is also a relation between $m \times n$ matrices and a plain old vector from $\mathbb{F}^{mn}$ where you write the elements of the matrix out into the components of the vector one at a time. If the relationship:

1. Is bijective (i.e. it maps every vector from the domain to exactly one vector from the codomain and vice versa), and

2. Preserves all of the vector space properties on each side (e.g. $(v_1 + v_2)^T = v_1^T + v_2^T$),

then the relation is called an isomorphism, and the two vector spaces are said to be isomorphic, and when two things are isomorphic you can essentially treat them as the same thing - as long as you only care about properties that pass through the isomorphism.

So yes - column vectors and row vectors are vectors, and they are equivalent to normal Euclidean vectors, as long as you're just talking about them as vectors. They also happen to be matrices, so you can do matrix things with them, but you need more sophisticated language if you want to relate those things to each other consistently since "doing matrix things" isn't necessarily covered by the vector space axioms.

As a small side note, you may not have even realised but you've already been bamboozled by isomorphic vector spaces, because that's what happens when you connect things written as $(x, y)$ with arrows drawn on a blackboard. Technically, one of those is just a pair of numbers inside a set of brackets while the other is a geometric construction, but we tend to just take it as granted that they're the "same" vectors because all the operations we want to do with them behave nicely across the divide so there's no point in making a real distinction.

Answer 2

Euclidian vectors have no concept of 'transpose', if the components are equal, its the same vector,

That's right. A column vector and a row vector with identical corresponding entries $$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}\quad\text{and}\quad[x_{1} \;\;x_{2} \;\;x_{3}]$$ are the exact same vector in $\mathbb R^3,$ just formatted differently (column form versus row form).

yet row and column vectors are transpose of each other?

Don't forget, we are never capriciously alternating between row and column vectors: in any given problem/modelling/text, we don't concurrently deal with both formats, so there is no context for asking whether, as Euclidean vectors, $\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$ transposes $[x_{1} \;\;x_{2} \;\;x_{3}].$ Reiterating our above agreement: the concept of transpose is inapplicable in Euclidean space, and in the context of $\mathbb R^3,\;\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$ and $[x_{1} \;\;x_{2} \;\;x_{3}]$ are the same object, just at different parties.

A matrix is a data structure, and a $3\times1$ matrix can be framed as a (column) vector in $\mathbb R^3$ while a $1\times3$ matrix can be framed as a (row) vector in $\mathbb R^3.$ These two matrices are transposes of each other: $$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}\quad\text{and}\quad[x_{1} \;\;x_{2} \;\;x_{3}].$$

I understand how row and column vectors can be used to represent vectors, but is it actually a vector, or a way of us trying to give the idea of components by putting them in a matrix?

Get rid of the notion that a directed line segment (i.e., arrow) is the real vector and that its corresponding 3x1 matrix (i.e., column vector, i.e., slim vertical data structure containing its $x,y,z$ components) is just its representation: an Euclidean vector is an abstract object and both are valid manifestations of it; in $\mathbb R^7,$ which representation is more practical?

$ y=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = [x_{1},x_{2},x_{3}]\;?$

this gives $y=y^T$

You need to clarify your notation. For example, from page 39 of David Lay's Linear Algebra:

Following this convention, $$ y=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = (x_{1},x_{2},x_{3}) \ne [x_{1} \;\;x_{2} \;\;x_{3}]=y^T.$$

Answer 3

We could think of $\mathbb R^n$ as row vectors which is consistent with the tuple notation we use in high school or multi-variable calculus. However, our function notation places the symbol for the function to the left of its arguments and, since linear functions on finite dimensional vector spaces are represented by matrices, we need a vector $x$ to be a column vector for an expression like $Ax$ to make sense.

Therefore, in linear algebra, we use column vectors for points in our space and row vectors as points in the dual space.