Are the following remarks correct ?

We've got the vector space $\mathbb{R^n}$ (for the following it doesn't matter whether we identify it with $M_{1 \times n}$ or $M_{n \times 1}$). We define $e_1, e_2, \ldots, e_n \in \mathbb{R}^n$ by $$ e_1 := (1, 0, \ldots, 0), \\ e_2 := (0, 1, 0, \ldots, 0), \\ \vdots \\e_n := (0, \ldots, 0, 1). $$ Up to this point no basis is required. Now, with respect to the basis $\mathcal{C} := \{e_1, e_2, \ldots, e_n\}$, every vector $v \in \mathbb{R}^n$ satisfies $$ v = [v]_{\mathcal{C}}, $$ where $$ [v]_{\mathcal{C}} := (v_1, v_2, \ldots, v_n) \in \mathbb{R}^n \text{ such that } v = v_1 e_1 + v_2 e_2 + \cdots + v_n e_n. $$ Let $\mathcal{D} := \{d_1, d_2, \ldots, d_n\}$ be a basis of $\mathbb{R}^n$ different from $\mathcal{C}$. Then, in general, if $v \neq 0 \in \mathbb{R}^n$, $$ v \neq [v]_{\mathcal{D}}. $$ When we talk about the matrix $A \in M_{n \times n}(\mathbb{R})$ describing the linear operator $T : \mathbb{R}^n \to \mathbb{R}^n$ with respect to the basis $\mathcal{D}$, we talk about the unique matrix $A \in M_{n \times n}(\mathbb{R})$ such that for all $v \in \mathbb{R}^n$, $$ [T(v)]_{\mathcal{D}} = A [v]_{\mathcal{D}}. $$ The case of the vector space $\mathbb{R}^n$ is quite singular, since we can always describe a vector in this vector space as an $n$-tuple of real numbers without ever mentioning a basis.

In general, however, when we identify a vector space $V$ of dimension $n$ over $\mathbb{R}$ to $\mathbb{R}^n$, we must, in all rigour, specify a reference basis for $V$, and we can't reduce the notation $[v]_{\mathcal{D}} \in \mathbb{R}^n$ to simply $v \in \mathbb{R}^n$ for vectors $v \in V$.

Conventionally, however, we do drop the heavy notation $[\cdot]_{\mathcal{D}}$, when the reference basis is understood. In fact even when the reference basis is not mentioned, we drop the notation and it is understood that the reference basis is the standard basis for $V$ (e.g. $\{1, x\}$ for the vector space of polynomials of degree at most 1).

  • $\begingroup$ Precisely your troubles come from not distinguishing from $M_{1\times n}$ and $M_{n\times 1}$. Check the example in [ juanmarqz.wordpress.com/2012/02/16/… ] to see an example of how you must think. $\endgroup$ – janmarqz Feb 9 '14 at 1:43

These kind of questions literally plagued me for a long while. So, I feel your pain when you feel like you need some outside affirmation that your remarks are correct or not.

Before moving on however let's define some notation. We define the set $\mathbb{R}^n$ as the set of n-tuples of real numbers. That is elements of $\mathbb{R}^n$ look like $(a_1,a_2,\dots, a_n)$ where the $a_i$ are in the set $\mathbb{R}$. We define $M_{n\times 1}(\mathbb{R})$ as the set of $n\times 1$ matrices with elements from the set $\mathbb{R}$. Also let's set $M_{m\times n}(F)$ to be the set of $m\times n$ matrices with entries from the underlying set of the field $F$. Also the little operator $^t$ is just matrix transpose.

The first thing that will help you understand the problem you're having is that you must remember that a vector space is so much more than a set. A vector space is a an ordered triple $(V, F, \cdot)$ such that $V$ is an abelian group and $F$ is a field, and $\cdot$ (scalar product) is a map from the underlying sets of $V$ and $F$ to the underlying set of $V$. There are also compatibility axioms regarding the addition operations in $V$ and $F$, and compatibility axioms regarding scalar multiplication with multiplication in $F$. The things that we call 'vectors' are the elements of the underlying set of $V$.

The most detail that one usually delves into when specifying a vector space is naming the underlying set of $V$ and the underlying set of $F$. This is because often the group, field, and scalar operations are self-evident and it's a pain to typeset. For example when say that $\mathbb{R}$ is a vector space over itself, we get that the group operation and the field addition is the same: real addition, and we get that scalar multiplication coincides with the field multiplication: real multiplication.

Now you should note that we can begin to discuss bases and dimensions. These discussions do not depend on the representation of the vectors of your vector space---only the vectors themselves. And as the natural numbers can be built up set-theoretically, we can talk about a vector space being finite dimensional (it has a basis equinumerous with a finite ordinal).

If all this layering was not bad enough, when we deal with finite dimensional vector spaces we often opt to write the vectors of a vector space not in their set theoretic representation but rather as a coordinate vector. This is where my hell began, and I think it is where yours is too.

Let's assume we are dealing with a finite dimensional vector space $V$ over some field $F$. Let's say the dimension is $n$. Now we note that $M_{n\times 1}(F)$ also forms a vector space over $F$ (with matrix addition and pointwise multiplication as scalar multiplication). A coordinate representation of $V$ is a linear isomorphism from $V$ to $M_{n\times 1}(F)$. And the coordinate vector of a vector is the image of the vector under this linear isomorphism. Here the problems start because there are soooooo very many different linear isomorphisms from $V$ to $M_{n\times 1}(F)$. In fact, for each ordered basis of $V$ and each ordered basis of $M_{n\times 1}(F)$ there is an isomorphism which maps the first onto the other in the same order.

Although this might seem like a deterrent, it is extremely nice computationally---especially when we set up some rules-of-thumb. Some important rules of thumb: there is a canonical way to represent $\mathbb{R}^n$ as a coordinate vector. The 'canonical' coordinate representation of $\mathbb{R}^n$ takes the vector $(1,0,0...,0)$ and carries it onto the vector $[1,0,0,...,0]^t$ and we call the latter the coordinate vector of the former. Another rule of thumb: we assume that the ordered basis that we choose for $M_{n\times 1}(F)$ to be the images of the ordered basis of $V$ is in order: $[1,0,\dots,0]^t, [0,1,\dots,0]^t,\dots,[0,\dots,0,1]^t$.

What do I mean by 'it is nice computationally'? As I'm sure you've worked with, setting up a coordinate representation allows us to codify a linear transformation as a matrix in $M_{m\times n}(F)$. That is when we have two vector spaces $V$ and $W$ and a linear transformation $T:V\rightarrow W$ AND we have chosen a coordinate representation of $V$ with $M_{n\times 1}(V)$ and a coordinate representation of $W$ with $M_{m\times 1}(V)$ we can codify $T$ as an element of $M_{m\times n}(F)$. And we can realize the application of $T$ to an element in $V$ as matrix multiplication between an element of $M_{m\times n}(F)$ with the coordinate representation of the vector. This way of explanation also reveals that a matrix is dependent on the two ordered bases chosen for $V$ and $W$.

Another rule of thumb: when $V=W$ we assume that the basis we are using to represent the elements of $V$ as coordinate vectors does not change (but there is no reason a priori to assume it; it's just convention).

Now let's directly tackle your questions. It appears your operator $[\cdot]_\mathcal{C}$ is the linear isomorphism I talked about above which assigns an element of $\mathbb{R}^n$ to its coordinate vector representation. I personally do not codify $n\times 1$ matrices as ordered $n$-tuples. In my theory, you do not have $v=[v]_\mathcal{C}$. You do get things which look similar. To typify what I mean let's take an example. Let's take $(1,0,\dots, 0)\in\mathbb{R}^n$. Then $$[(1,0,\dots,0)]_\mathcal{C}=[1,0,\dots,0]^t$$

And as I said, in my theory, I do not codify $n\times 1$ matrices as $n$-tuples. So this function is not the identity. Everything else in your post, I agree with however.

I hope I have explained enough and that I have saved you from future turmoil.


Perhaps if I give an example you can see why the conventions we use are important. Lets take the vector space $V=\mathbb{R}^2$. There are many bases for $V$. The standard basis is $\{(1,0), (0,1)\}$ (in this order). But let's work with another basis. Let's set $\mathcal{B}=\{(2,0), (1,1)\}$ (in this order). This basis was not chosen for any particular reason rather than its different from the standard basis. Now let's take the vector $v=(3,1)$ in $\mathbb{R}^2$. What is $v$'s coordinate vector? That is, what is $[(3,1)]_\mathcal{B}$? We have that $[\cdot]_\mathcal{B}$ is the linear operator from $\mathbb{R}^2$ to $M_{2\times 1}(\mathbb{R})$ which takes the vector $(2,0)$ to the vector $[1,0]^t$ and the vector $(1,1)$ to the vector $[0,1]^t$. Thus


Thus when we work in the basis $\mathcal{B}$ the coordinate vector of $(3,1)$ is $[1,1]^t$.

This all works because there is a convention that operator $[\cdot]_\mathcal{B}$ carries $\mathcal{B}$ to $\{[1,0]^t,[0,1]^t\}$. But what if $[\cdot]_\mathcal{B}$ carried $\mathcal{B}$ to a different basis? Lets take $\{[2,0]^t,[1,1]^t\}$ for an example. Then we get something which does not follow convention:


You see that we don't get something very different. This is the reason you might think that $v=[v]_\mathcal{C}$ in your example (especially if you implement $n\times 1$ matices as $n$-tuples) because the conventions force us to take a basis for $\mathbb{R}^n$ onto the standard basis for $M_{n\times 1}(\mathbb{R})$. And when we have chosen the standard basis for $\mathbb{R}^n$ the coordinate vector representations of vectors look almost exactly alike.

  • $\begingroup$ Thank you very much for your post (I appreciate your efforts). If I understand properly, since you haven't defined a vector space $\mathbb{R}^n$, $\mathbb{R}^n$ is just a set for you. In coincides with the $l^n$ that Lion talked about in previous comments (see his answer). For Lion it seems like $\mathbb{R}^n$ is a vector space (Euclidean space), however, and as far as I can remember I've always seen $\mathbb{R}^n$ being talked about as a vector space in Linear Algebra. In the end, you say that when we talk about the vector space $\mathbb{R}^n$, we really talk about $M_{n \times 1}$. $\endgroup$ – Guest Feb 8 '14 at 16:43
  • $\begingroup$ $\mathbb{R}^n$ can be made into a vector space. And when we talk about it we usually assume it has been. $M_{n\times 1}$ is a different set that can also be made into a vector space. But we usually use $M_{n\times 1}$ to represent the elements of a vector space once we have chosen a basis for that vector space. $\endgroup$ – Robert Wolfe Feb 8 '14 at 21:02

I am confuse why you mentioned $v\neq[v]_{\mathcal{D}}$. Did you want to say the coordinates of $v$ are different between $\mathcal{C}$ and $\mathcal{D}$?

In fact, basis is only a frame to show a vector in $\mathbb{R}^n$. It is true that one vector probably have different coordinates (say, $v_1,v_2,\dots v_n$ you mentioned in $[v]_{\mathcal{C}}$). But the coordinate is only a description of the vector. I mean, the vector's length, angle etc. is independent of coordinate, say basis. When we mention a operator from $\mathbb{R}^n$ to $\mathbb{R}^n$, we care its boundedness, continuity, duality .etc (See the analysis in matrix analysis or functional analysis). Obviously, as you care, the form of both operator and vector is depend on the basis. But the properties of operator which we care about is independent of the basis. What is more, we can also change the form in different basis by using transition matrix (See references about algebra).

For example, consider the following operator from $\mathbb{R}^2$ to $\mathbb{R}^2$: \begin{equation} \left( \begin{array}{cc} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{array} \right) \end{equation} This operator transform vector $a=(1,0)^\top$ to $b=(\cos\theta,\sin\theta)^\top$ in $\mathcal{C}$ which you mentioned. Indeed, in another basis, such as $\{d_1=(0,1)^\top,d_2=(-1,0)^\top\}$, both $a$ and $b$ have different form. But no matter how you choose the basis, the meaning of the operator is to clockwise rotate the vector in $\theta$ degree. Thus the form and properties of the operator is invariable.

In summary, although the form of operator probably depend on the basis you choose, but the properties which we care about is independent of the basis (In geometry, it is so called geometric invariants).

This is what I think, please check it if I have something wrong.

  • $\begingroup$ Let me illustrate what I mean by $v \neq [v]_{\mathcal{D}}$. Let $v$ be a vector in $\mathbb{R}^2$, say $v := (1,2)$. On one hand, since $v = 1 e_1 + 2 e_2$, we have that $[v]_{\mathcal{C}} = (1,2)$ and in this case $v = [v]_{\mathcal{C}}$. However, if we take $\mathcal{D} := \{d_1 := (0,1), d_2 := (-1,0)\}$, then $v = 2 d_1 + (-1) d_2$ and $[v]_{\mathcal{D}} = (2,-1)$. Hence $[v]_{\mathcal{D}} \neq v$ because $v = (1,2)$ and $(2,-1) \neq (1,2)$. I understand that many properties we investigate are basis-independent, e.g. the characteristic polynomial ($\det(P^{-1}AP) = \det(A)$). $\endgroup$ – Guest Feb 5 '14 at 3:03
  • $\begingroup$ The thing is, I distinguish between the vector $v \in \mathbb{R}^2$ and its basis representation, for a given basis. The basis representation is also a vector in $\mathbb{R}^2$, but does not coincide with the vector itself, in general, which is why I wrote that the case of $\mathbb{R}^n$ (in the general case) is, in a way, singular. $\endgroup$ – Guest Feb 5 '14 at 3:08
  • $\begingroup$ If you want to represent a vector in algebraic form, you must have a basis first. In your example, as you mentioned $(2,-1)\neq(1,2)$, $(2,-1)$ is a coordinate in basis $d_i$ but $(1,2)$ is a coordinate in basis $e_i$. These coordinate obviously different since they are in two different frame, say basis. The relationship between two representations is the transition matrix. In your example, the transition matrix is: $T=(0,1;-1,0)$ (I use the scheme in MATLAB). You can check that $T[v]_\mathcal{C}^\top=[v]_\mathcal{D}^\top$. Hope this is helpful for you. $\endgroup$ – Lion Feb 5 '14 at 3:25
  • $\begingroup$ I understand that if a vector is in some vector space different from $\mathbb{R}^n$, then you must have a basis first because otherwise you could not trace back the vector. But, my main concern is : if we are working in $\mathbb{R}^n$, say $\mathbb{R}^2$, then writing $v := (1,2)$ says that the vector is $(1,2) \in \mathbb{R}^2$. This is not a basis representation, it is the vector in its vector space, independent of any basis. But, it just so happens that it is in fact the basis representation when the basis is $\mathcal{C}$. Does this make any sense or am I still wrong ? $\endgroup$ – Guest Feb 5 '14 at 3:32
  • $\begingroup$ I think you are confusing $\mathbb{R}^n$ with $l^n$. If we want to represent a vector in Euclidian space, the two representations $v:=e_1+2e_2$ and $v:=(1,2)$ is equivalent by given the basis $e_i$. The later one is just an abbreviation for the former one. So I think you said 'it is the vector in its vector space, independent of any basis' is wrong. The basis must be declared but we default they are $e_i$ in most time. If you want to investigate an n-tuple only which is independent of Euclidian space, I think that is $l^n$ space (see functional analysis). $\endgroup$ – Lion Feb 5 '14 at 4:01

well, I think the answers may become clearer when u see that the basis vectors $d_{1},d_{2},d_{3}$ when expressed in the basis D would be of the form $[1. 0 ..0]^{T},[0.1. 0 ..0]^{T},[0. 0 0 0.. 1]^{T}$. So in the basis D the vector can again be expressed as $v=v_{1}d_{1}+v_{2}d_{2}+v_{3}d_{3}+...v_{n}d_{n}$ and each vector $d_{i}$ will have the form that you had mentioned earlier for $e_{i}$. Of course the value of individual components $v_{1}, v_{2}$ will be different in each basis.The fact that the same vector is represented as the sum of different components $v_{1}, v_{2}.. v_{n}$ multiplied with vectors $d_{i}$ which seemingly have the same representation as $e_{i}$ points to the fact that a basis vector such as $[1. 0 ..0]^{T}$ doesn't independently suggest on its own.It represent the first vector in our set of basis vectors,but says nothing more about the basis vectors themselves.

The elements of a vector space have an existence of their own and their identity is invariant regardless of the basis vector used to represent. This manifests in a variety of ways. The scalar invariants of a matrix can be perceived as the result of the independent existence of operators regardless of the choice of basis


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