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Consider the $n$-tuple $(x_1,\ldots,x_n)$ with entries in some field $K$.

What is the reason for perceiving this tuple either as a row vector, $$ [x_1,\ldots,x_n]$$or as a column vector $$\left[\begin{array}{c}x_1\\\vdots\\x_n\end{array}\right]$$?

To clarify further: All the answers on this site, that I looked up, like this, this or this one deal with the question, which objects should be thought of as row respectively column vectors - opposed to that is what I'm asking: What's reason for making this distinction in the first place ?

Because the thing is, I could define matrix multiplication, like this $$ \left[\begin{array}{cc} a & b \\ c & d \end{array}\right](x_1,x_2):=(ax_1+bx_2,cx_1+dx_2), $$so I only ever need to deal with "tuple-vectors", not row or column vectors; no need to ever talk about row or column vectors. So the thing is, that everything related to coordinates could be done, in the case of vectors, in row or column terms. So if that is possible, why is nobody doing it like that ?

The only benefit that I see from making the row-column distinction - in contrast to using tuples $(x_1,\ldots,x_n)$ which are neither "row" nor "column"-type - is to be gained on a notational level, so that it is easier to remember, for example, how to do matrix-vector multiplication, by "moving" along the rows of the matrix while "moving" along the column of vector.
But that seems a shallow reason, to put up with distinguishing between "row" and "column"-vectors all the time. I hope there's something deeper than that.

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  • $\begingroup$ The first rule of vector spaces is: avoid picking a basis. As a side-effect you won't have row or column vectors :) $\endgroup$ Jan 16, 2014 at 9:57
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    $\begingroup$ @HagenvonEitzen Very true, but I can't avoid this forever - for example numerical linear algebra deals heavily with matrices (instead of linear maps). Another example would be in analysis, where one implicitly uses the standard bases, to write the jacobian matrix. So there are places, where it's traditional to pick a bases and thus I have to deal with the question why it even makes sense to think of a vector of "row" or "column" type. :( $\endgroup$
    – resu
    Jan 16, 2014 at 10:01
  • $\begingroup$ @HagenvonEitzen: Do you mean one should do linear algebra without every using bases, coordinates, or matrices? Indeed, then you don't have to deal with the silly model space $K^n$ at all, nor decide how to write its elements. $\endgroup$ Jan 16, 2014 at 10:01

2 Answers 2

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If you just have a tuple of numbers then, as you said, there is no difference between column and row. However, if you want to have nice matrix algebra, you have to distinguish between two.

The reason nice matrix algebra exists is the fact that, any $m \times n$ matrix can be understood as a linear map from $K^n$ to $K^m$. In this formalism, a column vector is a map from $K$ to $K^n$ i.e., a vector; but a row vector is a map from $K^n$ to $K$ i.e., a functional.

If you define the multiplication as you did, you will lose associativity, and that is bad.

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  • $\begingroup$ Ah, now this is an interesting point. Besides endowing the matrix algebra with "must-have" properties do you know of any other reasons to make the row-column distinction ? $\endgroup$
    – resu
    Jan 16, 2014 at 10:28
  • $\begingroup$ If you are working with the ``string of numbers'' -- no. However, usualy people work with some natural objects (vectors or functionals) and they want to pick a nice basis. But column and row vectors behave differently when you change coordinates. $\endgroup$
    – user68061
    Jan 16, 2014 at 10:31
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    $\begingroup$ I don't understand your "but" in the last sentence (by what is the behaviour of column and row vector contrasting the picking of a nice basis ?) $\endgroup$
    – resu
    Jan 16, 2014 at 10:34
  • $\begingroup$ if $f$ is a functional on the space $V$ and $e_1, ... , e_n$ is a basis for $V$ then the row of this functional is $e=(f(e_1), ..., f(e_n))$. Now if $A$ is an operator on $V$ then the new row is $(f(Ae_1), ..., f(Ae_n) = eA$. But if $v$ is a vector then its new coordinates are $A^{-1}v$. (A maps the old basis to the new one). $\endgroup$
    – user68061
    Jan 16, 2014 at 10:47
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First, there is usually not made any formal difference between the two points of view: there is only one notation $K^n$ for the space of $n$-tuples of scalars, not separate ones for those tuples written vertically or horizontally (or maybe in other forms still).

However, once one uses matrices to represent linear maps, one has to make the choice whether to organise them so that the coordinates (at arrival) of the images of the vectors of the source basis used are given by its columns or by its rows. The most common convention is to have the images correspond to the columns of the matrix, but some authors (in my experience mostly those who do not rely heavily on matrix calculations in the first place) use the convention to use the rows of the matrix. Supposing columns are used, the effect of a linear map on the coordinates of the vectors it operates on is given by left-multpication by the matrix of the linear map, and that operation requires the coordinates of the vector acted upon to be written as a column vector in the matrix product. So of $A$ is the matrix, with respect to chosen bases, of a linear map $f:V\to W$, and $v\in V$ has coordinates $(v_1,\ldots,v_n)$ on the chosen basis of $V$, then the coordinates of $f(v)$ in the chosen basis of$~W$ are the components of the product $$ A \cdot \begin{pmatrix}v_1\\v_2\\\vdots\\v_n\end{pmatrix}. $$ If we had written the coordinates as a row vector, then the only way to multiply by a matrix is to do so on the right (and by the transpose of $A$ if it is not square), but that simply does not give the right coordinates of $f(v)$. So writing column vectors is really just an easy way to remember what convention was used to represent linear maps by matrices.

Although at almost every point where a choice is made there are some authors that choose the opposite convention (and this even applies to the question whether the row or column index in a matrix is written first), I think that fortunately everybody agrees about the rule to multiply matrices (taking rows on the left multiplied by columns on the right). The above is based on the assumption that this convention is universal. Changing that convention would be a most certain way to be misunderstood by everyone. Moreover if one would define matrix mutliplication in a way consistent which what you write in the question, then as user68061 observed you loose associativity of multiplication, which means that matrix multiplication no longer corresponds to (linear) function composition, and it in fact become a completely pointless operation.

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  • $\begingroup$ So, what you actually are saying, is that I was right saying that the only benefit from the row-column distinction is "on a notational level, so that it is easier to remember, for example, how to do matrix-vector multiplication". It puzzles me, that this is the only reason... $\endgroup$
    – resu
    Jan 16, 2014 at 10:26
  • $\begingroup$ Hey, writing coordinates for vectors is just a purely notational device anyway (though a mighty useful one). If we don't adopt notations that make computations possible and easy, then most of us would soon be lost. $\endgroup$ Jan 16, 2014 at 10:30
  • $\begingroup$ I don't understand, why would we lose the fact that a matrix corresponds to a linear map ? The alternative multiplication as I defined in my question "works" precisely as the usual one, when multiplying from the right and that is all I need for the correspondence matrices <-> linear map, as far as I can see. $\endgroup$
    – resu
    Jan 16, 2014 at 10:36
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    $\begingroup$ No, the alternative definition does not work as the usual one with the matrix on the right: then the first term would be $ax_1+cx_2$ rather than $ax_1+bx_2$ as you wrote. If one does not mix a row and a column in the definition, then matrix multiplication will not be associative. If you do what you think you did (write matrix multiplication to mimick ordinary row*matrix multiplication) then you just get the right-multiplying convention I described, which is a consistent one; only you would confuse readers by still writing the matrix on the left. $\endgroup$ Jan 16, 2014 at 10:58
  • $\begingroup$ In fact, I now see that your "alternative" definition of operating with a matrix on a typle is just exactly what left-multiplication on a column does, except that columns vectors are written as tuples. So this definition does not really change anything. The confusing point is that you call it defining matrix multiplication, which suggests you would extend the definition similarly if the right operand (the tuple) had more then one "row". It is doing that extension that gives a non-associative product (whence I wrote " in a way consistent which what you write in the question" in my answer). $\endgroup$ Jan 16, 2014 at 11:33

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