I'm a student in an elementary linear algebra course. Without bashing on my professor, I must say that s/he is very poor at answering questions, often not addressing the question itself. Throughout the course, there have been multiple questions that have gone unanswered, but I must have the answer to this one question.

"Why are some vectors written as row vectors and others as column vectors?"

I understand that if I transpose one, it becomes the other. However, I'd like to understand the purpose behind writing a vector in a certain format.

Taking examples from my lectures, I see that when I'm trying to prove linear independence of a group of vectors, the vectors are written as column vectors in a matrix, and the row reduced form is found.

Other times, like trying to find the cross product or just solving a matrix, I see the vectors written as row vectors.

My professor is very vague on the notations and explanations, and it bugs me as a person who needs to know the reason behind every small thing, why this variation occurs in the format. Any input is greatly appreciated.

  • 4
    $\begingroup$ It depends on many things. For example, if one is writing vectors in a short space (like this one), one is tempted to use row vectors since they occupy less space than column vectors, however, estheticaly, column vectors use to give a clearer image of themselves. The true is that a vector is just a tuple of elements of some set (e.g. $\mathbb{R}^n$) so it doesn't matter how you represent them. On the other hand, if you want to see vectors as a matrix, it depends on what you're going to do with them to choose the right representation. $\endgroup$
    – Daniel
    Mar 20, 2015 at 18:33
  • 3
    $\begingroup$ To follow up on @Devilathor: the choice is arbitrary (rows can become columns, matrices will be changed accordingly) -- the crux is to be consistent all throughout. $\endgroup$
    – Clement C.
    Mar 20, 2015 at 18:35
  • 2
    $\begingroup$ Indeed, it's often just aesthetics; given a vector space of column vectors, it's isomorphic to the vector space of row vectors (the isomorphism being the transpose map). The distinction is nontrivial when you let matrices act upon vectors, however: the matrix must act from the left, with a column, and from the right, with rows. $\endgroup$
    – pjs36
    Mar 20, 2015 at 18:50
  • 3
    $\begingroup$ To multiply a matrix by a vector (which happens very frequently), the dimensions need to match up. It usually makes more sense to use a column vector $\mathbf {x}$ for an expression like $\mathbf {Ax}$. If $\mathbf {x}$ happens to be a row vector, you can take the transpose and write $\mathbf {Ax}^t$, but that gets cumbersome. $\endgroup$
    – Théophile
    Mar 20, 2015 at 18:51
  • 1
    $\begingroup$ @Théophile, would changing the Ax(transposed) to xA change the "mathematical meaning" of the expression? I know that matrix multiplication is not commutative many times. However, is Ax in this case, a dot product, a cross product, or something else? $\endgroup$
    – Skipher
    Mar 20, 2015 at 19:08

2 Answers 2


In one sense, you can say that a vector is simply an object with certain properties, and it is neither a row of numbers nor a column of numbers. But in practice, we often want to use a list of $n$ numeric coordinates to describe an $n$-dimensional vector, and we call this list of coordinates a vector. The general convention seems to be that the coordinates are listed in the format known as a column vector, which is (or at least, which acts like) an $n \times 1$ matrix.

This has the nice property that if $v$ is a vector and $M$ is a matrix representing a linear transformation, the product $Mx$, computed by the usual rules of matrix multiplication, is another vector (specifically, a column vector) representing the image of $v$ under that transformation.

But because we write mostly in a horizontal direction and it is not always convenient to list the coordinates of a vector from left to right. If you're careful, you might write

$$ \langle x_1, x_2, \ldots, x_n \rangle^T $$

meaning the transpose of the row vector $\langle x_1, x_2, \ldots, x_n \rangle$; that is, we want the convenience of left-to-right notation but we make it clear that we actually mean a column vector (which is what you get when you transpose a row vector). If we're not being careful, however, we might just write $\langle x_1, x_2, \ldots, x_n \rangle$ as our "vector" and assume everyone will understand what we mean.

Occasionally we actually need the coordinates of a vector in row-vector format, in which case we can represent that by transposing a column vector. For example, if $u$ and $v$ are vectors (that is, column vectors), then the usual inner product of $u$ and $v$ can be written $u^T v$, evaluated as the product of a $1\times n$ matrix with an $n \times 1$ matrix. Note that if $u$ is a (column) vector, then $u^T$ really is a row vector and can (and should) legitimately be written as $\langle u_1, u_2, \ldots, u_n \rangle$.

This all works out quite neatly and conveniently when people are careful and precise in how they write things. At a deeper and more abstract level you can formalize these ideas as shown in another answer. (My answer here is relatively informal, intended merely to give a sense of why people think of the column vector as "the" representation of an abstract vector.)

When people are not careful and precise it may help to say to yourself sometimes that the transpose of a certain vector representation is intended in a certain context even though the person writing that representation neglected to indicate it.

  • $\begingroup$ This explains it all as a notational convenience, taking for granted the rules of matrix multiplication. What is the deeper reason? $\endgroup$
    – algal
    Jul 31, 2017 at 7:51
  • 2
    $\begingroup$ @algal It's true that there is a much deeper theory of vectors. In that theory you never speak of column vectors or row vectors, because that terminology is not necessary in that context. But in the context of conventional matrix multiplication, it is possible to identify an $n\times1$ matrix as a "column vector." In other words, the very idea that a vector can come in a "column" is born from the notation of matrices. Why would you expect there to be a deeper reason? $\endgroup$
    – David K
    Jul 31, 2017 at 11:40
  • 1
    $\begingroup$ @DavidK I always look for a deeper reason, because it is the way I am most comfortable understanding things. (I find arbitrary things without a deeper reason hard to remember.) I guess I am trying to understand the relationship between the "deeper theory of vectors" (I guess you are alluding to mathematical vector spaces?), and the presentation in terms of rows vs columns vs matrixes. $\endgroup$
    – algal
    Aug 1, 2017 at 16:13

For short: Column vectors live in say ${\mathbb{R}^n}$ and row vectors live in the dual of ${\mathbb{R}^n}$ which is denoted by ${\left( {{\mathbb{R}^n}} \right)^ * } \cong Hom({\mathbb{R}^n},\mathbb{R})$. Co-vectors are therefore linear mappings $\alpha :{\mathbb{R}^n} \to \mathbb{R}$. If one uses basis in ${\mathbb{R}^n}$ and basis in ${\left( {{\mathbb{R}^n}} \right)^ * }$, then for $v \in {\mathbb{R}^n}$ and $\alpha \in {\left( {{\mathbb{R}^n}} \right)^ * }$ with representations:

$$\alpha = {\sum\limits_j {{\alpha _j} \cdot \left( {{e^j}} \right)} ^ * }$$ and $$v = \sum\limits_i {{v^i} \cdot {e_i}}$$ we get: $$\alpha (v) = \alpha (\sum\limits_i {{v^i} \cdot {e_i}} ) = \sum\limits_i {{v^i} \cdot \alpha ({e_i}} )$$ $$\sum\limits_i {{v^i}\alpha ({e_i})} = \sum\limits_i {{v^i}{{\sum\limits_j {{\alpha _j} \cdot \left( {{e^j}} \right)} }^ * }({e_i})} = \sum\limits_i {{{\sum\limits_j {{\alpha _j}{v^i} \cdot \left( {{e^j}} \right)} }^ * }({e_i})}$$ $$\sum\limits_i {\sum\limits_j {{\alpha _j}{v^i} \cdot \delta _i^j} } = \sum\limits_k {{\alpha _k}{v^k}} = \left( {{\alpha _1}, \cdots ,{\alpha _n}} \right) \cdot \left( {\begin{array}{*{20}{c}} {{v^1}} \\ \vdots \\ {{v^n}} \end{array}} \right)$$ Here $\alpha$ is a row vector and $v$ a column vector. Note that $${\left( {{e^j}} \right)^ * }({e_i}) = \delta _i^j = \left\{ {\begin{array}{*{20}{c}} 1&{i = j} \\ 0&{i \ne j} \end{array}} \right.$$ is the link between a pair of dual bases. Using Einstein-Index notation (as usual) we have simply: $$\alpha (v) = {\alpha _k}{v^k}$$ This is co- and contra variant notation. Same story for ${T_p}M$ and $T_p^ * M$ that is Tangent- and Co-Tangent space for manifolds taken at a point $p \in M$. But it's another story.

  • 18
    $\begingroup$ First of all, I think this whole post is not very helpful to the OP, since it answers his elementary question with not quite elementary arguments. Also, it still doesn't explain why "Column vectors live in $\mathbb{R}^n$", while "Row vectors live in the dual". Why isn't it the other way around? Because it arbitrary, and that basically sums up the answer to this question. Be consistent in your choice, and it really doesn't matter. $\endgroup$ Mar 20, 2015 at 19:29
  • $\begingroup$ @user2520938: Don't agree. This is useful and common sense notation in linear algebra first course. These are the basics. Let's see other opinions. $\endgroup$
    – Frieder
    Mar 20, 2015 at 19:48
  • 4
    $\begingroup$ To be clear, I think what you have written down is good and well written, but to me this just doesn't seem the right answer for this question. Also, my remark about it really being arbitrary still stands. $\endgroup$ Mar 20, 2015 at 20:00
  • $\begingroup$ @user2520938: Let us say: Sam cant't be lucky about his professor. My teacher in earlier times, said how important the difference between rows and columns is, if one is going to study vector calculus, differential geometry or physics. Here it is essential to know the difference very well. So I wanted to give Sam the chance to see, how important these differences are. $\endgroup$
    – Frieder
    Mar 20, 2015 at 20:13
  • $\begingroup$ @Frieder I probably will be studying higher level math very soon, and I appreciate your help. I didn't understand most of the notations, but I like reading stuff which I can't understand because I like thinking about what they could mean. Just here to say that your answer was not overlooked nor disregarded. $\endgroup$
    – Skipher
    Mar 20, 2015 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.