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Row Vectors vs. Column Vectors - What's the difference?

  1. Mar 27, 2012 #1
    That’s an old time question that it’s still a mistery to me. It’s a lot of time that I am trying to find an answer, but no text is clear on the topic and I am basically self-taught.

    What’s the difference between row vectors and column vectors?

    I came to this question when I found that the gradient was defined in two different ways on two different books. This was a problem and I started to look around: the more I was searching, the more it became a mistery, cause lot of books state that the gradient is the row vector of the first partial derivatives of a given function.

    I fixed this problem in the end (the gradient is not the row vector, but the column vector), but still I don’t get what’s the difference between row and columns, beyond a practical one in terms of computation.

    Does exist a "deep" theoretical difference between those two types of vectors or it's a metter of distinction between places (row vectors) and displacements (column vectors)?

    Thanks in advance!
  2. jcsd
  3. Mar 27, 2012 #2


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    There's no real difference. Vectors in your original vector space are typically thought of as column vectors simply so the calculation Ax for a matrix A is a linear transformation from your vector space to your vector space. If you want to do a linear transformation from V to R, (say you want to take an arbitrary vector x and take the dot product with the gradient of a function, which I will call g) then to be able to write this as gx you need g to be a row vector, which is probably why the one book defined the gradient as a row vector
  4. Mar 27, 2012 #3


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    It is fairly common to represent your vectors as columns then you could represent your "co-vectors" (members of the dual space, the space of linear functionals that take each vector to a number) as a row so that the operation of the functional on vector becomes a matrix product:
    [tex]\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= ax+ by+ cz[/tex].

    Of course, that is, as Office Shredder says, purely arbitrary- you could always represent the vectors by rows, the funtionals by columns and do the product the other way around.
  5. Mar 27, 2012 #4


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    But you do need to be clear about which way round you (or a textbook) IS doing it.
    [tex]\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex] is a scalar, but
    [tex]\begin{bmatrix}a \\ b \\ c\end{bmatrix}\begin{bmatrix}x & y & z\end{bmatrix}[/tex] is a 3x3 matrix with rank 1. In some applications (e.g. optimization) both of these are used frequently!
  6. Mar 28, 2012 #5
    Thanks a lot.

    I kinda had the feeling at a certain point that it was arbitrary, but I found a book that gave a sort of hint about a deep reason behind the use of one or the other. Interestingly enough, this deep reason never showed up, leaving me with nothing more than this doubt.
  7. Mar 28, 2012 #6


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    I suspect that the "deep reason", at least the reason for distinguishing between "row" and "column" was, as I said, to be able to differentiate between "vectors" and "co-vectors" and treat their interaction as a matrix multiplication.
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