I'm reading Artin's Algebra.

$$\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$$

It says that $a_{ij}$ is the matrix entry such that $i$ is the horizontal coordinate and $j$ is the vertical coordinate. It gives some examples: $a_{11}=2, a_{13}=0,a_{23}=5$. I don't understand why $a_{13}=0$ The first number in the first row is $2$, the first number in the third row does not exist. The only way to make this interpretation valid would be to consider the matrix (it would also work for $a_{23}$):

$$\begin{bmatrix} {2}&{1}\\ {1}&{3}\\ {0}&{5} \end{bmatrix}$$

It kinda makes sense, is it possible that when such a situation occur, we just spin the matrix?

  • 3
    $\begingroup$ I would say that $i$ is the vertical coordinate and $j$ is the horizontal one. When you change $i$ you go up or down in the matrix, hence the change is vertical. $\endgroup$ – Git Gud Nov 28 '14 at 12:14
  • $\begingroup$ Verbatim, The index i is called the row index, and j is the column index. and not what you wrote. Also, given all the examples there I can't really see how could you get such a confusion. $\endgroup$ – Ilya Nov 28 '14 at 12:15
  • $\begingroup$ I think people are missing the point of the question which isn't a misconception of what $a_{13}$ is, but rather an apparent contradiction between the definition of $a_{ij}$ and what the authors want to convey. $\endgroup$ – Git Gud Nov 28 '14 at 12:17
  • $\begingroup$ @Ilya Here. $\endgroup$ – Billy Rubina Nov 28 '14 at 12:23
  • 2
    $\begingroup$ @GitGud: the author defines $a_{ij}$ as usual, please see my comment above where I copied the book. 2Vyska: everything is correct there, and yet I don't see the words horizontal or vertical in the book. $\endgroup$ – Ilya Nov 28 '14 at 12:25

It says that $a_{ij}$ is the matrix entry such that $i$ is the horizontal coordinate and $j$ is the vertical coordinate

It does not.

  • $\begingroup$ Yes, but wouldn't you deduce it from that figure? $\endgroup$ – Billy Rubina Nov 28 '14 at 12:38
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    $\begingroup$ From that figure I would only deduce that $i$ is the row index, and $j$ it the column one. The horizontal coordinate is where on a given row am I situated: that would be $j$. $\endgroup$ – Ilya Nov 28 '14 at 12:41

The definition you gave from your book is ambiguous. Here are some other definitions.

The classic book A Survey of Modern Algebra by Birkhoff and MacLane, Third edition, page 159, says:

whose $i$th row consists of the components $a_{i1}, \ldots, a_{in}$

Some other books just give a standard template for a matrix:

$$ \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ a_{21} & \cdots & a_{2n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \\ \end{bmatrix} $$

Your definition is simply wrong. The first coordinate is the row, the second is the column.


$a_{13}$ is the rightmost column in the first row (above $a_{23} = 5$), which is clearly $0$. In all other texts, $i$ describes the row (vertical) and $j$ the column (horizontal), it seems like your text got that wrong.

  • $\begingroup$ Yes, but I am confronting that information with another information given in the book. $\endgroup$ – Billy Rubina Nov 28 '14 at 12:12
  • $\begingroup$ I edited some stuff above. The interpretation of $a_{ij}$ in all of linear algebra texts is different than what your text says, they most likely got the vertical/horizontal wrong. $\endgroup$ – GDumphart Nov 28 '14 at 12:14
  • $\begingroup$ Refer to this: en.wikipedia.org/wiki/Matrix_(mathematics)#Notation This is the convention in every single meaningful text. $\endgroup$ – GDumphart Nov 28 '14 at 12:16

Think of the entries of a matrix as vectors, every row being a vector.

So, $a_{ij}$ is the $j$-th coordinate of the $i$-th vector.

In this case $a_{13}$ is the 3rd coordinate of the 1st vector.


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