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Surely a $1\times 1$ matrix can only 'produce' vectors with 1 entry, and can take as input also only one entry vectors.

So, is there any use for $1\times 1$ matrices?

Since to me they do the same like scalars, only worse.

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  • $\begingroup$ "can take as input also only one entry vectors" Technically correct, but it can multiply any any horizontal $1 \times n$ vector. $\endgroup$ Jan 21, 2017 at 1:37
  • $\begingroup$ @6005 oh, I wasn't aware of that, I didn't even knew such things exist! $\endgroup$
    – SAJW
    Jan 21, 2017 at 1:39
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    $\begingroup$ Often you might use it to start an induction. Or you might think of the diagonalization of a matrix as representing a linear operator as the sum of $1\times 1$ linear operators. $\endgroup$ Jan 21, 2017 at 1:43
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    $\begingroup$ Don't make things less simpler than they really are! If you implement an algorithm to do operations on matrices, you do not have to convert it to scalars for the special case 1x1. Mathematics is full of examples of neutrality such as x+0, x-0, x*1, x/1, x^1, operations that don't do anything to the input they receive, but you need them! $\endgroup$
    – sergiol
    Jan 21, 2017 at 1:53
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    $\begingroup$ Why exclude such things? It would make statements about matrices unnecessarily complicated. $\endgroup$
    – copper.hat
    Jan 21, 2017 at 1:54

4 Answers 4

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What is the use of the number zero? It doesn't do anything when you add it, and it always does the same thing when you multiply it.

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    $\begingroup$ And you can't even divide by it! $\endgroup$ Jan 21, 2017 at 1:40
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    $\begingroup$ @mathematician: But Chuck Norris can! $\endgroup$
    – sergiol
    Jan 21, 2017 at 1:44
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Maybe the point of $1\times 1$ matrices is to not have to change the definition of matrix to exclude the $1\times 1$ case.

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    $\begingroup$ Also scalar multiplication can be thought of as multiplication by a $1 \times 1$ matrix $\endgroup$ Jan 21, 2017 at 1:37
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Maybe you come across a 1 dimension linear map in the wild and you want to make sure it's invertible and you've taken a linear algebra class and you don't want to draw or think about computing an inverse and you remember that the determinant of a 1 by 1 matrix is the scalar.

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The point of a $n\times n$ (or $m\times n$)matrix is usually to generalize to $n$ dimensions what you already knew in $1$-dimension. And the point of the $1\times 1$ matrix is to give you back what you knew in $1$-dimension. So you won't need to prove something twice, once for 1-dimension and once for multiple dimensions (ಠ‿ಠ)

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