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Many people say that they are the same, even I can't find much difference in them except that a vector/ matrix can be multiplied by any scalar, but to multiply it with a vector in $\Bbb{R^1}$ the vector or matrix should be of the order $1\times n$. What's and why is there a difference in this case?

Hisoka
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Physics freak
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  • I think you may be confusing the restrictions on dimensions for matrix multiplication with those for addition, for multiplication only the "inner" dimensions need to match. – Stephen Donovan May 24 '21 at 08:14
  • yes, i understand that only the 'inner dimensions' need to match, if any vector in R1 were a scalar it wouldn't have been necessary for the inner dimensions to match, but it is not the case – Physics freak May 24 '21 at 08:18
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    One context where the distinction can be important is when a $1 \times 1$ matrix appears as a factor in a matrix product. For example, if $x$ is a $n \times 1$ column vector, then $x^T x$ is a $1 \times 1$ matrix, so the product $(x^T x) x^T = x^T x x^T$ makes sense. If we did not maintain the distinction that $x^T x$ is a $1 \times 1$ matrix as opposed to a scalar, we might be tempted to slide the "scalar" to the other side of the product (after all, you can do this with actual scalars) to obtain $x^T (x^T x) = x^T x^T x$. But this no longer makes sense: $x^T x^T$ is illegal. –  May 24 '21 at 08:45
  • This is similar to Is $\mathbb{R}^1$ really the same as $\mathbb{R}$ – Mark S. May 24 '21 at 11:43

1 Answers1

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Referring to formal definitions, the main difference is that:

  • When you consider $\mathbb R$ as a field, the multiplication is an internal operation. That is maps an ordered pair $(x,y) \in \mathbb R^2$ to $z= x \cdot y \in \mathbb R$.
  • When you consider $V = \mathbb R$ as a one-dimensional vector space, then the multiplication is formally a scalar multiplication defined between two distinct objects: the field $\mathbb R$ and $V$.
mathcounterexamples.net
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