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I'd like to preface this by saying I'm just starting to learn and understand linear algebra so apologies if there are any mistakes in the post (or if the question doesn't make sense).

I read somewhere that the dot product of two vectors is the same as the number inside the vector of the product of the transpose of one with the other, i.e., $\vec a \cdot \vec b = \vec a ^ T \vec b$

First of all is this true? If it is, why?

Like, I understand the calculations behind it, but is there a geometric reasoning behind this?

proof-of-correctness
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  • What do you mean by $\vec a ^ T\times \vec b$? – joriki Jan 16 '23 at 20:13
  • Like, have you checked whether it is true, like, for , like, some simple examples? – uniquesolution Jan 16 '23 at 20:13
  • @joriki the cross product of the transpose of a with b (the T stands for the transpose, should've made that clearer in the post, my bad!) – proof-of-correctness Jan 16 '23 at 20:15
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    Related. I don't think you mean the cross product of a row vector & a column vector; I think you mean the $1\times1$ matrix $a^Tb$, which involves no anticommutative cross product. – J.G. Jan 16 '23 at 20:16
  • @uniquesolution What I meant there is that I'm not sure if the way I wrote it is correct, but like it is true for the way I'm thinking about it. Which is: when I take the product of $\vec a$'s transpose with $\vec b$, computationally its doing sort of the same thing as the dot product and the answer comes out as a vector with one value inside it, which is the dot product. Why does this happen? Now, I'm not sure how to write that question in a proper mathematical way. – proof-of-correctness Jan 16 '23 at 20:19
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    It will depend on what you take as the definition of dot product. It's not uncommon to define it as a sum of products, which is essentially the definition of matrix multiplication as well. So there's nothing geometric on that account. But if you define dot product geometrically (i.e., as $\vec{a}\cdot\vec{b}=ab\cos\theta$) then the question is why this definition coincides with the sum-of-products definition. – Semiclassical Jan 16 '23 at 20:37
  • How do you define the dot product? If it's $\lvert\vec a\rvert\lvert\vec b\rvert\cos\theta$, then it's not clear geometrically why this is linear, that is, why $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$. Is this close to your question? – Akiva Weinberger Jan 16 '23 at 21:43
  • That said, it's not too hard to show that $\lvert\vec b\rvert\cos\theta$ equals the length of the projection of $\vec b$ onto $\vec a$, from which linearity can be seen more clearly. (And once we have linearity, the formula in terms of multiplying coordinates can be shown quickly.) – Akiva Weinberger Jan 16 '23 at 21:50

1 Answers1

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Addendum: OP originally stated the post as looking at the equality

$$|\vec a \cdot \vec b| = |\vec a^T \vec b|$$

After my post, they sought a geometric reasoning in the comments for the correct claim, which is found here on MSE.

The equality claimed is not true in general. Both cross products and dot products can be framed in terms of an angle $\theta$ between them, and, in the case of the cross product, a unit normal vector $\mathbf{\hat{n}}$ for the plane spanned by the two (with direction determined via the right-hand rule):

$$\newcommand{\n}[1]{\left\| #1 \right\|} \newcommand{\abs}[1]{\left| #1 \right|} \newcommand{\vect}[1]{\mathbf{\vec{#1}}} \begin{align*} \vect x \cdot \vect y &= \n{x} \cdot \n{y} \cdot \cos \theta \\ \vect x \times \vect y &= \Big( \n{x} \cdot \n{y} \cdot \sin \theta \Big) \mathbf{\hat{n}} \end{align*}$$

Clearly, these are not equal in magnitude unless $\sin \theta = \cos \theta$.

My best guess is that what is meant to be referred to is not $\vect x\,^{\mathsf{T}} \times \vect y$ (a cross product) but instead $\vect x\,^{\mathsf{T}} \vect y$, the usual product of matrices in general. We have the general equality, for $\vect x,\vect y$ vectors of the same dimension,

$$\vect x\,^{\mathsf{T}} \vect y = \vect x \cdot \vect y$$

which some take as the definition of the dot product.

PrincessEev
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