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The dot product of two vectors is defined as following: $$ \langle \vec v, \vec u \rangle = \left< \begin{pmatrix} v_1 \\ v_2 \\ \dots \\ v_n \end{pmatrix}, \begin{pmatrix} u_1 \\ u_2 \\ \dots \\ u_n \end{pmatrix} \right> = v_1 \cdot u_1 + v_2 \cdot u_2 + \dots + v_n \cdot u_n $$

Still the multiplication of transposition of $\vec v$ and u gives: $$ \vec v^T \cdot \vec u = (v_1, v_2, \dots, v_n) \cdot \begin{pmatrix} u_1 \\ u_2 \\ \dots \\ u_n \end{pmatrix} = v_1 \cdot u_1 + v_2 \cdot u_2 + \dots + v_n \cdot u_n $$

so the result is the same!

It may be just a silly observation but I'm just surprised because I have never seen using it.

Are these two notations the same thing or is there something important in their definitions that don't allow interchanging them?

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    $\begingroup$ It's used a lot. For example, there would be no other way to make sense of the denominators here and here. $\endgroup$ – Omnomnomnom Aug 21 '15 at 14:25
  • $\begingroup$ Thank you, but I should say in my linear algebra course I have never seen the relationship of my question (even if it's quite trivial). $\endgroup$ – Blex Aug 24 '15 at 15:17
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Strictly speaking, a row vector represents a linear form, i.e. a lineap map from $\mathbf R^n\to\mathbf R$. So they're different in essence.

However, to the vector $\vec v$, you can associate the linear map \begin{align*}\varphi_{\vec v}\colon\mathbf R^n&\longrightarrow\mathbf R\\\vec u&\longmapsto \langle\vec v,\vec u\rangle \end{align*} and in this association, the column vector that represents $\vec v$, becomes its transpose. So it is quite normal the results are the same.

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As others have noted, they are identical. If you would like to see the equality in practice, consider the following:

Theorem: If $A$ is an $m\times n$ matrix with real coefficients, then there exists an $n\times m$ matrix $B$ such that $$(Ax)\cdot y = x\cdot(By)$$ for all $x\in\mathbb{R}^n$ and $y\in\mathbb{R}^m$.

Proof: By your equality: $$(Ax)\cdot y = (Ax)^Ty = x^TA^Ty = x\cdot(A^Ty).$$ So, $B=A^T$.

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  • $\begingroup$ thank you for proving this example! $\endgroup$ – Blex Aug 24 '15 at 15:15
  • $\begingroup$ @Blex Happy to help! $\endgroup$ – user171308 Aug 24 '15 at 15:16

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