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?

  • 1
    $\begingroup$ It's used a lot. For example, there would be no other way to make sense of the denominators here and here. $\endgroup$ Aug 21, 2015 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, 2015 at 15:17

2 Answers 2


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.


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, $B=A^T$.

  • $\begingroup$ thank you for proving this example! $\endgroup$
    – Blex
    Aug 24, 2015 at 15:15
  • $\begingroup$ @Blex Happy to help! $\endgroup$
    – user171308
    Aug 24, 2015 at 15:16

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