The action of Lorentz transformations on 4-vectors in special relativity

So I am studying special relativity and have been introduced to basic tensor calculus used in the theory. Recently, I came across a statement that is confusing me: $$\Lambda^\mu_{\,\,\nu} x^\nu = x^\nu \Lambda^\mu_{\,\,\nu}$$ where $$\Lambda^{u} _{v}$$ is the Lorentz transformation matrix and $$x^u$$ is a 4-vector. Now what I don't understand is why this is the case? More specifically why is it possible to swap the order of the 4-vector and Lorentz matrix, I thought that matrix multiplication was not commutative and so this should be wrong.

• Are you sure about the placement of the indices? Specifically, the $u$ index should appear in the bottom for the Lorentz transformation, not at the top. Otherwise, that's not a valid expression. – enumaris Oct 24 '18 at 20:05
• @enumaris oh sorry yeah I edited the question – daljit97 Oct 24 '18 at 20:09
• ... and also are you sure there is no difference between $\Lambda^\mu_{\,\,\nu}$, $\Lambda^{\,\,\mu}_\nu$ and $\Lambda^\mu_\nu$? – ZeroTheHero Oct 24 '18 at 20:13
• @ZeroTheHero I improved the "notation". I wasn't aware of the appropriate Mathjax syntax for the notation – daljit97 Oct 24 '18 at 20:17

You are not swapping the order of the 4-vector and the Lorentz matrix, this notation is contracted. What this equation is saying is that: $$\sum_u\Lambda^{v} _{u} x^u =\sum_u x^u \Lambda^{v} _{u}$$
• Ok so $$\Lambda^\mu_{\,\,\nu} x^\nu = \Lambda^\mu_{\,\,\nu} x^0 + \Lambda^\mu_{\,\,\nu} x^1 +\Lambda^\mu_{\,\,\nu} x^2 +\Lambda^\mu_{\,\,\nu} x^3$$ and NOT $$\Lambda^\mu_{\,\,\nu} \begin{pmatrix}x^0 \\ x^1 \\ x^2 \\ x^3 \end{pmatrix}$$? Am I right? – daljit97 Oct 25 '18 at 12:31
• Almost, the right answer is:$$\Lambda^\mu_{\,\,\nu} x^\nu = \Lambda^\mu_{\,\,0} x^0 + \Lambda^\mu_{\,\,1} x^1 +\Lambda^\mu_{\,\,2} x^2 +\Lambda^\mu_{\,\,3} x^3$$ – Hugo V Oct 25 '18 at 12:38