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$$\begin{bmatrix} Z^{11}& Z^{12} & Z^{13} \\ Z^{21} & Z^{22} &Z^{23} \\ Z^{31} & Z^{32} & Z^{33} \\ \end{bmatrix}\begin{bmatrix} \vec{v} \cdot e_1 \\ \vec{v} \cdot e_2 \\ \vec{v} \cdot e_3 \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$$

The above relation is equivalent too:

$$ Z^{ij} \vec{v} \cdot e_j = v_i$$

And, I can write: $\vec{v} \cdot Z^{ij} e_j = v_i$ and I could use defnition of dual basis to make it $\vec{v} \cdot e^i= v_i$.

What would the linear algebra action equivalent to the simplification that I have done in the indical tensor notation here?

Note : $e_i$ are basis vectors. More context in this answer

Clemens Bartholdy
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  • What are you looking for, beside $e^i=Z^{ij}e_j$? – J.G. May 09 '21 at 21:03
  • @J.G. I had made a typo there. Basically what I want is to represent the simplification I made using tensor notation into matrix notation. – Clemens Bartholdy May 09 '21 at 21:04
  • So something like $\vec{v}\cdot Z^{ij}e_j=v_i=\vec{v}\cdot e^i\implies Z^{ij}e_j=e^i$? – J.G. May 09 '21 at 21:05
  • Yes, those steps in matrix notation @J.G. The first into second step is meaning that we do the action of matrix on a column with entries e_1, e_2,e_3. The dot product each entry of the resulted vector with $v$.. but I am a bit confused on how to show this using matrices – Clemens Bartholdy May 09 '21 at 21:06
  • Actually this post was for me to complete a step in this answer. So maybe that will put more context on what I'm looking for @J.G. – Clemens Bartholdy May 09 '21 at 21:10
  • Thanks for that context. It sounds like the tricky part is proving if $\vec{v}\cdot\vec{a}=\vec{v}\cdot\vec{b}$ for all vectors $\vec{v}$ then $\vec{a}=\vec{b}$. Equivalently, if $\vec{v}\cdot\vec{w}=0$ for all vectors $\vec{v}$ then $\vec{w}=\vec{0}$. I don't think focusing on matrix notation is the best approach to that. – J.G. May 09 '21 at 21:15
  • @J.G. Yes that's more or less the idea but the reason I want to understand how to do it in the matrix notation, so I can get an insight on how to visualize the tensor expression in matrix form (also for teaching purposes because I think, tensors are more relatable if we show the same expressions work in linear algebra) – Clemens Bartholdy May 09 '21 at 21:19
  • I would say I think the most tricky part is figuring out how to show the action pushing in the contravariant metric tensor to multiply the basis in the equation in matrix format – Clemens Bartholdy May 09 '21 at 21:20

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