# Tensor transformation Formula Proof

Ok so basically I am trying to prove that the following expression:

Can be written using matrices like this:

Any suggestions on how to approach this?

• It's pretty much defined that way. I'm not clear what you're asking. It's a little bit weird to write the $\cdot$ instead of just putting the symbols next to each other, but I think that's not your question. – Brick Dec 24 '20 at 16:08
• Why does the $\lambda$ matrix become a $Q$ matrix? That change in notation seems pointless and confusing. – G. Smith Dec 24 '20 at 17:40

To prove that you need to know this,

$$a_{ij}b_{jn} = (\textbf{a} \cdot \textbf{b})_{in} = \textbf{a} \cdot \textbf{b}$$

Note that the position of index $$j$$ .

$$a'_{mn} = v_{mi} v_{nj} a_{ij}$$

and you want to show

$$\textbf{a}' = \textbf{v} \cdot \textbf{a} \cdot \textbf{v}^T$$

So,

$$a'_{mn} = v_{mi} v_{nj} a_{ij} =$$

$$a'_{mn} = v_{nj} v_{mi} a_{ij} =$$

$$a'_{mn} = v_{nj} (\textbf{v} \cdot \textbf{a})_{mj} =$$

$$a'_{mn} = (\textbf{v} \cdot \textbf{a})_{mj} v_{nj} =$$

$$a'_{mn} = (\textbf{v} \cdot \textbf{a})_{mj} v^T_{jn} =$$

$$a'_{mn} = (\textbf{v} \cdot \textbf{a} \cdot \textbf{v}^T)_{mn} \implies$$

$$\textbf{a}' = \textbf{v} \cdot \textbf{a} \cdot \textbf{v}^T$$

• oh thank u very much <3 – stefan .gkotsis Dec 24 '20 at 16:18