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

enter image description here

Can be written using matrices like this:

enter image description here

Any suggestions on how to approach this?

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  • $\begingroup$ 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. $\endgroup$ – Brick Dec 24 '20 at 16:08
  • $\begingroup$ Why does the $\lambda$ matrix become a $Q$ matrix? That change in notation seems pointless and confusing. $\endgroup$ – G. Smith Dec 24 '20 at 17:40
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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 $

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

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