Tensor transformation Formula Proof

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 $