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they all contain real numbers, matrix $A \in\mathbb{R}^{n\times n} $and vectors $v,w \in\mathbb{R}^{n}$

A solution I saw but do not understand is $\langle Av,w\rangle = (Av)^{t}\cdot w = ... =v^{t}\cdot(A^{t}w) = \langle v,A^{t}w\rangle$

My question is how do I transform from an inner dot product to normal dot prodct, and what causes the transpose when doing so.

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    How are you defining the inner product $\langle{\cdot, \cdot}\rangle$, if not in terms of the dot product? – 2017-01-28

2 Answers 2

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The popular convention treats vectors in $\mathbb{R}^n$ as column vectors. With this convention, the standard inner product of two vectors

$$ v = \begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}, w = \begin{pmatrix} w_1 \\ \vdots \\ w_n \end{pmatrix} $$

can be written as

$$ \left< v, w \right> = v^T \cdot w = (v_1, \dots, v_n)\begin{pmatrix} w_1 \\ \vdots \\ w_n \end{pmatrix} = \sum_{i=1}^n v_i w_i $$

where the expression $v^T \cdot w$ is the product of the $1 \times n$ matrix $v^T$ with the $n \times 1$ matrix $w$ and we identify the result (a $1 \times 1$ matrix) with a number. In particular, we can write things such as

$$ \left< Av, w \right> = (Av)^T \cdot w = (v^T \cdot A^T) \cdot w = v^T \cdot (A^T \cdot w) = \left< v, A^T \cdot w \right>.$$

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    Note that it is important for the transpose to be written on the left in $v^T \cdot w$ because $v \cdot w^T$ is an $n \times n$ matrix and not a $1 \times 1$ matrix. – 2017-01-28
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The Euclidean inner (dot) product is $ = v^T w$

and $(Av)^T = v^TA^T$

And everything flows from there.