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i want to convert the following to a quadratic form in terms of vector $b$: $$trace(A^\top (\sum{b_i^2K_i})^\top T(\sum{b_i^2K_i}A))$$ where $T,K_i$ are symmetric and $n\times n$.

A is $n\times m$ and $b_i$ is scalar.

Can i write it in a quadratic form in terms of the vector $b$? or even $b^2$?

if so, Then what would be the $H$ matrix in the quadratic term?

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    Do you mean $b$ is the variable and you want to find a matrix representation of the quadratic form? – 2017-02-20
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    me too, but how can i formulate it? – 2017-02-20

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$$\mbox{tr} \left( \mathrm A^{\top} \left( \sum_{i=1}^m b_i^2 \mathrm K_i\right) \mathrm T \left( \sum_{j=1}^m b_j^2 \mathrm K_j\right) \mathrm A \right) = \sum_{i=1}^m \sum_{j=1}^m b_i^2 b_j^2 \, \mbox{tr} \left( \mathrm A^{\top} \mathrm K_i \mathrm T \mathrm K_j \mathrm A \right) = \begin{bmatrix} b_1^2\\ b_2^2\\ \vdots\\ b_m^2\end{bmatrix}^{\top} \begin{bmatrix} \mbox{tr} \left( \mathrm A^{\top} \mathrm K_1 \mathrm T \mathrm K_1 \mathrm A \right) & \mbox{tr} \left( \mathrm A^{\top} \mathrm K_1 \mathrm T \mathrm K_2 \mathrm A \right) & \cdots & \mbox{tr} \left( \mathrm A^{\top} \mathrm K_1 \mathrm T \mathrm K_m\mathrm A \right)\\ \mbox{tr} \left( \mathrm A^{\top} \mathrm K_2 \mathrm T \mathrm K_1 \mathrm A \right) & \mbox{tr} \left( \mathrm A^{\top} \mathrm K_2 \mathrm T \mathrm K_2 \mathrm A \right) & \cdots & \mbox{tr} \left( \mathrm A^{\top} \mathrm K_2 \mathrm T \mathrm K_m \mathrm A \right)\\ \vdots & \vdots & \ddots & \vdots\\ \mbox{tr} \left( \mathrm A^{\top} \mathrm K_m \mathrm T \mathrm K_1 \mathrm A \right) & \mbox{tr} \left( \mathrm A^{\top} \mathrm K_m \mathrm T \mathrm K_2 \mathrm A \right) & \cdots & \mbox{tr} \left( \mathrm A^{\top} \mathrm K_m \mathrm T \mathrm K_m \mathrm A \right)\\ \end{bmatrix} \begin{bmatrix} b_1^2\\ b_2^2\\ \vdots\\ b_m^2\end{bmatrix}$$

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    i just found that a moment ago! ;-) but now i'm sure it is correct. Thanks! – 2017-02-20
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    If $\mathrm T$ is positive semidefinite, then $$\mbox{tr} \left( \mathrm A^{\top} \mathrm K_i \mathrm T \mathrm K_j \mathrm A \right) = \langle \mathrm T^{\frac 12} \mathrm K_i \mathrm A, \mathrm T^{\frac 12} \mathrm K_j \mathrm A \rangle$$ – 2017-02-20
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    yes it is psd!! – 2017-02-20