Kronecker product identity when multiplied by two vectors?

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I am interested in least-squares optimization for problems with space-time separable prior state covariances and am trying to break down the quadratic cost function into respective space-time contributions. From Wikipedia, we have that the mixed Kronecker matrix-vector product can be written as:

\begin{align} \left( \mathbf{A} \otimes \mathbf{B} \right) \mathbf{v} = \operatorname{vec} (\mathbf{B} \mathbf{V} \mathbf{A}^T) \end{align} where $\mathbf{V} = \operatorname{vec}^{-1}(\mathbf{v})$ is the inverse of the vec() operator (formed by reshaping the vector $\mathbf{v}$).

Is there a similar simplification for the product (yielding a scalar)

\begin{align} \mathbf{v}^\top \left( \mathbf{A} \otimes \mathbf{B} \right) \mathbf{v} \end{align}

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Yes. In general, $$ \operatorname{vec}(A)^\top\operatorname{vec}(B) = \operatorname{tr}(A^\top B). $$ It follows that $$ v^\top (A \otimes B)v = \operatorname{vec}(V)^\top\operatorname{vec}(BVA^\top) = \operatorname{tr}(V^\top BVA^\top). $$