I am working on one specific problem, which mounts to prove the following equation. Say $a\in\mathbb{R}^n,b\in\mathbb{R}^n$ and $A\in \mathbb{R}^{n\times n}$. In what situation does the following equality hold?
$$ tr(ba^\intercal A)=tr(a^\intercal b A) $$
Much appreciated!
PS: More about the setting (linear mixed effect model) $$ Y=\boldsymbol{X}_1\beta + \boldsymbol{X}_2u + \boldsymbol{e},u\sim \mathcal{N}(0,\sigma_u^2I_q),\boldsymbol{e}\sim\mathcal{N}(0,\sigma_e^2I_n), $$ and based on Mixed Model Equations (MME), we have \begin{aligned} \widehat{\boldsymbol{\beta}}&=(\boldsymbol{x}_{1}^{\intercal}\boldsymbol{V}^{-1}\boldsymbol{x}_{1})^{-1}\boldsymbol{x}_{1}^{\intercal}\boldsymbol{V}^{-1}\boldsymbol{y},\\ \boldsymbol{\widehat{u}}&=(\boldsymbol{x}_{2}^{\intercal}\boldsymbol{x}_{2}+\lambda I_q)^{-1}\boldsymbol{x}_{2}^{\intercal}(\boldsymbol{y}-\boldsymbol{x}_{1}\widehat{\boldsymbol{\beta}}) \end{aligned} with $\boldsymbol{V}=\sigma_e^2 I_n+\sigma_u^2\boldsymbol{X}_2\boldsymbol{X}_2^\intercal,\lambda=\sigma_e^2/\sigma_u^2$. And I am intended to prove that $$ \mathbb{E}(\mathbf{1}^\intercal Y\mid\boldsymbol{X},u)= \mathbb{E}(\mathbf{1}^\intercal \widehat{Y}\mid\boldsymbol{X},u) $$ with $\widehat{Y}=\boldsymbol{X}_1\widehat{\beta} + \boldsymbol{X}_2\widehat{u}$ where $\boldsymbol{X}_2$ is the random intercept (e.g. the first row can be $(1,0,\cdots,0)$ etc.), which mounts to my question at beginning. Any other methods all also appreciated!
You want $tr ((ba^t-b^ta Id)A)=0$. This says that $A$ is orthogonal to $(ab^t-b^taId)$ in the ($n^2$ dimensional) space of matrices. As soon as $n>1$ and $a \neq 0, b\neq 0$ the matrix $(ab^t-b^taId)$ is non-zero (proof: if $b^ta\neq 0$ it has rank $\geq n-1$ and if $b^ta=0$ it has rank 1 unless $a=0$ or $b=0$), so there will be a $n^2-1$ dimensional subspace of matrices $A$ for which this is true.