Maximisation of a trace suppose there are matrices $A$ and $B$, and a vector, $v$. When maximising over some matrix, $X$:
$$
\max_{X}\{trace\left(v^T( XAX + XB)v\right)\}= \max_{X}\{trace\left(( XAX + XB)vv^T\right)\}
$$
Say the optimal $X$ is found to be $X^*$. Does it hold the maximising $X^*$ is the same as the maximising $X^*$ for:
$$
\max_{X}\{trace\left(XAX + XB\right)\}
$$ 
I feel like due to the positive semi definiteness of the $vv^T$ it should be but cant seem to find a hard proof.
 A: I will constuct an extreme counterexample. 
Let 
$$
A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad v = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.
$$
For the sake of simplicity, I will consider the domain of $X$ (which is the set of all possible $X$) to be very small, namely the two element set:
$$
D_X=\text{Domain of }X = \{ \begin{bmatrix} 5 & 0 \\ 0 & 0 \end{bmatrix},  \begin{bmatrix} 0 & 0 \\ 0 & 3 \end{bmatrix} \}.
$$
This restriction is here only for illustrative purposes.
As you can see, now we have 
$$
X_1^*= \underset{X \in D_X}{\operatorname{argmax}}\operatorname{trace}\left(( XAX + XB)vv^T\right) = \underset{X \in D_X}{\operatorname{argmax}}\operatorname{trace}\left(X\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right) = \begin{bmatrix} 0 & 0 \\ 0 & 3 \end{bmatrix},
$$
$$
X_2^*= \underset{X \in D_X}{\operatorname{argmax}}\operatorname{trace}\left(( XAX + XB)\right) = \underset{X \in D_X}{\operatorname{argmax}}\operatorname{trace}\left(X\right) = \begin{bmatrix} 5 & 0 \\ 0 & 0 \end{bmatrix}.
$$
This counterexample shows that in general you are not allowed to make the desired simplification.
