Let $t>0$ be a scalar, $P \in \mathbb{R}^{n\times n}$ is a symmetric positive-definite matrix, and $B \in \mathbb{R}^{n\times m}$ for some integers $n$ and $m$. I have the matrix inequality $tP^2-B B^\top \ge 0 $ and I want to make it linear with respect to $P$ and $B$. Using Schur compliment, I have $$tP^2-B B^\top \ge 0 \Leftrightarrow \begin{bmatrix}tP^2 & B \\ B^\top & I_{m\times m}\end{bmatrix} \ge 0.$$ But I do not know how to transform it to avoid the $P^2$ term. Any ideas?
Update: I also have some other LMIs involving $P$ and $B$, thus I cannot redefine $W:=P^2$ and solve an LMI for $W$.
In order to make your initial matrix inequality linear in both $P$ and $B$, one could introduce an intermediate symmetric positive definite variable $X$, such that your initial matrix inequality can also be written as
$$ t\,X \ge t\,P^2 \ge B\,B^\top. \tag{1} $$
This nested inequality is equivalent to the following two inequalities
$$ X \ge P^2 \Leftrightarrow \begin{bmatrix}X & P \\ P & I\end{bmatrix} \ge 0, \tag{2a} $$
$$ t\,X \ge B\,B^\top \Leftrightarrow \begin{bmatrix}t\,X & B \\ B^\top & I\end{bmatrix} \ge 0. \tag{2b} $$