If $A$ is positive definite relative to a subspace $L \subset \mathbb{R}^n$, i.e. $$\forall x \in L, x \ne 0, \quad \langle x, Ax \rangle > 0 $$ then can we find a positive definite matrix $B$ (on the entire $\mathbb{R}^n$) such that $$\forall x \in L, \quad \langle x, Bx \rangle = \langle x, Ax \rangle? $$
To approach this question, I have verified its correctness in the trivial cases of $L$ and when $L$ is a $1$-dimensional subspace. Indeed, let $L = \langle \bar{x} \rangle$ for $\bar{x} \ne 0$. The positive definiteness of $A$ relative to $L$ reduces to the single condition $$\langle \bar{x}, A \bar{x} \rangle > 0. $$ Consider the diagonal matrix $B$ with the entries on the main diagonal being $b_1,\ldots,b_n > 0$. It is obvious that $B$ is positive definite, now we will choose $b_i$ to suit our need. Since $\bar{x} \ne 0$, there exists $i$ such that $\bar{x}_i \ne 0$. We need
$$\sum\limits_{i=1}^n b_i \bar{x}_i^2 = \langle \bar{x}, B\bar{x} \rangle = \langle \bar{x}, A \bar{x}\rangle = \alpha > 0, $$
which is equivalent to
$$b_i = \frac{\alpha - \sum\limits_{j \ne i} b_j \bar{x}_j^2}{\bar{x}_i^2}. $$
To ensure that this works, we only need to select $b_j > 0$ sufficiently small such that the right hand side is positive, and such matrix $B$ satisfies our requirement.
I couldn't extend this method to the cases of $L$ being multi-dimensional. My question is, is the statement even true in general, and is there a better way to construct the matrix $B$?
Thank you for your attention. Any help would be greatly appreciated.
Yes, you can.
Choose a subspace $M\subset \mathbb{R}^n$ such that $L\oplus M=\mathbb{R}^n$ (this can always be done). Given positive definite quadratic forms $Q_L:L\to \mathbb{R}$ and $Q_M:M\to \mathbb{R}$, one has a positive definite quadratic form $Q:L\oplus M\to \mathbb{R}$ given as $(l,m)\mapsto Q_L(l)+Q_M(m)$. Moreover, $Q|_L=Q_L$.