Equivalent Positive definite matrix transformation

Assume that we have matrix $$$$\begin{bmatrix} X_1 & X_2 & X_3 \\ X_2^\top & X_4 & X_5 \\ X_3^\top &X_5^\top & X_6 \end{bmatrix} \succ 0$$$$, where $$X_i$$ are matrix blocks and $$X_2, X_3, X_5$$ are not square matrices. I would like to form an equivalent transformation from this inequality to obtain a positive definite matrix from this matrix, but swapping the position of $$X_4$$ and $$X_6$$, which means it has this form

$$$$\begin{bmatrix} X_1 & \cdot & \cdot \\ \cdot & X_6 & \cdot \\ \cdot & \cdot & X_4\\ \end{bmatrix} \succ 0$$$$

How can I fill the "$$\cdot$$" in the matrix above with $$X_2, X_3, X_5$$? I guess I should use Schur compliments to obtain the result but still do not know how to to. Thank you for helping me.

When all blocks are 1-dimensional, use the matrix $$P:=\begin{pmatrix} 1&0&0\\0&0&1\\0&1&0 \end{pmatrix}=P^\top=P^{-1}:$$
$$P^\top\begin{pmatrix} a&b&c\\ d&e&f\\ g&h&k \end{pmatrix}P=\begin{pmatrix} a&c&b\\ g&k&h\\ d&f&e\\ \end{pmatrix}.$$ Similarly, in the general case, your matrix is congruent to $$\begin{pmatrix} X_1 & X_3& X_2\\ X_3^\top & X_6& X_5^\top\\ X_2^\top &X_5& X_4 \end{pmatrix}$$ via the permutation $$(u_1,\dots,u_p,v_1,\dots,v_q,w_1,\dots,w_r)\mapsto(u_1,\dots,u_p,w_1,\dots,w_r,v_1,\dots,v_q)$$ of the basis (where $$p,q,r$$ are the respective widths of $$X_1,X_2,X_3$$).
• You seem to believe that the product of two (symmetric) positive definite matrices is again (symmetric) positive definite, which is false. $P$ only needs to be invertible: see Matrix congruence. Commented Oct 19, 2022 at 11:20