Assume that we have matrix \begin{equation} \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 \end{equation}, 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{equation} \begin{bmatrix} X_1 & \cdot & \cdot \\ \cdot & X_6 & \cdot \\ \cdot & \cdot & X_4\\ \end{bmatrix} \succ 0 \end{equation}

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.


1 Answer 1


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 $X_1,X_2,X_3$).

As a consequence, this new matrix is still positive definite.

  • $\begingroup$ Hi Anne, Thank you for your quick answer and it is easy to understand for me. I would like to clarify my question a little bit: the X2, X3, X5 are also matrices and they are not square. In this case, how can I obtain the same results. Thank you for your help. $\endgroup$ Commented Oct 19, 2022 at 10:38
  • $\begingroup$ I have just check the matrix P as above is not positive definite since it has an eigenvalue equal to -1. How can we multiple a positive definite matrix with a undefinite matrix but still maintain that the matrix is positive definite? $\endgroup$ Commented Oct 19, 2022 at 11:05
  • $\begingroup$ 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. $\endgroup$ Commented Oct 19, 2022 at 11:20
  • $\begingroup$ Hi Anne, thank you very much for your answer and point out my misunderstanding. Could you recommend some chapters or lectures directly relating to the condition that the positive definite matrix multiplied by which matrices/conditions that it is still positive definite? $\endgroup$ Commented Oct 19, 2022 at 12:15
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    $\begingroup$ Thank you for your time Anne. I see your points now. Wish you a nice day! $\endgroup$ Commented Oct 19, 2022 at 14:43

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