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Hei guys,

I would like to check with you a prove I have

Let the matrix $X_1 = \begin{bmatrix} A_1 & B_1\\ B_1 ^ T & A_1\\ \end{bmatrix}$ be positive definite. Based on this I would like to show that the matrix $X_2 = \begin{bmatrix} X_1 & B_2\\ B_2 ^ T & A_1\\ \end{bmatrix}$, where $B_2 = \begin{bmatrix} 0\\ B_1\\ \end{bmatrix}$ is also positive definite.

The prove I have looks like this:

  1. Based on the assumptions on $X_1$ and the Schur Lemma we get that $A_1, A_1 - B_1 ^ T \times A_1 ^ {-1} \times B_1$ and $A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T$ are positive definite.
  2. In order for $X_2$ to be positive definite one has to show (based on the same lemma) that $X_1 - B_2 \times A_1 ^ {-1} \times B_2 ^ T$ is positive definite. This is equivalent with showing that the matrix:

$\begin{bmatrix} A_1 & B_1\\ B_1 ^ T & A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T\\ \end{bmatrix}$ is positive definite.

For me this is true on if the matrix $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ is positive definite.

I would like to ask you if is possible $X_2$ to be positive definite with the $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ being positive definite?

Thanks, Bogdan.

Bogdan
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2 Answers2

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The result seems to follow from the Sylvester criterion for positive definiteness and determinant identity for Schur complement. You need $A_1-B_1 A^{-1} B^T_1$ to be positive definite (which is positive definite, because $X_1$ is, as you say), but $-B_1 A^{-1}_1B^T_1$ is not positive definite if $X_1$ is positive definite. In other words, $-B_1 A^{-1}_1B^T_1$ is never positive definite if $A_1$ is.

DVD
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  • So $X_2$ is positive definite by construction ? – Bogdan May 13 '13 at 09:24
  • You were doing fine, I just added details, except the matrix in your last question is negative definite. – DVD May 13 '13 at 09:37
  • Just to make sure I understood correctly:

    $X_2$ is positive definite only if $−B_1 \times A_1 ^ {-1}×B_1 ^ T$ is also positive definite? And there is no other way for $X_2$ to be so.

     – Bogdan May 13 '13 at 09:46
  • But $A_1 - B_1 A_1 ^ {-1} B_1 ^T$ is positive definite because $X_1$ is. I dont understand your point – Bogdan May 13 '13 at 11:01
  • Can you give more intuition why this is not possible ... i simply do not see it. – Bogdan May 13 '13 at 12:08
  • $det(-B_1 A^{-1} B_1^T)=-(detB_1)^2/detA_1$. So if $detA_1>0$ then $det(-B_1 A^{-1} B_1^T) \le 0$. – DVD May 13 '13 at 12:22
  • So this means that X_2 is positive definite simply by construction? – Bogdan May 13 '13 at 14:16
  • Well, construction and a proof... – DVD May 13 '13 at 18:03
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I have to say I got lost somewhere on the way, but was thinking about the following alternative proof:

  1. Assuming that $X_1$ is positive definite we would get that $\begin{bmatrix} x & y \end{bmatrix} \times \begin{bmatrix} A & B\\ B^T & A\\ \end{bmatrix} \times \begin{bmatrix} x ^ T\\ y ^ T \end{bmatrix} \geq 0$. This is equivalent with: $x A x^T + 2x B y^T + y A y^T \geq 0, \forall x, y$ (vectors with proper sizes).

  2. For $X_2$ to be positive definite (based on the same definition) would imply that: $u X_1 u^T + 2u B_2 v^T + v A v^T \geq 0, \forall u, v$. Choosing $u = (x, y), v = z$ we get that $x A x^T + 2x B y^T + y A y^T + 2y B z^T + z A z^T \geq 0$.

I fail to see why the second inequality is true $\forall x, y, z$ assuming that the first one is true $\forall x, y$.

This makes me to think that $X_2$ in general is not positive definite, but rather for some specific B with some specific properties.

Bogdan
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