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Given a positive-definite and symmetric matrix $A$,which can be written as: $A=\begin{bmatrix} d & u^{T}\\ u & H \end{bmatrix}=\begin{bmatrix} \sqrt d & 0\\ \frac{u}{\sqrt d} & I_{n-1} \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & K \end{bmatrix}\begin{bmatrix} \sqrt d & \frac{u^{T}}{\sqrt{d}}\\ 0 & I_{n-1} \end{bmatrix}$

where $K=H-\frac{1}{d}uu^{T}$

how can I show that the matrix $K\in \mathbb{R}^{n-1,n-1}$ is also positive-definite?

evinda
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1 Answers1

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Note that the matrix

$$P = \begin{bmatrix}\sqrt{d} & \frac{u^T}{\sqrt{d}}\\ 0 & I_{n-1}\end{bmatrix}$$

is invertible, with inverse

$$P^{-1} = \begin{bmatrix}\frac{1}{\sqrt{d}} & -\frac{u^T}{d}\\ 0 & I_{n-1} \end{bmatrix},$$

thus from

$$A = P^T \begin{bmatrix}1 & 0\\ 0 & K\end{bmatrix} P$$

you obtain

$$\begin{bmatrix}1 & 0\\ 0 & K\end{bmatrix} = (P^T)^{-1} A P^{-1},$$

and the positive definiteness of $K$ follows from that of $A$ considering vectors with first component $0$.

Daniel Fischer
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    Do we know that $P$ is invertible??How do we conclude it? – evinda Jan 11 '14 at 12:04
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    On the one hand, since $P$ is a triangular matrix, its determinant is particularly easy to compute, it's the product of the diagonal elements, which is $\sqrt{d} \neq 0$. So $P$ is invertible. On the other hand, computing the product of $P$ and what I called $P^{-1}$ leads to the identity matrix: another confirmation of $P$'s invertibility. – Daniel Fischer Jan 11 '14 at 12:12