MHB How can I show that K is positive-definite?

[evinda]

Hi! :)

I have also an other question about the proof of the Cholesky decomposition.
We write A like that:
$A=\begin{bmatrix}
d & u^{T}\\
u & H
\end{bmatrix}=\begin{bmatrix}
\sqrt d & 0\\
\frac{u}{\sqrt d} & I_{n-1}
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 & K
\end{bmatrix}\begin{bmatrix}
\sqrt d & \frac{u^{T}}{\sqrt{d}}\\
0 & I_{n-1}
\end{bmatrix}$

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

and then we suppose that $K$ is symmetric and positive-definite,to use the assumption step(that is valid for $(n-1)x(n-1)$ symmetric and positive-definite matrices.
But...then I have to prove that $K$ is positive-definite.How can I do this?

 

[evinda]

Hi! :)

I have also an other question about the proof of the Cholesky decomposition.
We write A like that:
$A=\begin{bmatrix}
d & u^{T}\\
u & H
\end{bmatrix}=\begin{bmatrix}
\sqrt d & 0\\
\frac{u}{\sqrt d} & I_{n-1}
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 & K
\end{bmatrix}\begin{bmatrix}
\sqrt d & \frac{u^{T}}{\sqrt{d}}\\
0 & I_{n-1}
\end{bmatrix}$

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

and then we suppose that $K$ is symmetric and positive-definite,to use the assumption step(that is valid for $(n-1)x(n-1)$ symmetric and positive-definite matrices.
But...then I have to prove that $K$ is positive-definite.How can I do this?

Do I have to use the condition $x^{T}Ax>0$ ?

 

[Opalg]

Hi! :)

I have also an other question about the proof of the Cholesky decomposition.
We write A like that:
$A=\begin{bmatrix}
d & u^{T}\\
u & H
\end{bmatrix}=\begin{bmatrix}
\sqrt d & 0\\
\frac{u}{\sqrt d} & I_{n-1}
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 & K
\end{bmatrix}\begin{bmatrix}
\sqrt d & \frac{u^{T}}{\sqrt{d}}\\
0 & I_{n-1}
\end{bmatrix}$

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

and then we suppose that $K$ is symmetric and positive-definite,to use the assumption step(that is valid for $(n-1)x(n-1)$ symmetric and positive-definite matrices.
But...then I have to prove that $K$ is positive-definite.How can I do this?

The matrix $P = \begin{bmatrix} \sqrt d & 0\\ \frac1{\sqrt d}u & I_{n-1} \end{bmatrix}$ is invertible, because its determinant is the product of its diagonal entries, namely $\sqrt d.$ But $A = P \begin{bmatrix} 1 & 0\\ 0 & K \end{bmatrix}P^*,$ and $A$ is positive definite. Therefore $\begin{bmatrix} 1 & 0\\ 0 & K \end{bmatrix} = P^{-1}A(P^{-1})^*$ is positive definite. Hence $K$, which is a corner of that matrix, is also positive definite.