# how can I show that K is positive-definite?

#### evinda

##### Well-known member
MHB Site Helper
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

##### Well-known member
MHB Site Helper
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

##### MHB Oldtimer
Staff member
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.