# Positive Definite Matrix Problem

Suppose that the Symmetric Matrix

$$B=\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)$$

of order $n+1$ is positive definitive.

(a) Show that the scalar $\alpha$ must be positive and the $n*n$ matrix $A$ must be positive definitive.

(b) What is the Cholesky factorization of $B$ in terms of $\alpha$, $a$, and the Cholesky factorization of $A$?

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For (b) you should do an explicit computation. Then you should find the relations. Try. – Lucien Sep 14 '12 at 12:50

## 4 Answers

(a) Since $B$ is positive definite, for every column vector $v$ of order $n+1$, $0<v^TBv$. In particular for $v:=[1, 0, \dots, 0]^T$, we get $0<v^TBv=a$.

Now to show that $A$ is positive definite we let $w$ be any column vector of order $n$ and show that $0<w^TAw$. Set $v:=[0, v_1, v_2, \dots, v_n]^T$ (so $v$ is a column vector of order $n+1$). Since $B$ is positive definite, $0<v^TBv$, but $v^TBv=w^TAw$.

(b) I'm afraid i'm not conversant with the Cholesky factorization, so you're on your own...

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I also find a solution for my problem:

$$B=\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)$$

is a symmetric matrix of order $n+1$ positive definite.
We will show that $\alpha$ is positive and $A$ is a $n\times n$ positive definite matrix.
$B$ is positive definite, so

$$\exists\;v=\left( \begin{array}{c} v_1 \\ \tilde{0} \end{array} \right)$$

a non-zero $(n+1)\times1$ vector such that $v^TBv$ is positive ($\tilde{0}$ denote $n$ zero elements of $v$).
Consequently, we have

$$\left( \begin{array}{cc} v_1 & \tilde{0} \end{array} \right)\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)\left( \begin{array}{c} v_1 \\ \tilde{0} \end{array} \right)=\left( \begin{array}{cc} v_1 \alpha & v_1 a^T \end{array} \right)\left( \begin{array}{c} v_1 \\ \tilde{0} \end{array} \right)=v_1 \alpha v_1=v_1^2\alpha >0$$

Because $B$ is positive definite the $v^T B v$ should be positive which in this case forces $\alpha$ to be positive.
Again, considering that

$$\exists\;v=\left( \begin{array}{c} 0 \\ \tilde{v} \end{array} \right)$$

a non-zero $(n+1)\times1$ vector such that $v^T B v$ is positive ($\tilde{v}$ denote a non-zero $n\times1$ vector).
We can write our formula again:

$\left( \begin{array}{cc} 0 & \tilde{v}^T \end{array} \right)\left( \begin{array}{cc} \alpha & a^T \\ a & A \end{array} \right)\left( \begin{array}{c} 0 \\ \tilde{v} \end{array} \right)=\left( \begin{array}{cc} \tilde{v}^Ta & \tilde{v}^TA \end{array} \right)\left( \begin{array}{c} 0 \\ \tilde{v} \end{array} \right)=\tilde{v}^TA \tilde{v}>0$

again, because $B$ is positive definite, $v^TB v$ should be positive and it will force $A$ to be positive definite.

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Given the characterization of a positive definite matrix that says that Its leading principal minors are all positive, if we indicate with $M_n$ the leading principal minor of order $n$ of a generic matrix $M$, we have

\begin{align} &B_1=\alpha>0\\ &B_k=A_{k-1}>0\qquad\forall k=2,\ldots,n+1 \end{align}

so that also all principal minors of $A$ are positive, and this means that $A$ is a positive definite matrix.

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$$\begin{pmatrix} \sqrt\alpha & 0 \\ \frac{1}{\sqrt\alpha}a & L \end{pmatrix} \begin{pmatrix} \sqrt\alpha & \frac{1}{\sqrt\alpha}a^T \\ 0 & L^T \end{pmatrix} = \begin{pmatrix} \alpha & a^T \\ a & LL^T + \frac 1{\alpha}aa^T \end{pmatrix}$$ $L$ is obtained from rank-one downdate from the cholesky factor of $A$.

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