Prove that if the matrix $\begin{bmatrix}a & b\\b & c\end{bmatrix}$ is nonnegative definite, then it has a factorization $LL^{T}$ Prove that if the matrix $\begin{bmatrix}a & b\\b  & c\end{bmatrix}$ is nonnegative definite, then it has a factorization $LL^{T}$ in which $L$ is lower triangular.
I am using the following theorem:
Theorem: If $A$ is real, symmetric, and pisitive definite matrix, then it has a unique factorization, $A=LL^{T}$, in which $L$ is lower triangular with a positive diagonal.
Clearly $A$ is real, is symmetric since $\begin{bmatrix}a & b\\b  & c\end{bmatrix}^{T}=\begin{bmatrix}a & b\\b  & c\end{bmatrix}$ and is positive definite or is of the form P, di is positive definite is of the form $\begin{bmatrix}0 & 0\\0  & 0\end{bmatrix}$ for Find the precise conditions on $a,b,c$ so that $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$ is nonnegative definite.  but 
P has a $LL^{T}$ factorization trivially.
Is this reasoning right? Thank you very much.
 A: The (semi)definedness of the matrix tells you that $a\ge0$, $c\ge0$ and $ac-b^2\ge0$.
In particular, if $a=0$, then also $b=0$, so the matrix is
$$
\begin{bmatrix} 0 & 0 \\ 0 & c\end{bmatrix}=
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{c}\end{bmatrix}
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{c}\end{bmatrix}^T
$$
Suppose $a>0$. We can go with Gauss-Jordan elimination; let's first assume $d=ac-b^2>0$:
\begin{align}
\begin{bmatrix} a & b \\ b & c \end{bmatrix}
&\to
\begin{bmatrix} 1 & b/a \\ b & c \end{bmatrix}
&& R_1\gets\tfrac{1}{a}R_1
\\[6px]&\to
\begin{bmatrix} 1 & b/a \\ 0 & d/a \end{bmatrix}
&& R_2\gets R_2-bR_1
\\[6px]&\to
\begin{bmatrix} 1 & b/a \\ 0 & d \end{bmatrix}
&& R_2\gets aR_2
\end{align}
This means that
\begin{align}
\begin{bmatrix} a & b \\ b & c \end{bmatrix}
&=\begin{bmatrix} a & 0 \\ b & a^{-1} \end{bmatrix}
  \begin{bmatrix} 1 & b/a \\ 0 & d \end{bmatrix} \\[6px]
&=\begin{bmatrix} \sqrt{a} & 0 \\ b/\sqrt{a} & \sqrt{d}/\sqrt{a} \end{bmatrix}
\begin{bmatrix} \sqrt{a} & 0 \\ 0 & 1/\sqrt{ad} \end{bmatrix}
\begin{bmatrix} 1 & b/a \\ 0 & d \end{bmatrix} \\[6px]
&=\begin{bmatrix} \sqrt{a} & 0 \\ b/\sqrt{a} & \sqrt{d}/\sqrt{a} \end{bmatrix}
  \begin{bmatrix} \sqrt{a} & b/\sqrt{a} \\ 0 & \sqrt{d}/\sqrt{a} \end{bmatrix}
\end{align}
If $d=0$ the above decomposition is good as well:
$$
\begin{bmatrix} \sqrt{a} & 0 \\ b/\sqrt{a} & 0 \end{bmatrix}
\begin{bmatrix} \sqrt{a} & b/\sqrt{a} \\ 0 & 0 \end{bmatrix}
=\begin{bmatrix} a & b \\ b & c\end{bmatrix}
$$
because $b^2/a=c$.
A: $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}\in S_2(\mathbb{R})$ is $\geq 0$ iff $a\geq 0,c\geq 0,ac-b^2\geq 0$. We seek $L=\begin{pmatrix}p&0\\q&r\end{pmatrix}$ s.t. $LL^T=A$, that is equivalent to $p^2=a,pq=b,q^2+r^2=c$.
We may assume that $p\geq 0$ (change $L$ with $-L$); then $p=\sqrt{a}$.
Case 1. $a>0$. Then $q=b/\sqrt{a}$ and $r^2=\dfrac{ac-b^2}{a}\geq 0$. We may choose $r\geq 0$.
Case 2. $a=0$. Then, necessarily $b=0$. Thus $p=0,q^2+r^2=c$. We may choose $q=0,r=\sqrt{c}$.
$\textbf{Conclusion.}$ When $A\geq 0$, there is a non-unique lower triangular matrix $L$with $\geq 0$ diagonal s.t. $LL^T=A$.
