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.

• I think you have forgotten the case $\mathrm{rank}(A)=1.$ $\mathrm{rank}(A)=2$ and nonnegative definite means positive definite, so you can use your theorem. $\mathrm{rank}(A)=0$ means $A=0$, that is the trivial case you mentioned. $\mathrm{rank}(A)=1$ can happen, too. – Reinhard Meier Apr 14 '18 at 17:48

$A=\begin{pmatrix}a&b\\b&c\end{pmatrix}\in S_2(\mathbb{R})$ 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. $LL^T=A$.

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 $b^2/a=c$.