2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Reinhard Meier Apr 14 '18 at 17:48
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $b^2/a=c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.