Find the $LDL^{T}$ factorization of $A$ when in the range of the positive definite

I am trying to find the $LDL^{T}$ factorization of the following matrix $$A = \begin{bmatrix} 1 & b \\ b & 4 \end{bmatrix}$$ when $b$ is in the range of positive definiteness.

I have already determined that $b$ is in the range of positive definiteness when $b < 2$. However, I don't really understand how to find the $LDL^{T}$ factorization. I know how to do it with a matrix with all numbers, but when I applied the same strategy to this particular matrix I got a big mess when I calculated $LDL^{T}$.

The only thing I can think to do is plug in a value less than 2 and do it the way I normally do it, but I have a feeling any time I change the value for $b$ I will get a different $LDL^{T}$ factorization.

Can anyone give me any advice?

• It isn't actually too hard to write the factorization in terms of $b$, where did you get stuck? – BaronVT Mar 18 '15 at 16:42
• Find eigenvectors corresponding to the eigenvalues. – copper.hat Mar 18 '15 at 16:44
• Or, more simply, apply one step of the Cholesky decomposition algorithm. See en.wikipedia.org/wiki/…. – copper.hat Mar 18 '15 at 16:49
• Well, I will show you my process for say $b = -2$ $$\det\left( \begin{bmatrix} 1 & -2 \\ -2 & 4 \end{bmatrix} \right) = 8 \quad\to\quad D = \begin{bmatrix} 1 & 0 \\ 0 & 8 \end{bmatrix}$$ For $L^{T}$ I put it into upper triangular form and then transpose it for $L$ $$L^{T} = \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix} \quad L = \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix}$$ and thus $$\begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & 8 \end{bmatrix} \cdot \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -2 & 4 \end{bmatrix}$$ – user224534 Mar 18 '15 at 16:50
• Oh jeez, I just looked back at my calculation. I did $$L^{T}DL$$ no wonder I was getting such a crazy answer – user224534 Mar 18 '15 at 16:55

$A = \begin{bmatrix} 1 & 0 \\ b & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 4-b^2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ b & 1 \end{bmatrix}^T$.