1. Cholesky Decomposition

Hi, I'm currently taking linear algebra and my professor gave us a homework on Cholesky decompositions. However, when I look at his solution, I don't understand what he's doing when he is doing iterate. I understand the first iteration where he is calculating the L1 column, but I don't understand how he is getting S in the next iterations. Here is his solution:

Thank you so much for the help!

2. Re: Cholesky Decomposition

Looking at this I see that

$L_1 = \dfrac {1}{\sqrt{S_{1,1}}} \text{ 1st column of S} = \begin{pmatrix}2 \\ -3 \\ 4 \end{pmatrix}$

Then they are using $L_1$ to zero out entries in columns 2 and 3 of S and then looking that the 2x2 matrix with row and column 1 removed.

in other words

the second column of $S$ has $3 L_1$ added to it to become $\begin{pmatrix}0 \\ 4 \\ -4 \end{pmatrix}$

the third column of $S$ has $-4 L_1$ added to it to become $\begin{pmatrix}0 \\ -4 \\ 20 \end{pmatrix}$

and now we start over using the lower 2 rows and columns of the resulting matrix above, i.e. $S=\begin{pmatrix}4 &-4 \\ -4 &20 \end{pmatrix}$

Now repeat all the above steps

$L_2 = \dfrac {1}{\sqrt{S_{1,1}}} \text{ 1st column of S} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}$

the second column of $S$ has $2 L_2$ added to it to become $\begin{pmatrix}0 \\ 16 \end{pmatrix}$

and again we take $S$ to be the reduced matrix formed by removing the first row and column. Thus

$S=16$

and $L_3 = \dfrac{1}{\sqrt{S_{1,1,}}} \text{ 1st column of S} = 4$

3. Re: Cholesky Decomposition

Thank you so much! If I may ask, how did you figure that out? Did you already know this method of solving it because I can't find this method online.

4. Re: Cholesky Decomposition

Originally Posted by STEMSTRUGGLES
Thank you so much! If I may ask, how did you figure that out? Did you already know this method of solving it because I can't find this method online.
I just looked at what they had done. I didn't realize about the square root until the last term when they got 4 from 16.