Question

Matrix algebra: Is C=LDL^(-1) the same as C^(-1)=LD^(-1)L^(-1)?

Answer 1

hmmm let me think abt this. What course r u in?

Answer 2

Doing a basic calculus/algebra course. Not sure how to re-express for C^(-1)

Answer 3

umm to me it makes sense let me check my linear algebra book

Answer 4

this topic is on eigenvalues-eigenvectors... So if you make up some matrix values for L & D, and then find the eigenvalues and eigenspace for C^(-1), try confirming the answer by multiplying both sides Cv=Dv by C^(-1) ----> you'll find that its not proven

Answer 5

whtvr dont listen to me hehe i totally guessed that

Answer 6

yeah, so i was hoping someone could point me in the right direction in re-expressing the above equation in C^(-1)

Answer 7

umm i doubt neone on now wld know this stuff but let me check

Answer 8

pls do, thanks! :)

Answer 9

lol noone is on. Like i am taking linear algebra now and like only a few ppl know this topic. And they r not on now :(

Answer 10

lol. I was sort of hoping this would be quick to tackle! oh well :(

Answer 11

C=LDL^(-1)
C^(-1) = [LDL^(-1)]^(-1)
= [L^(-1)]^(-1)D^(-1)L^(-1)
= LD^(-1)L^(-1)

Answer 12

Thanks, that is the same answer I got. However, now if you plug in some made up values: e.g. \[L=\left[\begin{matrix}1 & 1 \\ 2 & -4\end{matrix}\right]\] and \[D = \left[\begin{matrix}4 & 0 \\ 0 & -2\end{matrix}\right]\]
.. And find the eigenvalues and corresponding eigenspace for C^(-1):

Try confirming the answer by multiplying both sides \[Cv = Dv\] by C^(-1), to obtain an eigenvalue-eigenvector equation for the matrix C^(-1)

Answer 13

So the eigenvalues I found are -0.5 and 0.25:
Hence, the eigenspaces are \[s={\left\{ t\left(\begin{matrix}1/2 \\ 1\end{matrix}\right) \right\}}\] and \[s = \left\{ t\left(\begin{matrix}1/2 \\ 1\end{matrix}\right) \right\}\] respectively.

Now, if you plug in those values into \[v = C^{-1}Dv\],
you will get:\[\left(\begin{matrix}-1/4 \\ 1\end{matrix}\right) = \left(\begin{matrix}-1/16 \\ 1/4\end{matrix}\right)\] and \[\left(\begin{matrix}1/2 \\ 1\end{matrix}\right) = \left(\begin{matrix}1/32 \\ 1/16\end{matrix}\right)\]

So if I were to express ab eigenvalue-eigenvector equation for the matrix C^(-1), am I right to say that it is: \[C^{-1}\lambda v = \alpha v\]??

Answer 14

im really sorry i dont think i can help you in this.. not with my own knowledge
so i think you if you re post the question you may find somebody who can help