# Thread: 2-Norm and Cholesky decomposition

1. ## 2-Norm and Cholesky decomposition

Hello everybody,

i try to prove this statement:

if a Matrix M has a cholesky decomposition $\displaystyle M=L*L^t$, then the equation holds:
$\displaystyle (\| M\|_2) = (\| L\|_2)^2$.

Do you know, how i can prove this?
I Try this: $\displaystyle (\| M\|_2)^2 = \lambda_{max} (M^t * M)=\lambda _{max} (M^2)=(\lambda _{max} (M))^2$
therefore $\displaystyle \|M \|_2 = \lambda_max (M)=\lambda_{max} (L*L^t)=\|L^t\|$.

but why is $\displaystyle \|L^t\| = \|L\|$??

is this correct? and why the last equation holds?

Regards.

2. we have also the decomposition M= B*D*B, whereas D is a diagonal matrix. But i don't see the solution

3. The eigenvalues of $\displaystyle LL^t$ and $\displaystyle L^tL$ are same (in general the eigenvalues of $\displaystyle AB$ and $\displaystyle BA$ are the same).

4. Ok, thanks for your help!! How can i prove your hint'?
is this correct?:

AB=A*B*A*A^(-1), that is AB and BA are äquivalent matrices.

but therefore they have the same eigenvalues:

since AB-\lambda* I = A*(BA-\lambda*I)*A^(-1) => det(AB-\lambda*I)=det(BA-\lambda*I)

is this prove correct?

Thanks a lot.

5. The result is clear if for example $\displaystyle A$ is invertible. If neither $\displaystyle A$ nor $\displaystyle B$ are invertible, then put $\displaystyle A_n :=A+\frac 1n I$, where $\displaystyle I$ is the identity matrix. For n sufficiently large, $\displaystyle A_n$ is invertible. Now, use the fact that the determinant is continuous.

6. Originally Posted by Sogan
is this correct?:

AB=A*B*A*A^(-1), that is AB and BA are äquivalent matrices.

but therefore they have the same eigenvalues:

since AB-\lambda* I = A*(BA-\lambda*I)*A^(-1) => det(AB-\lambda*I)=det(BA-\lambda*I)

is this prove correct?

Thanks a lot.
It's correct if $\displaystyle A$ is invertible.

7. Ok Thank you very much.

8. Take $\displaystyle \lambda \in\mathbb C$. We have, since $\displaystyle a_n$ is invertible for $\displaystyle n\geq n_0$:
$\displaystyle \det(AB-\lambda I ) =\lim_{n\to +\infty}\det(A_nB-\lambdaI)=\lim_{n\to+\infty}\det(BA_n-\lambda I)=\det(BA-\lambda I)$.