# more matrix algebra

• January 26th 2009, 03:15 AM
tsal15
more matrix algebra
Ok, so, given that, $C^{-1} = LD^{-1}L^{-1}$, and $C$ is a real valued non-singular $n$ x $n$ matrix, $D$ is the diagonal matrix of $C$'s $n$ distinct real non-zero eigenvalues, and $L$ have as columns a linearly independent set of eigenvectors in the order given by eigenvalues down D's diagonal, how do I find eigenvalues and the corresponding eigenspaces for $C^{-1}$ ?

Is it just the normal method $[A - \lambda I_n]$? (I can't see a result out of this method...)

Your help is much appreciated :)
• January 26th 2009, 11:09 PM
Isomorphism
Quote:

Originally Posted by tsal15
Ok, so, given that, $C^{-1} = LD^{-1}L^{-1}$, and $C$ is a real valued non-singular $n$ x $n$ matrix, $D$ is the diagonal matrix of $C$'s $n$ distinct real non-zero eigenvalues, and $L$ have as columns a linearly independent set of eigenvectors in the order given by eigenvalues down D's diagonal, how do I find eigenvalues and the corresponding eigenspaces for $C^{-1}$ ?

Is it just the normal method $[A - \lambda I_n]$? (I can't see a result out of this method...)

Your help is much appreciated :)

Do you realise that $C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}$?

Now read this matrix equation as $C^{-1}$ acting on each column of L.... Voila! You have $C^{-1}$'s eigen vectors :D
• January 27th 2009, 04:58 AM
tsal15
Quote:

Originally Posted by Isomorphism
Do you realise that $C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}$?

Now read this matrix equation as $C^{-1}$ acting on each column of L.... Voila! You have $C^{-1}$'s eigen vectors :D

So the eigenvalues = $LD^{-1}
$
?
• January 27th 2009, 05:04 AM
tsal15
Quote:

Originally Posted by Isomorphism
Do you realise that $C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}$?

Now read this matrix equation as $C^{-1}$ acting on each column of L.... Voila! You have $C^{-1}$'s eigen vectors :D

also is the eigenspace = x_n = {t(v_n)}?
• January 27th 2009, 05:09 AM
tsal15
Quote:

Originally Posted by Isomorphism
Do you realise that $C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}$?

Now read this matrix equation as $C^{-1}$ acting on each column of L.... Voila! You have $C^{-1}$'s eigen vectors :D

I realised what you said, but I still need more explanation on how to do this question. Sorry Isomorphism. But how does one find eigenvalues and eigenspaces with no values....this is the complex part for me....

Thanks for all your help :)
• January 27th 2009, 08:30 AM
Isomorphism
Quote:

Originally Posted by tsal15
I realised what you said, but I still need more explanation on how to do this question. Sorry Isomorphism. But how does one find eigenvalues and eigenspaces with no values....this is the complex part for me....

Thanks for all your help :)

Let $l_1, l_2, ... l_n$ be columns of L, then observe that $C^{-1} L = L D^{-1} = C^{-1} [l_1 \, l_2 \, ... \, l_n] = [l_1 \, l_2 \, ... \, l_n] D^{-1}$ Let $d_1, d_2,.. d_n$ be the eigen values of C, then $C^{-1} [l_1 \, l_2 \, ... \, l_n] = [l_1 \, l_2 \, ... \, l_n] \text{diag}(\frac1{d_1},\frac1{d_2},..,\frac1{d_n} )$. This clearly means $C^{-1} l_i = \frac1{d_i} l_i$...