# Thread: more matrix algebra

1. ## more matrix algebra

Ok, so, given that, $\displaystyle C^{-1} = LD^{-1}L^{-1}$, and $\displaystyle C$ is a real valued non-singular $\displaystyle n$ x $\displaystyle n$ matrix, $\displaystyle D$ is the diagonal matrix of $\displaystyle C$'s $\displaystyle n$ distinct real non-zero eigenvalues, and $\displaystyle L$ 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 $\displaystyle C^{-1}$ ?

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

Your help is much appreciated

2. Originally Posted by tsal15
Ok, so, given that, $\displaystyle C^{-1} = LD^{-1}L^{-1}$, and $\displaystyle C$ is a real valued non-singular $\displaystyle n$ x $\displaystyle n$ matrix, $\displaystyle D$ is the diagonal matrix of $\displaystyle C$'s $\displaystyle n$ distinct real non-zero eigenvalues, and $\displaystyle L$ 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 $\displaystyle C^{-1}$ ?

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

Your help is much appreciated
Do you realise that $\displaystyle C^{-1} = LD^{-1}L^{-1} \implies C^{-1} L = L D^{-1}$?

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

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

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

So the eigenvalues = $\displaystyle LD^{-1}$?

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

Now read this matrix equation as $\displaystyle C^{-1}$ acting on each column of L.... Voila! You have $\displaystyle C^{-1}$'s eigen vectors
also is the eigenspace = x_n = {t(v_n)}?

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

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

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