# Math Help - more matrix algebra

1. ## 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...)

2. 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...)

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

3. 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

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

4. 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
also is the eigenspace = x_n = {t(v_n)}?

5. 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
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....

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 $C^{-1} l_i = \frac1{d_i} l_i$...