more matrix algebra

**tsal15** · January 26th 2009, 03:15 AM

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

**Isomorphism** · January 26th 2009, 11:09 PM

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

**tsal15** · January 27th 2009, 04:58 AM

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 = $LD^{-1}<br />$ ?

**tsal15** · January 27th 2009, 05:04 AM

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)}?

**tsal15** · January 27th 2009, 05:09 AM

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....

Thanks for all your help

**Isomorphism** · January 27th 2009, 08:30 AM

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

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