Thread: the algebra of Diagonalized matrices

1. the algebra of Diagonalized matrices

if $C = LDL^{-1}$ , where $D$ is the Diagonal Matrix, how do I rearrange the expression to have $C^{-1}$ in terms of $L, L^{-1}$ and $D$. (also, what specific diagonal matrix can I give? is the $I_n$ ok?)

Many Thanks

2. Originally Posted by tsal15
if $C = LDL^{-1}$ , where $D$ is the Diagonal Matrix, how do I rearrange the expression to have $C^{-1}$ in terms of $L, L^{-1}$ and $D$. (also, what specific diagonal matrix can I give? is the $I_n$ ok?)

Many Thanks

Multiply both sides by $LD^{-1}L^{-1}$

$C LD^{-1}L^{-1} =LD^{-1}L^{-1}LDL^{-1}$

$C LD^{-1}L^{-1} =LL^{-1}DD^{-1}LL^{-1}$

$C LD^{-1}L^{-1} = I$

Now multiply both sides by $C^{-1}$

$CC^{-1} LD^{-1}L^{-1} = IC^{-1}$

$C^{-1} = LD^{-1}L^{-1}$

If $D$ is the identity matrix then $D^{-1} = D$, and hence:

$C^{-1} = LDL^{-1}$

3. Originally Posted by Mush
If $D$ is the identity matrix then $D^{-1} = D$, and hence:

$C^{-1} = LDL^{-1}$
Having said that, theoretically, $C^{-1} = C$, right?

So, D = I is not feasible?

4. This question goes on to say that I need to specify the diagonal matrix.

Could the Diagonal matrix be:

$\begin{bmatrix} \lambda_1 & 0 & n \\0 & \lambda_2 & n \\ n & n & \lambda_n \end{bmatrix}$

I mentioned this question in an earlier post. I'm just wondering if I can get some further assistance, if possible

5. Originally Posted by tsal15
This question goes on to say that I need to specify the diagonal matrix.

Could the Diagonal matrix be:

$\begin{bmatrix} \lambda_1 & 0 & n \\0 & \lambda_2 & n \\ n & n & \lambda_n \end{bmatrix}$

I mentioned this question in an earlier post. I'm just wondering if I can get some further assistance, if possible

A matrix of eigenvalues normally works, yes.

6. Originally Posted by tsal15
if $C = LDL^{-1}$ , where $D$ is the Diagonal Matrix, how do I rearrange the expression to have $C^{-1}$ in terms of $L, L^{-1}$ and $D$. (also, what specific diagonal matrix can I give? is the $I_n$ ok?)

Many Thanks
Just a note, using the property that

$\left( AB \right)^{-1} = B^{-1}A^{-1}$

gives your result pretty quickly. To go on further, if $C$ has n distict eigenvalues, say $\lambda_n$, then to each we find an associatede eigenvector, say $\bold{e}_n$, then the diagonal matrix is given by

$D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_1 & \cdots & 0 \\
0 & 0 & \ddots& 0 \\
0 & \cdots & 0 & \lambda_n \end{bmatrix}$

and L by

$L = \begin{bmatrix} \bold{e}_1 & \bold{e}_1 & \cdots & \bold{e}_n\end{bmatrix}$.

Check with

$C = \begin{bmatrix} 2 & 1 \\
1 & 2 \end{bmatrix}$