I have been given an assignment where I need to find the inverse of an upper triangular matrix. I have been given these two formulas and am to use Lemma $2$. Can someone explain to me how this works. I understand Lemma $1$ its pretty simple but the different powers in lemma two lost me. Why would it go from first power to squared to suddenly fourth and then to logarithms? Below are the two lemmas I have been given.
I Triangular matrices for linear systems with special structure
$1$. Triangular matrices and back substitution
$A$ - lower triangular square matrix $n \times n$; $a_{ij} = 0$ for $i<j$
$\det A \neq 0$ nonsingularFirst $a_{ii}=1$ for all $i$
Split $A=I-L$, $l_{ij}$ for $i \leq j$
$L^n=0$ easy to verify
Lemma $\mathbf 1$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^{-1} =I +L +L^2 +\ldots +L^{n-1}.$$
Proof: Let $B$ be equal to the right-hand side of equation ($1$). Then $$B(I-L)=I-L^n=I$$ because $L^n=0$ and therefore $B=A^{-1}$.
Lemma $\mathbf 2$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^{-1} =(I+L^{2^{[\log n]-1}}) (I+L^{2^{[\log n]-2}}) \cdot \ldots \cdot (I+L^4) (I+L^2) (I+L). \tag{2}$$ Proof: Expand the product in the right-hand side of ($2$). Since $L^n=0$, we are left with $I +L +L^2 +\ldots +L^{n-1}$ which is equal to $A^{-1}$ by Lemma $1$. $$\tag{Q.E.D.}$$
I am working with an original matrix of size $1473 \times 1473$. Was told to do a Cholesky decomposition on it and then invert it using Lemma $2$.
Maybe even a name for these two lemmas so I can do more reading would help too.
Any help is greatly appreciated.