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Is there a special method to find the the inverse for a matrix which would classified as a lower or left triangular matrix for a matrix L which is n by n. Additionally where the upper part of the matrix would also be all zeros.

where none of the diagonals are equal to zero{(1,1), (2,2),...,(n,n)} ≠ 0 , or a,c,j and j in the example below. so the determinant ≠ zero.

For example when n=4 finding the inverse of the matrix $L$ where $$L=\begin{pmatrix} a & 0 & 0 & 0\\ b & c & 0 & 0\\ d & e & f & 0 \\ g & h & i & j \\ \end{pmatrix} $$

But could also work when n = 5,6,..., at least 10 (for the sake of simplicity)

Does there exist a method to find the inverse of any sized matrix in this form?

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  • $\begingroup$ Is it for numerical application? $\endgroup$ – Surb Feb 11 '15 at 7:30
  • $\begingroup$ Solve $L u_k = e_k$. $\endgroup$ – copper.hat Feb 11 '15 at 7:32
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    $\begingroup$ Forward substitution, my friend. $\endgroup$ – user1551 Feb 11 '15 at 8:24
  • $\begingroup$ @user1551 could you possibly elaborate? $\endgroup$ – John Feb 12 '15 at 1:52
  • $\begingroup$ @John Let the $i$-th row of $L^{-1}$ be $r_i$ and the $i$-th row of $I$ be $e_i$. By comparing the first rows of both sides of $LL^{-1}=I$, we get $ar_1=e_1$ and hence $r_1=\frac1ae_1$. Compare the second rows, we get $br_1+cr_2=e_2$ and so $r_2 = \frac1c(e_2-br_1)$. Continue in this way, we can solve for $r_i$ one at a time. This is called forward substitution, because the index $i$ runs from $1$ up to $n$. Alternatively, by comparing the $j$-th columns of both sides of $L^{-1}L=I$ from $j=n$ down to $1$, you can obtain all columns of $L^{-1}$. This is called backward substitution. $\endgroup$ – user1551 Feb 12 '15 at 4:55
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We can write $L = D(I + N)$ where $D$ is diagonal and $N$ is strictly lower triangular and nilpotent ($N^n = 0$): $N_{ij} = L_{ij}/D_{ii}$. Then $L^{-1} = (I+N)^{-1} D^{-1}$.
$D^{-1}$ is diagonal with $(D^{-1})_{ii} = 1/D_{ii}$, and $(I+N)^{-1} = I + \sum_{j=1}^{n-1} (-1)^j N^j$.

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In case anybody is looking for an actual closed form (that is, not forward substitution or a Neumann expansion), because this came up in my research at one point:

You can also work out a closed form for each element of the inverse of a triangular matrix. From Cramer's rule, we know $[A^{-1}]_{ij} = \frac{1}{det(A)}C_{ji}$, where $C_{ji}$ is the matrix of cofactors of $A$. The $ji$-cofactor is given by $(-1)^{i+j}$ times the $ji$-minor of $A$, which is defined as the determinant of the matrix obtained by eliminating the $j$th row and $i$th column of $A$. In the case of a (say, lower) triangular matrix, eliminating one row and column results in a block lower triangular matrix, where

$|C_{ji}|= \left|\begin{matrix}T_0 & \mathbf{0} & \mathbf{0} \\ A_0 & H & \mathbf{0} \\ A_1 & A_2 & T_1\end{matrix}\right|$.

Here, $T_0$ and $T_1$ are lower triangular and $H$ is lower Hessenberg. By nature of block determinants, $|C_{ji}| = |T_0||T_1||H|$. The relative size of these matrices depends on $i$ and $j$. For $i=j$, $C_{ji}$ is triangular. For $i=1$ and $j=n$, the entire matrix is Hessenberg. The determinant of triangular matrices $T_0$ and $T_1$ is just the product of the diagonals, and there is a nice form for the determinant of Hessenberg matrices given here:

https://www.math.uni-bielefeld.de/ahlswede/pub/tamm/hessen.ps

There they assume a unit upper diagonal, so for the general case it will need to be modified, but doing so is fairly straightforward.

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