# Getting the inverse of a lower/upper triangular matrix

For a lower triangular matrix, the inverse of itself should be easy to find because that's the idea of the LU decomposition, am I right? For many of the lower or upper triangular matrices, often I could just flip the signs to get its inverse. For eg: $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -1.5 & 0 & 1 \end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 1.5 & 0 & 1 \end{bmatrix}$$ I just flipped from -1.5 to 1.5 and I got the inverse.

But this apparently doesn't work all the time. Say in this matrix: $$\begin{bmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 3.5 & -2.5 & 1 \end{bmatrix}^{-1}\neq \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ -3.5 & 2.5 & 1 \end{bmatrix}$$ By flipping the signs, the inverse is wrong. But if I go through the whole tedious step of gauss-jordan elimination, I would get its correct inverse like this: $\begin{bmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 3.5 & -2.5 & 1 \end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ 1.5 & 2.5 & 1 \end{bmatrix}$ And it looks like some entries could just flip its signs but not for others.

Then this is kind of weird because I thought the whole idea of getting the lower and upper triangular matrices is to avoid the need to go through the tedious process of gauss-jordan elimination and can get the inverse quickly by flipping signs? Maybe I have missed something out here. How should I get an inverse of a lower or an upper matrix quickly?

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you can only flip the signs for atomic triangular matrices. The first one is atomic, the second one is not. –  Nana Jun 25 '11 at 6:56

As I said in my comment, you can only flip the signs for atomic triangular matrices to get their inverses. To explore more see this.

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An atomic triangular matrix is an identity matrix that has only one of its entry a number not zero? Like in my first example, everything is similar to identity matrix and only one of the entry is a number not zero? So if they are non-atomic triangular matrices, I still have to go through that tedious process of elimination to get its inverses? –  xenon Jun 25 '11 at 7:06
Yes, but at least you have some entries to be zero...:) –  Nana Jun 25 '11 at 7:17
The "negation" bit actually applies to a wider class of unit triangular matrices. In particular, Gauss transforms, which are triangular modifications of the identity that have possibly nonzero entries in only one column and above (or below) the $1$, can be inverted by flipping the signs. –  Ｊ. Ｍ. Sep 2 '11 at 16:48

Ziyuang's answer handles the cases, where $N^2=0$, but it can be generalized as follows. A triangular $n\times n$ matrix $T$ with 1s on the diagonal can be written in the form $T=I+N$. Here $N$ is the strictly triangular part (with zeros on the diagonal), and it always satisfies the relation $N^{n}=0$. Therefore we can use the polynomial factorization $1-x^n=(1-x)(1+x+x^2+\cdots +x^{n-1})$ with $x=-N$ to get the matrix relation $$(I+N)(I-N+N^2-N^3+\cdot+(-1)^{n-1})=(I-N^n)=I$$ telling us that $(I+N)^{-1}=I+\sum_{k=1}^{n-1}(-1)^kN^k$.

Yet another way of looking at this is to notice that it also is an instance of a geometric series $1+q+q^2+q^3+\cdots =1/(1-q)$ with $q=-N$. The series converges for the unusual reason that powers of $q$ are all zero from some point on. The same formula can be used to good effect elsewhere in algebra, too. For example, in a residue class ring like $\mathbf{Z}/2^n\mathbf{Z}$ all the even numbers are nilpotent, so computing the modular inverse of an odd number can be done with this formula.

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Under the assumption of diagonal entries being 1, what you mean is $(I+N)^{-1}=(I-N)$, where $N$ is a nilpotent matrix, yielding $N^2=0$.

If diagonal entries are not all 1's (and none is 0), denote the matrix with only those diagonal entries as $D$, it will be reduced to $N^2=ND^{-1}-DN=D^{-1}N-ND$. Waiting for further simplification.

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In case of a lower triangular matrix with arbitrary non-zero diagonal members, you may just need to change it in to: $T = D(I+N)$ where $D$ is a diagonal matrix and $N$ is again an strictly lower diagonal matrix. Apparently, all said about inverse in previous comments will be the same.

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