# proof: inverse of lower triangular identity matrix

As you know that is enough negating below of diagonal to inverse of lower triangular identity matrix.

example

$$A = \left(\begin{matrix} 1 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 2 & 0 & 0 & 1 \\ \end{matrix}\right)$$

basically inverse of A

$$A' = \left( \begin{matrix} 1 & 0 & 0 & 0 \\ -3 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ -2 & 0 & 0 & 1 \\ \end{matrix}\right)$$

I just need to prove it.

my question is not related any software. it is general linear algebra question. It's not enough to say that "if $A'=$ (inversion of $A$), multiplication $A'$ and $A$ should be $I$ (identity matrix)". we cannot say for all case. I need a general proof.

it's related topic with Gauss Elimination - LU decomposition

thank you for any help.

• I don't understand what you mean by "cannot say for all case". – Kushal Bhuyan Nov 16 '15 at 10:07
• my friend tried to prove it like "if A′= (inversion of A), multiplication A′ and A should be I (identity matrix)". i just want to say that it is not a actual proof. because we should check for all lower triangular identity matrices – Murat Cabuk Nov 16 '15 at 10:22
• Your observation only works for so-called "Gauss transforms", which are rank-1 corrections to the identity matrix that turn up in LU decomposition. Golub and Van Loan should have a proof of this. – J. M. isn't a mathematician Nov 16 '15 at 11:05
• yest J. M., you are right. it's related topic with Gauss Elimination - LU decomposition. – Murat Cabuk Nov 16 '15 at 11:32

I don't think your statement is correct. For example, $$\pmatrix{ 1\\ 1&1\\ 0&1&1\\ 0&0&1&1 }^{-1} = \left( \begin{array}{rrrr} 1\\ -1&1\\ 1&-1&1\\ -1&1&-1&1 \end{array} \right)$$ However, if only the first column is non-zero, then we can write our matrix in the form $$A = \pmatrix{ 1&0\\ x&I_3 }$$ where $x \in \Bbb R^3$ and $I_3$ is the size $3$ identity matrix. We then note using block-matrix multiplication that $$\pmatrix{ 1&0\\ x&I_3 } \pmatrix{ 1&0\\ -x & I_3} = \pmatrix{1&0\\0&I_3} = I_4$$ so that indeed, we can find the inverse by negating whatever is below the diagonal.
• with following equality, you just showed us A'.A=I $$\pmatrix{ 1&0\\ x&I_3 } \pmatrix{ 1&0\\ -x & I_3} = \pmatrix{1&0\\0&I_3} = I_4$$ thak you anyway – Murat Cabuk Nov 16 '15 at 12:11
• In what way is this not a proof? If $AA'=I$, then $A'$ is the inverse of $A$. – Ben Grossmann Nov 16 '15 at 13:06