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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.

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  • $\begingroup$ I don't understand what you mean by "cannot say for all case". $\endgroup$
    – Kushal Bhuyan
    Nov 16, 2015 at 10:07
  • $\begingroup$ 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 $\endgroup$
    – Murat Cabuk
    Nov 16, 2015 at 10:22
  • $\begingroup$ 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. $\endgroup$
    – J. M. ain't a mathematician
    Nov 16, 2015 at 11:05
  • $\begingroup$ yest J. M., you are right. it's related topic with Gauss Elimination - LU decomposition. $\endgroup$
    – Murat Cabuk
    Nov 16, 2015 at 11:32

1 Answer 1

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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.

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  • $\begingroup$ thank you Omnomnomnom. Actullay I showed my prof exactly same thing like your last equality. However, He didn't accept. Because this answer is not a proof actually. $\endgroup$
    – Murat Cabuk
    Nov 16, 2015 at 11:38
  • $\begingroup$ 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 $\endgroup$
    – Murat Cabuk
    Nov 16, 2015 at 12:11
  • $\begingroup$ In what way is this not a proof? If $AA'=I$, then $A'$ is the inverse of $A$. $\endgroup$
    – Ben Grossmann
    Nov 16, 2015 at 13:06
  • $\begingroup$ I have a strong feeling that you don't know what I mean by "block-matrix multiplication" $\endgroup$
    – Ben Grossmann
    Nov 16, 2015 at 13:26
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    $\begingroup$ I would be very surprised if you actually "already presented the same thing", and the prof didn't accept it as a proof. What is "prove $M1 + M2$" supposed to mean? Also, I am saying the reverse: if $AA' = I$, then $A' = A^{-1}$. This is always true for matrices. $\endgroup$
    – Ben Grossmann
    Nov 16, 2015 at 16:41

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