0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't understand what you mean by "cannot say for all case". $\endgroup$ – Kushal Bhuyan Nov 16 '15 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 '15 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. isn't a mathematician Nov 16 '15 at 11:05
  • $\begingroup$ yest J. M., you are right. it's related topic with Gauss Elimination - LU decomposition. $\endgroup$ – Murat Cabuk Nov 16 '15 at 11:32
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '15 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 '15 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 '15 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 '15 at 13:26
  • $\begingroup$ I understand what you mean but believe me it's not a proof. I already presented same thing to prof. M1 = A - I M2 = A’ - I A x A’ = (M1 + I) x (M2 x I) = (M1 x M2) + (M1 x I) + (I x M2) + (I x I) = I + M1 + M2 + I= 0 + 0 + I Also my equalities are also not a proof. why because I should prove M1 + M2 . How you explane that "if A′= (inversion of A), multiplication A.A′ and A should be I (identity matrix)" . You just talked about result. really thank you for your time. no body give any answer. $\endgroup$ – Murat Cabuk Nov 16 '15 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.