# Thread: Prove A is singular

1. ## Prove A is singular

Let A be an nxn marix, if A is row equivalent to a matrix B and there is a non-zero matrix C such that BC = 0n, prove that A is singular.

2. Originally Posted by 450081592
Let A be an nxn marix, if A is row equivalent to a matrix B and there is a non-zero matrix C such that BC = 0n, prove that A is singular.

A row equivalent to B $\displaystyle \Longrightarrow\,\exists$ elementary matrices $\displaystyle E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B$ , so:

$\displaystyle 0_n=BC=(E_1....E_kA)C$ , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$ you get a contradiction.

Tonio

3. Originally Posted by tonio
A row equivalent to B $\displaystyle \Longrightarrow\,\exists$ elementary matrices $\displaystyle E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B$ , so:

$\displaystyle 0_n=BC=(E_1....E_kA)C$ , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$ you get a contradiction.

Tonio

if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$ you get a contradiction.

what does does this mean? are we not trying to prove that A is singular? there shouldnt be a contradction?

4. Originally Posted by 450081592
if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$ you get a contradiction.

what does does this mean? are we not trying to prove that A is singular? there shouldnt be a contradction?

Read again and carefully: we supposed A is NON-SINGULAR, i.e. we supposed A is invertible, and we got a contradiction...and this is what you wanted, didn't you?

Tonio

5. Originally Posted by tonio
Read again and carefully: we supposed A is NON-SINGULAR, i.e. we supposed A is invertible, and we got a contradiction...and this is what you wanted, didn't you?

Tonio
oh ic, so we are trying to prove that A is not an elementaery matrix, right?

6. Originally Posted by tonio
A row equivalent to B $\displaystyle \Longrightarrow\,\exists$ elementary matrices $\displaystyle E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B$ , so:

$\displaystyle 0_n=BC=(E_1....E_kA)C$ , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$ you get a contradiction.

Tonio

$\displaystyle E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B$ , this line, isn't it Ek...E2E1A = B? does it make a different? And what does multiplying the right hand from the left mean, multiply that at the right side, or left side?

7. Originally Posted by 450081592
oh ic, so we are trying to prove that A is not an elementaery matrix, right?

No. We're trying to prove that A is singular. Of course, this includes A not being elementary since elem. matrices are non-singular, but our quest is way more general.

Tonio

8. Originally Posted by 450081592
$\displaystyle E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B$ , this line, isn't it Ek...E2E1A = B? does it make a different? And what does multiplying the right hand from the left mean, multiply that at the right side, or left side?

I decided $\displaystyle E_1...E_kA$. but of course you can order the elem. matrices as you wish, and yes: multiplying from the left means at the left side, since matrix multiplication isn't commutative.

Tonio

9. Originally Posted by tonio
A row equivalent to B $\displaystyle \Longrightarrow\,\exists$ elementary matrices $\displaystyle E_1,...,E_k\,\,s.t.\,\,E_1...E_kA=B$ , so:

$\displaystyle 0_n=BC=(E_1....E_kA)C$ , and if A weren't singular and since elementary matrices aren't singular, multiplying the right hand from the left by $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$ you get a contradiction.

Tonio

ok, but why do I multiply [tex]A^{-1}E_K^{-1}...E_1^{-1}[/math? where is it from, is it the inverse of B? If I only multiply this on the left side, won't it change the equation?

10. Originally Posted by 450081592
ok, but why do I multiply $\displaystyle A^{-1}E_K^{-1}...E_1^{-1}$? where is it from, is it the inverse of B? If I only multiply this on the left side, won't it change the equation?

Do it! Multiply the expression $\displaystyle BC=(E_1...E_kA)C=0$ from the left by $\displaystyle A^{-1}E_k^{-1}...E_1^{-1}$ and you'll get a contradiction....don't ever forget the given data!

Tonio

11. ok I got it, we multiply by the inverse of B, that we get C =0 which contradicts the hypothesis, thanks so much