0
$\begingroup$

Assuming $A$ is non-singular I can show that the system $AX=b$ has a solution for each $b$. Now I assume that the system $AX=b$ has a solution for each $b$. To prove $A$ is non-singular that is $A$ is invertible. I know that $A$ is invertible if and only if $A$ is row-equivalent to identity matrix. Now how to show that $A$ is equivalent to a identity matrix of same size?

$\endgroup$
  • 5
    $\begingroup$ Questions like this highly depend on the theorems allowed. $\endgroup$ – Kenny Lau Aug 11 '18 at 14:46
  • $\begingroup$ Using the dimension of the image should be the most elegant way to finish the proof. $\endgroup$ – Peter Aug 11 '18 at 15:14
1
$\begingroup$

Since the system has a solution for every $b$, therefore we can choose specific $b$’s. Namely the standard basis vectors $e_i=(0,0, \ldots , 1, \ldots ,0)^T$. So there is a solution $X_1$ for $AX=e_1$, solution $X_2$ for $AX=e_2$, and so on.

Thus you have a matrix $M$ with $X_i$’s as column vectors such that $AM=I$. Hopefully you can take it from here.

$\endgroup$
0
$\begingroup$

A quick one: suppose $\;A\;$ is an $\;n\times n\;$ matrix over some field $\;\Bbb F\;$ , then

$\;AX=b\;$ has a unique solution for each $\;b\in\Bbb F^n\;$ iff $\;A\;$ is onto (as a linear operator) iff it is one-to-one iff it is an isomorphism iff it is regular iff it is non-singular.

Check carefully your definitions now.

$\endgroup$

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.