0
$\begingroup$

Suppose the system of linear equations $AX=B$ has a unique solution for some $B$ . Prove that rref of $A$ is the same as $I_n$. ($A$ is a square matrix)

My try : Because the system has a unique solution therefore it's possible to write $x_1 = a , x_2 = b , \dots , x_n = z$ . In the matrix form, for the coefficient matrix, we have :

\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} I don't know whether that's enough for proving that statement or not .

$\endgroup$
  • $\begingroup$ If $\mathbf{x}_0$ is a solution to $AX=0$ and $\mathbf{x}_B$ is a solution to $AX=B$, then $\mathbf{x}_B+\mathbf{x}_0$ is a solution to $AX=B$. So if $AX=B$ has a unique solution, then $AX=0$ must also have a unique solution. Do you know anything about when homogeneous systems have unique solutions? $\endgroup$ – Arturo Magidin Jun 6 at 20:26
  • $\begingroup$ @ArturoMagidin Unfortunately no . $\endgroup$ – S.H.W Jun 6 at 20:32
  • $\begingroup$ Are you familiar with the concepts of rank and nullity? $\endgroup$ – paulinho Jun 6 at 20:48
  • $\begingroup$ @paulinho I'm only familiar with the rank. It is the number of leading 1's in ref form. $\endgroup$ – S.H.W Jun 6 at 20:50
  • $\begingroup$ Counterexample: $\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}X=\begin{bmatrix}1\\1\\2\end{bmatrix}$. This has a unique solution, but the coefficient matrix isn’t square, so its rref can’t possibly equal the identity matrix. Seems like there are some missing conditions from the claim you’re trying to prove. $\endgroup$ – amd Jun 7 at 0:47
0
$\begingroup$

If there is a unique solution, then there must be n pivots, or else there will be other variables defined uniquely by a pivot which leads to a n-dimensional subspace of pivots. That is, a unique solution only exists when the rank is maximized.

$\endgroup$

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.