Suppose $Ax = b$ and $Cx = b$ have the same (complete) solutions for every b (A, B, C, x are suitable matrices). Is it true that A = C?
My attempt: I could just get $(A-C)x=0$ and I am unable to draw any further conclusion from this.
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3$\begingroup$ Do you mean that $\{x : Ax=b\} = \{x : Cx=b\}$ for all $b$? [I'm not sure what your phrasing "the same (complete) solutions" means.] $\endgroup$– MichaelCommented Dec 16, 2019 at 15:39
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$\begingroup$ @Michael Yes, that is what I meant. Sorry if that was confusing. $\endgroup$– user_9Commented Dec 16, 2019 at 15:58
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3 Answers
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Yes, it is indeed true that $A=C$.
Observe $(A-C)x=0$, for every $x$, so obviously $A-C=0$.
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This question appears in Gilbert Strang's "Introduction to Linear Algebra" (4th ed.) Section 3.4 Challenge Problems 36 (p. 167) and also in the Preface.
Since it's one of the challenge problems, I don't think it is so obvious.
One possible answer that I thought intuitive is this way, which I found here. Since we can freely choose $\mathbf{b}$, let's set it to be the first column of $A$. Then in that case, one solution is $\mathbf{x}^T = [1\ 0\ ...\ 0]^T$. Since $C \mathbf{x} = \mathbf{b}$ for such $\mathbf{b}$ and $\mathbf{x}$, the first column of $C$ is identical to the first column of $A$. Similarly, we can show every column of $A$ and $C$ is identical and thus $A=C$.
The link also shows the nullspace of $A$ and $C$ is identical. That is trivial by setting $\mathbf{b}$ to $\mathbf{0}$, but I'm not sure if we need that to show $A=C$.
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Yes. For every vector $v$, let $b=Av$. Then $x=v$ is a solution to the equation $Ax=b$ and hence by assumption, it is also a solution to the equation $Cx=b$. Therefore $Cv=b=Av$. Since $v$ is arbitrary, we obtain $A=C$.
answered Dec 16, 2019 at 16:32