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Prove the following statement or give a counterexample if it is false.

Let $A$ be an $n\times n$ matrix. If there exists a $b\in\mathbb R^{n}$ such that $Ax=b$ has a unique solution then $A$ is invertible.

What i tried:

We can see that there exists a $b=0$ such that the equation becomes $Ax=0$ and the unique solution is the trival solution $x=0$, hence $A$ is invertible since only the trival solution exists.

Is my proof correct? Also is this question the contrapositive to the theorem in this question:

Proving the existence of an inverse of a matrix. (Linear algebra).

Beacuse if it is i could use the same method to solve it?

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ys wong
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  • yes, but I think "If there exists a $b∈R^{n}$ such that $Ax=b$ has a unique solution then $A$ is invertible" have to be for every $b∈R^{n}$ .... – R.N Sep 18 '15 at 08:45
  • @RaziehNoori No, it is enough to demand that a single such $b$ exists. – 5xum Sep 18 '15 at 08:47
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    And yeah is this question the contrapostive to the theorem in this question. http://math.stackexchange.com/questions/1418693/proving-the-existence-of-an-inverse-of-a-matrix-linear-algebra If so i could have used the same method to solve it. – ys wong Sep 18 '15 at 08:47
  • @5xum may I know how? – R.N Sep 18 '15 at 08:53
  • @RaziehNoori Take $y$ such that $Ay=0$ and $x$ such that $Ax=0$. Then, $A(x+y) = \dots$. – 5xum Sep 18 '15 at 08:56
  • Then $A(x+y)=b$ Am i correct? – ys wong Sep 18 '15 at 09:00
  • i think you mean is if for that special $b$, $x$ be answer and $Ax=0$ then $A(x+y)=b$ and uniqness will give $x+y=x$ hence $y=0$ – R.N Sep 18 '15 at 09:01
  • excellent @5xum very vey good – R.N Sep 18 '15 at 09:02

1 Answers1

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Your proof is wrong. Your proof would be correct if the statement was

If for every $b\in\mathbb R^n$, the equation $Ax=b$ has a unique solution, then $A$ is invertible.

But you don't have that. You have one specific vector $b$ for which there exists a single $x$ such that $Ax = b$.

My hint:

Take any $y$ such that $Ay = 0$ and see what $A(x+y)$ is.

Also, your statement is not the contraposition of the linked question. The linked question states

If $A$ is not invertible, then there exists some $b$ such that $Ax=b$ has no solution.

The contraposition of this is (since $p\implies q$ is contraposed as $\neg q \implies \neg p$)

If there exists no $b$ such that $Ax=b$ has no solution, then it is not true that $A$ is not invertible

Getting rid of double negations, that would mean that the contraposition is

If for all $b$, the equation $Ax=b$ has some solution, then $A$ is invertible.

Which is different than the statement you have here.

5xum
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