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$Ax=b$

$m$ number of Rows

$n$ number of columns

true or false

A) If $n > m$, given any $b$ you can always solve $Ax=b$.

The answer: False. Counterexample: A is the zero matrix.

We have unknowns more than equations, so we can always solve $Ax=b$. Why the answer is false? And even if $A$ is the zero matrix we have the matrices x = zeros then $b$ zeros so there is always answer!! Can anyone explain why the answer is false? How can he proved false by $A$ is the zero matrix?

B) If $n < m$, the only solution of $Ax=0$ is $x=0.$ The answer: False. Counterexample: let $A$be the zero matrix.

I understand why it is false but I'm wondering, is there any special case makes this statement true?

Edit: I made a matrix $A= 2$ rows $\times 1$ column contains $(1,0)$ and assume Ax=0 as the statement, then the only solution is when x=0 because we can't use row echelon form any more. is what I did correct to make the statement true?

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For A), the statement is that given any $b \in \mathbb{R}^n$, there is a solution to the matrix equation $Ax = b$. But if $A$ is the zero matrix, then there is only one $b$ for which there is a solution, namely $b=0$. Since there are some $b$ for which $Ax = b$ cannot be solved, the original statement is false.

For B), the condition that the only solution to $Ax=0$ is $x=0$ means that there are no free variables in the reduced row echelon form of $A$. Or said differently, when you apply Gauss-Jordan elimination, you must get $n$ pivots (or equivalently, $n$ non-zero rows). Said still another way, the only solution to $Ax = 0$ is $x=0$ exactly when the rank of $A$ is equal to $n$.

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  • $\begingroup$ Can you give an example of a matrix its rank equal to n? thank you so much $\endgroup$ – Rayanh Apr 21 '13 at 15:41
  • $\begingroup$ The $n \times n$ identity matrix has rank $n$. $\endgroup$ – Michael Joyce Apr 21 '13 at 19:15
  • $\begingroup$ I mean when n < m as the condition because nxn doesn't satisfy the condition. Can you give an example of a matrix its rank is equal to n and n < m ? Thank you very much for your answer it was really helpful. I will wait your example and thanks again in advance. $\endgroup$ – Rayanh Apr 21 '13 at 19:58
  • $\begingroup$ Sure, just add $m-n$ rows of zeroes to the $n \times n$ identity matrix. This will be an $m \times n$ matrix with rank $n$. Do you see why? The most general example of an $m \times n$ matrix of rank $n$ is by using $n$ linearly independent column vectors in $\mathbb{R}^m$. $\endgroup$ – Michael Joyce Apr 22 '13 at 1:11
  • $\begingroup$ I'll tell you what I did $\endgroup$ – Rayanh Apr 22 '13 at 2:14
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Let $f$ the canonical linear application associated to the matrix $A$ hence $$f\colon \mathbb{R}^n\to \mathbb{R}^m$$

A) given any $b$ we can solve $f(x)=b$ if $f$ is surjective i.e. $\mathrm{rank}(f)=\mathrm{rank}(A)=m$

B) the only solution to $f(x)=0$ is $x=0$ if $f$ is injective i.e. $\ker(f)=\ker(A)=\{0\}$ so by the Rank-nullity theorem $\mathrm{rank}(f)=\mathrm{rank}(A)=n$

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  • $\begingroup$ Maybe tell him what the canonical linear application is $\endgroup$ – RougeSegwayUser May 15 '13 at 6:36

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