$m$ 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?


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 $n$.

  • $\begingroup$ 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

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 $\mathrm{rank}(f)=\mathrm{rank}(A)=n$

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

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.