Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

True or False: For a basis $\{v_1, \dots, v_n\}$ of $\Bbb R^n$, $\{Av_1, \dots, Av_n\}$ is a basis of $\Bbb R^n$ if $\det(A) \neq 0$.

The answer is true, but I don't know why. My guess is that if $\det(A) \neq 0$, there exist $A^{-1}$ and therefore $A$ has trivial solution ($x=0$). It means nullspace of $A$ has one dimension? I'm stuck in here. If you know the logic which is used in here, please explain to me.

share|cite|improve this question
up vote 3 down vote accepted

No, the nullspace of $A$ has dimension $0$ ($Ax=0$ has only the trivial solution).

Hint: suppose some linear combination $a_1 A v_1 + \ldots + a_n A v_n = 0$. If $\det A \ne 0$, what can you say about $a_1 v_1 + \ldots + a_n v_n$?

share|cite|improve this answer
I understand what you mean. I can multiply $A^{-1}$ to both sides then get $a_1v_1+...+a_nv_n=0$ and because {$v_1,...,v_n$} is a basis they are linearly independent($a_1=...=a_n=0$) So $a_1Av_1+...+a_nAv_n=0$ means that {$Av_1,...,Av_n$} also from a basis, is that right? – niagara Dec 3 '12 at 2:32
you have just verified linear independence. I suppose you now need to justify that the $Av_i$ span the space, which should follow from rank-nullity. – John Martin Dec 3 '12 at 2:34
To span the space, any vector can be express as a linear combination of {$Av_1,...,Av_n$}. I think since {$v_1,...,v_n$} is a basis so it is possible, in which part should I use rank-nullity? – niagara Dec 3 '12 at 2:48
In order to write $x = c_1 A v_1 + \ldots + c_n A v_n$, what do you suppose $c_1 v_1 + \ldots + c_n v_n$ should be? – Robert Israel Dec 3 '12 at 3:39

Suppose $c_1Av_1+...+c_nAv_n=0$. Since the determinant of $A$ is nonzero, $A$ is invertible. Premultiply each side by $A^{-1}$ to obtain $c_1v_1+...+c_nv_n=0$. Since $v_1,...,v_n$ forms a basis, these are linearly independent so that $c_1=...=c_n=0$. This means that $Av_1,...,Av_n$ are linearly independent so that $Av_1,...,Av_n$ is a basis.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.