# On multiplying elements of a basis by a matrix with nonzero determinant

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.

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## 2 Answers

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 $a_1 v_1 + \ldots + a_n v_n$?

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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.

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