Suppose $\{v_1,v_2,\ldots,v_n\}$ is a linearly independent set of vectors of the vector space $V$ and $A$ is an $n×n$ matrix. Let
$[u_1 u_2 \ldots u_n]^T=A [v_1 v_2 \ldots v_n]^T$
Show that $\{u_1,u_2,\ldots,u_n\}$ is linearly independent if and only if $A$ is invertible.

I have proved that if $A$ is invertible, then {${u_1,u_2,\ldots,u_n}$} is linearly independent, but I can not show the converse.
And I want to mention this problem is an exercise from a book that posed this in the vector space section before introducing basis and dimension of spaces. So please help maintaining this.

  • $\begingroup$ I think the best method is to use Gaussian elimination. $\endgroup$
    – M. Vinay
    Jun 8, 2014 at 16:25
  • 1
    $\begingroup$ Have you tried to show that when $\{u_i\}$ are not linearly independent, then $A$ is not invertible? $\endgroup$ Jun 8, 2014 at 16:28

1 Answer 1


If the $u_i$ are linearly independent, then $[u_1 u_2 \ldots u_n]^T$ is invertible, and since the $v_i$ are linearly independent so is $[v_1 v_2 \ldots v_n]^T$. So we can write: $A=[u_1 u_2 \ldots u_n]^T([v_1 v_2 \ldots v_n]^T)^{-1}$. Since $A$ is a product of invertible matrices, it is also invertible.


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