I'm currently practicing for my linear algebra exam and there was a question that I could not understand.The question is :

Let $\{v_1, v_2, v_3\}$ be a linearly independent set of vectors in $\Bbb R^3$. Let $A\in M_{3\times 3}(\Bbb R)$ be non-invertible. Prove or disprove that the set of vectors $\{Av_1, Av_2, Av_3\}$ is linearly independent.

I approached the problem by using the fact that the linearly independent vectors satisfy the following:

$c_1v_1 + c_2v_2 + c_3v_3 = 0$ and since $Ax=0$ would only have one solution in this case $c_1 = c_2 = c_3 = 0$, but I couldn't reach to a solution.

Another way that I used to solve it, was using the fact that $\det(A) = 0$ and $\det([v_1 v_2 v_3])\ne 0$. However, this also didn't help me to find the proof.

And finally I used some random values for $A$ and the vectors what I found was that they are linearly dependent, but this solution is not a valid proof, I think.

I would be very happy if you could help me.

  • $\begingroup$ Disprove that the vectors $Av_1,Av_2,Av_3$ are independent by considering the trivial case of where $A$ is the zero matrix. You have that $Av_1,Av_2,Av_3$ are all the zero vector and are clearly not independent of one another. $\endgroup$
    – JMoravitz
    Apr 9, 2020 at 13:27
  • $\begingroup$ What is more interesting is to show that for all $A$ you have that $A$ non-invertible implies that $Av_1,Av_2,Av_3$ must be dependent. $\endgroup$
    – JMoravitz
    Apr 9, 2020 at 13:28

1 Answer 1


Supoose on the contrary that $\{Av_1, Av_2, Av_3\}$ is linearly independent.

Then, $$\det\left( \begin{bmatrix} Av_1 & Av_2 & Av_3\end{bmatrix}\right) \ne 0$$

This is equivalent to saying that

$$\det(A)\det\left( \begin{bmatrix} v_1 & v_2 & v_3\end{bmatrix}\right) \ne 0$$

but $\det(A)=0$, which is a contradiction.

Remark about your attempts:

  • You want to verify whether $\sum_{i=1}^3c_i(Av_i)=0$ implies $c_i=0, \forall i$.
  • $Ax=0$ will only have one solution is not a true statement unless $A$ is invertible.

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