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 '20 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 '20 at 13:28

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.

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.