# Linearly independent vectors and singular matrix

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.

• 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. – JMoravitz Apr 9 '20 at 13:27
• 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. – 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.

• 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.