# Thread: Prove n x 1 vectors Linearly independent

1. ## Prove n x 1 vectors Linearly independent

Let $A$ be an $n\times n$-matrix. Three column vectors $X_1,X_2,X_3$ are given, and we have that $$A^2X_1=A^2X_2=A^2X_3=0$$ We know that $AX_1,AX_2, AX_3$ are linearly independent. Prove that the column vectors $X_1,X_2,X_3,AX_1,AX_2,AX_3$ are also linearly independent.

It is a hard problem for me, and my first thought was that $A^2X_1=0 \Rightarrow AX_1=0$. That is probably not correct and not leading to anything.

I then concluded that $\det(AX_1,AX_2, AX_3)\not=0$ which can lead me somewhere, since it's a proof with linear independence and $\det\not=0 \Leftrightarrow$ *lin. independent*.

2. ## Re: Prove n x 1 vectors Linearly independent

Let me post this correctly

Let $\displaystyle A$ be an $\displaystyle n\times n$-matrix. Three column vectors $\displaystyle X_1,X_2,X_3$ are given, and we have that $\displaystyle A^2X_1=A^2X_2=A^2X_3=0$ We know that $\displaystyle AX_1,AX_2, AX_3$ are linearly independent. Prove that the column vectors $\displaystyle X_1,X_2,X_3,AX_1,AX_2,AX_3$ are also linearly independent.

It is a hard problem for me, and my first thought was that $\displaystyle A^2X_1=0 \Rightarrow AX_1=0$. That is probably not correct and not leading to anything.

I then concluded that $\displaystyle \det(AX_1,AX_2, AX_3)\not=0$ which can lead me somewhere, since it's a proof with linear independence and $\displaystyle \det\not=0 \Leftrightarrow$ *lin. independent*.

3. ## Re: Prove n x 1 vectors Linearly independent

Originally Posted by jacob93
my first thought was that $\displaystyle A^2X_1=0 \Rightarrow AX_1=0$.
This does not have to be the case.

Originally Posted by jacob93
I then concluded that $\displaystyle \det(AX_1,AX_2, AX_3)\not=0$ which can lead me somewhere, since it's a proof with linear independence and $\displaystyle \det\not=0 \Leftrightarrow$ *lin. independent*.
$\displaystyle AX_1,AX_2,AX_3$ is not a square matrix unless n = 3, so the determinant is not defined.

The statement is obvious if you try proving it by contradiction.