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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.

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  • $\begingroup$ It is spelled 'prove'. $\endgroup$ – Amzoti Nov 24 '13 at 14:34
  • $\begingroup$ You definitely have $n\geq6$ (if what you want to prove is to be true) but $\det(AX_1,AX_2,AX_3)$ only makes sense if $n=3$. So that is certainly not the way to go. $\endgroup$ – Marc van Leeuwen Nov 24 '13 at 14:40
  • $\begingroup$ The problem doe not say anything about $n \ge 6$, why must that be the case? $\endgroup$ – jacob Nov 24 '13 at 14:48
  • $\begingroup$ If the dimension would be less than$~6$, how can you expect to prove that six vectors are linearly independent? They would never be. However the main point of my remark is that by writing down a determinant you are making an assumption about the dimension, which as you rightly say the question says nothing about. You can only use determinants is the number of vectors is exactly equal to the dimension (determinants are defined for square matrices only). $\endgroup$ – Marc van Leeuwen Nov 24 '13 at 15:00
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Hint: write down the general form of a linear relation between those $6$ vectors. Assuming you have such a relation, apply $A$ to it. Using the given linear independence, conclude that three of the coefficients must be$~0$. Then look at your original linear relation in the light of this new information, and conclude that the remaining three coefficients must also be$~0$. Conclude.

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