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$A$ is an $n\times n$ matrix, $n \ge 3$. Assume $A^2 = 0$. Which one is true and which one is false and explain:

  1. The linear system $Ax = 0$ has only trivial solution.

  2. The linear system $Ax = 0$ has more than only the trivial solution.

  3. The linear system $(I - A)x$ has only trivial solution.

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  • $\begingroup$ Yes you understood correctly. Can you tell me why is it equivalent to A = 0 ? And about 3. I proved it by muliplying in I+A snd got I , is it a good explanation to show there's only trivial solution? $\endgroup$
    – user84636
    Jul 9, 2013 at 23:47
  • $\begingroup$ Sorry, I meant A^2 = 0, very sorry $\endgroup$
    – user84636
    Jul 9, 2013 at 23:54
  • $\begingroup$ What's the correct answer then? Sorry. .. $\endgroup$
    – user84636
    Jul 9, 2013 at 23:55
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    $\begingroup$ Yes, multiplying by $I+A$ is the way for 3. $\endgroup$
    – Julien
    Jul 9, 2013 at 23:57
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    $\begingroup$ If $Ax=0$ has only the trivial solution, then for $x\ne0$, $Ax\ne0$. Then $A(Ax)=0$, a contradiction. So 1. is false. Item 2. is just the negation of item 1.. $\endgroup$ Jul 10, 2013 at 0:06

1 Answer 1

3
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Hint: $$0=\det(A^2)=\det(A)^2$$ What does that tell you about the rank of $A$? is $A$ invertable?

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  • $\begingroup$ There's something that's confusing me about 3. If I look at (I- A) x = 0 and multiply it by A from the left side I get (A- A^2) x = 0 so it means Ax = 0 so more than trivial solution but by the other way we got only trivial solution! What's wrong? $\endgroup$
    – user84636
    Jul 10, 2013 at 5:39
  • $\begingroup$ @user84636 - you can't multiply $x$ by $A$ on the right.. $\endgroup$
    – nbubis
    Jul 10, 2013 at 5:54
  • $\begingroup$ Okay but why does it matter that I can't? $\endgroup$
    – user84636
    Jul 10, 2013 at 6:58
  • $\begingroup$ @user84636 - then I didn't understand your previous comment.. $\endgroup$
    – nbubis
    Jul 10, 2013 at 15:23
  • $\begingroup$ What's the answer to 3 and why the way I wrote in the previous comment is not right? $\endgroup$
    – user84636
    Jul 10, 2013 at 17:56

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