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Let $\{a_1, a_2,...,a_n\}$ and $\{b_1,b_2,...,b_n\}$ be two bases of $\Bbb R^n$. Let P be square matrix of order $n$ with real entries such that $Pa_i=b_i, i=1,2..,n$. Suppose that every eigenvalue of P is either $-1$ or $1$. Let $Q=I+2P$. Then which of the fallowing is true?

1) $\{a_i+2b_i | i=1,2,...,n\}$ is also a basis of $\Bbb R^n$.

2) Q is invertible

I got that 2) is true. but not getting the 1)

I know that, I have to show that $\sum_{i=1}^{n} \alpha _i(a_i+2b_i)=0$ when all constants $\alpha _i, i=1,2,...,n$ It seems difficult for me to show this. I write all $b_1,b_2,...,b_n$ as linear combination of $a_1,a_2,...,a_n$ and putted the values in the sum. which gives more difficult expression to give all constants are zero. Please help me. Thank you

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    $\begingroup$ What if $b_i = -a_i / 2$? $\endgroup$ – user251257 Oct 2 '16 at 17:31
  • $\begingroup$ then set will contain a zero vector, and it will become linearly dependent $\endgroup$ – aryan Oct 2 '16 at 17:34
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    $\begingroup$ could that be a basis? $\endgroup$ – user251257 Oct 2 '16 at 17:35
  • $\begingroup$ Nice counter example. $\endgroup$ – mvw Oct 2 '16 at 17:35
  • $\begingroup$ Nopz, It will not be a basis. wait sir i'm editing my question. $\endgroup$ – aryan Oct 2 '16 at 17:36
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As you can consider $Q$ as an injective operator which preserve linear independence and $Q_{ai} =I_{ai} + 2P_{ai}=ai+2bi$.

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  • $\begingroup$ Welcome on the MathSE! This site supports $\latex$. :-) Just write $Q_{ai}$ and you get $Q_{ai}$. I am not sure if I latexized your post correctly, feel free to make it ready. :-) Good luck here! $\endgroup$ – peterh Jun 14 '18 at 19:47

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