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

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

As you can consider $Q$ as an injective operator which preserve linear independence and $Q_{ai} =I_{ai} + 2P_{ai}=ai+2bi$.

| cite | improve this answer | |
  • $\begingroup$ 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 - Reinstate Monica Jun 14 '18 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.