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Let ${a_{1},a_{2},...a_{n}}$ and ${b_{1},b_{2},...b_{n}}$ be two bases of $\mathbb{R}^{n}.$ Let P be an $n \times n$ matrix with real entries such that $Pa_{i}=b_{i}$ for $i=1,2, ...,n.$ Suppose that every eigenvalue of $P$ is either $+1$ or $-1.$ Let $Q=I+2P.$ Then which of the following statements are true?

  1. $\{a_{i}+2b_{i} \mid i=1,2,...,n\}$ is also a basis of $V$.

  2. $Q$ is invertible.

  3. Every eigenvalue of $Q$ is either $3$ or $-1.$

  4. If $\det{P} > 0$ then $\det{Q} > 0.$

If I consider $a_{1}=2$, $b_{1}=-1$ as bases of $\mathbb{R}$, then $a_1+2b_1$ is not a basis of $\mathbb{R}$. So (1) is clearly incorrect.

As eigenvalues of $Q$ are $3$ or $-1$ so clearly $Q$ is always invertible and further (4) is also correct.

Am I correct in my reasoning according to me 2,3,4 are correct

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  • $\begingroup$ In your counterexample,$ P $ has eigenvalue $-1/2$. $\endgroup$ – Corey Jun 30 '16 at 17:03
  • $\begingroup$ But, if it had been a valid counterexample then 3. would not be true. $\endgroup$ – Corey Jun 30 '16 at 17:06
  • $\begingroup$ ohhh, that means $a_i+2b_i$ is never zero because P has eigenvalue 1 or -1 $\endgroup$ – slow but keen learner Jun 30 '16 at 17:07
  • $\begingroup$ @Corey I will take care of that. Yes V=$\mathbb{R^{n}}$ $\endgroup$ – slow but keen learner Jun 30 '16 at 17:25
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we begin by 3)

we have $\sigma(P)=\{1,-1\}$ and for all polynomials $f$ $$ f(\sigma(P))=\sigma(f(P)) $$ so if we take $f=1+2 x$ $Q=F(P)$ and then $$ \sigma(Q)=\{f(1),f(-1) \}=\{3,-1\} $$ so in particular $0\not\in\sigma(Q)$ and then $Q$ is invertible so this give as 2)

but 2) implies that $det(Q)\neq0$,

and we know that $\det(P)=\{ \textrm{Product of all eigenvalues (counting multiplicity)}\}$, so $\det(P)>0$ implies that multiplicity of $-1$ is even, and then the multiplicity of $f(-1)$ is even too, so $\det(Q)>0$ which give 3)

if $a_k$ is a basis of $\mathbb{R}^n$ then as $Q$ is invertible we have $\{Q(a_k)\}$ is also a basis then $(a_k+2b_k)$ is also a basis of $\mathbb{R^n}$

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  • $\begingroup$ Thanx very much for your time. I will be simply pleased if you can explain last line in detail. $\endgroup$ – slow but keen learner Jun 30 '16 at 17:16
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    $\begingroup$ a matrix is invertible if and only if it transform every basis of $\mathbb{R}^n$ to a basis of \mathbb{R^n}$ $\endgroup$ – Hamza Jun 30 '16 at 17:19

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