Eigenvalues of $Q=I+2P$ 
I have tried to do it evaluated option (a). I think it is correct.Can not get the other options.Please help me.
 A: Here are some hints to help you get started.
Part c) Suppose $v$ is an eigenvector of $P$ with eigenvalue $\lambda$, i.e. $Pv = \lambda v$. Then $Qv = (I+2P)v = v+2Pv = v+2\lambda v = (1+2\lambda)v$, i.e. $v$ is an eigenvector of $Q = I+2P$ with eigenvalue $1+2\lambda$. So if $-1$ and $+1$ are the possible eigenvalues of $P$, what are the possible eigenvalues of $Q$?
Part b) A matrix is invertible if and only if all the eigenvalues are non-zero. Use your answer to part c to answer this. 
Part d) The determinant of a matrix is the product of its eigenvalues (counting multiplicity). So $\det P$ is the product of a bunch of $-1$'s and $+1$'s. Hence, $\det P > 0$ if $P$ has an even number of eigenvalues that are $-1$ and $\det P < 0$ if $P$ has an odd number of eigenvalues that are $-1$. Can you make a similar statement about $Q$? How are the eigenvalues of $P$ and $Q$ related?
Part a) If $Q$ is invertible (see part b), then since $\{a_1,\ldots,a_n\}$ is a basis, so is $\{Qa_1,\ldots,Qa_n\}$. But $Qa_i = (I+2P)a_i = \cdots$.
A: Let $B_1=\{a_1, a_2, ... , a_n\}$ and $B_2=\{b_1, b_2, ... , b_n\}$ are bases of $\mathbb{R^n}$.
Given that $P=(p_{ij})_{n \times n}$ such that $Pa_i = b_i$, $i=1,2, .  .  . n$.
Since every eigen value of $P$ is either $-1$ or $1$, so $P$ is an orthogonal matrix 
i.e., $\det P \lt 0 $ or $\det P\gt 0$ 
Let $Q=I+2P$
Eigen values of $Q$ are either $(1-2)=-1$ or $(1+2)=3$. (Option (c) is correct)

Since $P$ is non-singular (since every eigen value of $P$ is either $-1$ or $1$, so $\det P\neq 0$), so $Q=I+2P$ is also non-singular.
Hence $Q$ is invertable. (Option (b) is correct) 

If $\det P \gt 0$, then all eigen values of $P$ is either $1$, or $-1$ and then all eigen values of $Q$ is either $3$, or $-1$ respectively. Again $\det P = $product of all eigen values of $P$ so then for each cases $\det Q \gt 0$ also holds.  (Option (d) is correct) 

Since $B_1$, and $B_2$ are bases, so $B_1$, and $B_2$ are all linearly independent. 
hence linear combinition of vectors of $B_1$, and $B_2$ are also linearly independent.
Therefore $\{a_i + 2 b_i : i=1,2, .  . ., n\}$ is also a basis of $V$. (Option (a) is correct)
