Problem about orthonormal basis Let $(V,g)$ a vector space with inner product $g.$ Let $ \beta = \{ a_1, \ldots a_n \}, \beta ' = \{ b_1, \ldots b_n \}$ two orthonormal basis for $V$ with same orientation (i.e. $\det ([ I]_{\beta ' \beta}) >0,$ where $[ I]_{\beta ' \beta}$ is the change of basis matrix from $\beta '$ to $\beta$ ). I need to prove that the basis $\beta$ and $ \gamma = \{ a_1+2 b_1, \ldots ,a_n +2 b_n \}$  have the same orientation. 
I began the problem doing this:
Computing $[I]_{\gamma \beta}$ I obtained that: 
$[I]_{\gamma \beta} = 2[ I]_{\beta ' \beta}+ I_n. $ I want to prove that $\det ( [I]_{\gamma \beta} )>0.$
Since $\beta, \beta '$ are both orthonormal basis, then $[ I]_{\beta ' \beta }$ is diagonalizable with eigenvalues $\pm 1$. 
I don't know if this is useful.
I'm stuck; can someone help me with this problem?
 A: I will answer your question in terms of eigenvalues. Let $A$ be an arbitrary $n\times n$ matrix with real entries. Then it may have complex eigenvalues. However, if $\lambda$ is a complex eigenvalue, then also $\overline\lambda$ is an eigenvalue of $A$ and the (algebraic) multiplicities of the two eigenvalues are equal. Since $\det A$ is the product of the eigenvalues of $A$ (counting multiplicities), $\det A$ it is negative if and only if $A$ is invertible and has an odd number of negative eigenvalues (counting multiplicities).
Let's consider the matrix $A = [I]_{\beta\beta'} + 2I_n$. Since $\beta$ and $\beta'$ are orthonormal bases, the matrix $[I]_{\beta\beta'}$ is orthogonal. Hence, its eigenvalues lie on the unit circle. The matrix $A = [I]_{\beta\beta'} + 2I_n$ has the eigenvalues $\lambda+2$, where $\lambda$ runs through the eigenvalues of $[I]_{\beta\beta'}$. Hence, its real eigenvalues are all $\ge 1$ and so $\det A > 0$. Now,
$$
[I]_{\gamma\beta} = 2[I]_{\beta'\beta} + I_n = [I]_{\beta'\beta}(2I_n + [I]_{\beta\beta'}) = [I]_{\beta'\beta}A.
$$
Hence, $\det [I]_{\gamma\beta} = \det [I]_{\beta'\beta}\det A = \det A > 0$.
