Let \(V\) be a real \(n\)-dimensional inner-product space. Two bases are said to have the same orientation when the corresponding change-of-basis matrix has positive determinant.
Suppose \((a_i)\) and \((b_i)\) are orthonormal bases with the same orientation. Prove that \[ (a_i+2b_i)_{i=1}^n \] is again a basis of \(V\), with the same orientation as \((a_i)\).
Solution
Let \(Q=(q_{ji})\) be the change-of-basis matrix determined by \[ b_i=\sum_{j=1}^n q_{ji}a_j. \]
1. The matrix \(Q\) is orthogonal and\[ \det Q=1. \]
Because \((a_i)\) is orthonormal, \[ \inner{b_i}{b_k} = \sum_{j=1}^n q_{ji}q_{jk}. \] Since \((b_i)\) is also orthonormal, this equals \(\delta_{ik}\). Therefore \[ Q^TQ=I. \] Thus \(Q\) is orthogonal, so \[ (\det Q)^2=1. \] The two bases have the same orientation, so their change-of-basis determinant is positive. Hence \(\det Q=1\).
2. Relative to the basis \((a_i)\), the vectors\[ c_i=a_i+2b_i \]are the columns of\[ C=I+2Q. \]
Using the definition of \(Q\), \[ \begin{aligned} c_i &= a_i+2b_i\\ &= \sum_{j=1}^n (\delta_{ji}+2q_{ji})a_j. \end{aligned} \] Thus the \(i\)th coordinate column of \(c_i\) is the \(i\)th column of \(I+2Q\).
3. Every complex eigenvalue \(\lambda\) of \(Q\) satisfies\[ \abs{\lambda}=1, \]and the nonreal eigenvalues occur in conjugate pairs.
If \(Qv=\lambda v\) for a nonzero \(v\in\CC^n\), then orthogonality of \(Q\) implies \[ \norm{Qv}=\norm{v}. \] Hence \[ \abs{\lambda}\norm{v} = \norm{\lambda v} = \norm{Qv} = \norm{v}, \] so \(\abs{\lambda}=1\). Since \(Q\) has real entries, its characteristic polynomial has real coefficients, and therefore every nonreal root occurs together with its complex conjugate.
4. The multiplicity of the eigenvalue \(-1\) of \(Q\) is even.
An orthogonal matrix is normal, hence diagonalizable over \(\CC\). Its determinant is the product of its eigenvalues, counted with multiplicity. By step 3, every nonreal conjugate pair \(\lambda,\overline\lambda\) contributes \[ \lambda\overline\lambda = \abs{\lambda}^2 =1 \] to that product. The eigenvalue \(1\) also contributes \(1\). If \(m\) is the multiplicity of \(-1\), step 1 therefore gives \[ 1=\det Q=(-1)^m. \] Thus \(m\) is even.
5. One has\[ \det(I+2Q)>0. \]
Because \(Q\) is diagonalizable over \(\CC\), the eigenvalues of \(I+2Q\) are \(1+2\lambda\), where \(\lambda\) ranges over the eigenvalues of \(Q\). For a nonreal conjugate pair \(\lambda,\overline\lambda\), the corresponding factors satisfy \[ (1+2\lambda)(1+2\overline\lambda) = \abs{1+2\lambda}^2 >0. \] An eigenvalue \(\lambda=1\) contributes the positive factor \(3\). An eigenvalue \(\lambda=-1\) contributes the factor \(-1\), and step 4 shows that the number of these factors is even. Hence their total product is positive. Therefore \[ \det(I+2Q)>0. \]
6. The vectors \((a_i+2b_i)_{i=1}^n\) form a basis with the sameorientation as \((a_i)\).
By step 2, their coordinate matrix relative to \((a_i)\) is \(I+2Q\). Step 5 gives \[ \det(I+2Q)>0. \] In particular, this matrix is invertible, so its columns form a basis. Its positive determinant says precisely that this basis has the same orientation as \((a_i)\).
7. Q.E.D.
Step 6 is the required conclusion.