Show that \(a_1,\ldots,a_n\in\mathbb{R}^m\) are linearly independent over \(\mathbb{C}\) iff they are linearly independent over \(\mathbb{R}\).
Show that if \(A\in M_n(\mathbb{R})\), then an eigenvalue \(\lambda\) of \(A\) is real iff it has a real corresponding eigenvector.
(Notation: \(M_n(\mathbb{R})\) denotes the set of \(n\times n\) real matrices.)
Solution
1. The vectors \(a_1,\ldots,a_n\in\mathbb R^m\) are linearly independent over \(\mathbb C\) if and only if they are linearly independent over \(\mathbb R\).
If they are linearly independent over \(\mathbb C\), then they are automatically linearly independent over the subfield \(\mathbb R\).
Conversely, suppose they are linearly independent over \(\mathbb R\) and \[ \sum_{j=1}^n z_j a_j=0 \] with \(z_j\in\mathbb C\). Write \[ z_j=x_j+iy_j, \qquad x_j,y_j\in\mathbb R. \] Because every \(a_j\) has real coordinates, \[ 0=\sum_j(x_j+iy_j)a_j =\sum_jx_ja_j+i\sum_jy_ja_j \] implies \[ \sum_jx_ja_j=0, \qquad \sum_jy_ja_j=0 \] in \(\mathbb R^m\). Real linear independence gives \(x_j=y_j=0\) for every \(j\), hence \(z_j=0\) for every \(j\). Thus the vectors are linearly independent over \(\mathbb C\).
2. Let \(A\in M_n(\mathbb R)\) and let \(0\ne v\in\mathbb R^n\) satisfy\[ Av=\lambda v \]for some \(\lambda\in\mathbb C\). Then \(\lambda\in\mathbb R\).
Since \(A\) and \(v\) are real, \(Av\in\mathbb R^n\). Choose an index \(j\) with \(v_j\ne0\). The \(j\)-th coordinate of the eigenvalue equation gives \[ (Av)_j=\lambda v_j. \] Both \((Av)_j\) and \(v_j\) are real and \(v_j\ne0\), so \[ \lambda=\frac{(Av)_j}{v_j}\in\mathbb R. \]
3. Conversely, if \(\lambda\in\mathbb R\) is an eigenvalue of \(A\), then \(A\) has a nonzero real eigenvector for \(\lambda\).
Choose a nonzero complex eigenvector \[ z=x+iy\in\mathbb C^n, \qquad x,y\in\mathbb R^n, \] with \[ Az=\lambda z. \] Since \(A\) and \(\lambda\) are real, \[ Ax+iAy=\lambda x+i\lambda y. \] Equality of real and imaginary parts gives \[ Ax=\lambda x, \qquad Ay=\lambda y. \] Because \(z\ne0\), at least one of \(x,y\) is nonzero. That nonzero vector lies in \(\mathbb R^n\) and is an eigenvector of \(A\) for \(\lambda\).
4. Therefore an eigenvalue of a real matrix is real if and only if it has a real corresponding eigenvector.
The two implications are 2 and 3.