Question

Let \(A\in M_n(\mathbb{R})\) be an \(n\times n\) real matrix, \(n\geq2\). Consider the following two statements:
(I) If \(\lambda\in\mathbb{C}\) is an eigenvalue of \(A\), then its complex conjugate \(\bar\lambda\) is also an eigenvalue.
(II) If \(v=(v_1,v_2,\ldots,v_n)\in\mathbb{C}^n\) is an eigenvector corresponding to eigenvalue \(\lambda=x+iy\), \(y\neq0\), then
\[ \text{Re}(v)=(\text{Re}(v_1),\ldots,\text{Re}(v_n))\quad\text{and}\quad\text{Im}(v)=(\text{Im}(v_1),\ldots,\text{Im}(v_n)) \]
are linearly independent vectors over \(\mathbb{R}\).

Which of the following statements is correct?

• A. Only statement (I) is correct
• B. Only statement (II) is correct
• C. Both statements (I) and (II) are correct
• D. Neither statement (I) nor statement (II) is correct

Hint:

Real-coefficient polynomials have roots in conjugate pairs, which settles statement (I); for statement (II), split \(Av=\lambda v\) into real and imaginary parts and check that \(\text{Re}(v)\) and \(\text{Im}(v)\) cannot be proportional when \(y\neq0\).

Answer 1

The Correct Option is C

Step 1: Check statement (I) using the characteristic polynomial.
The eigenvalues of \(A\) are the roots of the characteristic polynomial \(p(\lambda)=\det(A-\lambda I)\). Because every entry of \(A\) is real, every coefficient of \(p(\lambda)\) is also real.

Step 2: Use the fact that real-coefficient polynomials have conjugate root pairs.
If \(p(\lambda)=0\) for some \(\lambda\in\mathbb{C}\), take the complex conjugate of both sides. Since every coefficient is real, conjugation passes through each term unchanged, giving \(p(\bar\lambda)=\overline{p(\lambda)}=\overline{0}=0\). So \(\bar\lambda\) is also a root of \(p\), that is, also an eigenvalue of \(A\). Statement (I) is TRUE.

Step 3: Set up statement (II) in real and imaginary parts.
Write \(v=u+iw\), where \(u=\text{Re}(v)\) and \(w=\text{Im}(v)\) are real vectors, and \(\lambda=x+iy\) with \(y\neq0\). From \(Av=\lambda v\),
\[ A(u+iw)=(x+iy)(u+iw)=(xu-yw)+i(yu+xw). \]
Since \(A\) is real, \(Au\) and \(Aw\) are both real vectors, so matching real and imaginary parts on the two sides gives the pair of real equations
\[ Au=xu-yw \qquad\text{and}\qquad Aw=yu+xw. \tag{$\ast$} \]

Step 4: Suppose, for contradiction, that \(u\) and \(w\) are linearly dependent.
Since \(v\neq0\) (an eigenvector is never the zero vector), \(u\) and \(w\) are not both zero. First suppose \(w=0\) with \(u\neq0\). The second equation in \((\ast)\) gives \(0=Aw=yu+x\cdot0=yu\). Since \(y\neq0\), this forces \(u=0\), contradicting \(u\neq0\). Now suppose \(u=0\) with \(w\neq0\). The first equation gives \(0=Au=x\cdot0-yw=-yw\), and since \(y\neq0\) this forces \(w=0\), a contradiction. So neither vector alone can be zero.

Step 5: Rule out a nonzero proportional relation.
So assume \(u\neq0\), \(w\neq0\), and \(w=cu\) for some nonzero real number \(c\). Substitute into the second equation of \((\ast)\) and also use the first:
\[ Aw=A(cu)=c\,Au=c(xu-yw)=c(xu-y(cu))=(cx-c^2y)u. \]
But also directly from \((\ast)\), \(Aw=yu+xw=yu+x(cu)=(y+cx)u\). Since \(u\neq0\), equate the two expressions for \(Aw\):
\[ cx-c^2y=y+cx\ \Rightarrow\ -c^2y=y\ \Rightarrow\ y(c^2+1)=0. \]
Since \(y\neq0\), this needs \(c^2=-1\), impossible for a real number \(c\).

Step 6: Conclude statement (II).
Every possible way \(u\) and \(w\) could be linearly dependent leads to a contradiction. So \(u=\text{Re}(v)\) and \(w=\text{Im}(v)\) must be linearly independent over \(\mathbb{R}\). Statement (II) is TRUE.

Final Answer:
Both statement (I) and statement (II) are correct. \[ \boxed{\text{Both statements (I) and (II) are correct}} \]