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: Statement (I).
For a real matrix, the characteristic equation $\det(A-\lambda I)=0$ expands into a polynomial whose coefficients come only from sums and products of the real entries of $A$, so every coefficient is real. A standard algebra fact says that if a polynomial with real coefficients has a complex root $\lambda$, its conjugate $\bar\lambda$ is automatically a root too. Applying this to the characteristic polynomial proves statement (I) is TRUE for every real matrix.

Step 2: Statement (II), attack it through a $2\times2$ real block.
Let $v=u+iw$ with $u,w\in\mathbb{R}^n$, and $\lambda=x+iy$, $y\neq0$. Comparing real and imaginary parts of $Av=\lambda v$ gives $Au=xu-yw$ and $Aw=yu+xw$. This says the pair $(u,w)$ spans a 2-dimensional real subspace on which $A$ acts exactly like the $2\times2$ real matrix $\begin{pmatrix}x & y\\ -y & x\end{pmatrix}$.

Step 3: Use that this $2\times2$ matrix has no real eigenvector.
The matrix $\begin{pmatrix}x & y\\ -y & x\end{pmatrix}$ has eigenvalues $x\pm iy$, which are not real because $y\neq0$. If $u$ and $w$ were linearly dependent, say $w=cu$ with $u\neq0$, then $A$ restricted to that line would have to send $u$ to a real multiple of itself, meaning $u$ is a real eigenvector for a real eigenvalue, contradicting that the eigenvalues $x\pm iy$ are non-real.

Step 4: Handle the case where one of $u,w$ is exactly zero.
If $w=0$, the relation $Aw=yu+xw$ reduces to $0=yu$, and since $y\neq0$ this forces $u=0$ too, which cannot happen because $v=u+iw\neq0$. The symmetric argument rules out $u=0$. So $u,w$ are both nonzero and, by Step 3, cannot be proportional either.

Step 5: Conclude.
So $u=\text{Re}(v)$ and $w=\text{Im}(v)$ are linearly independent over $\mathbb{R}$, proving statement (II) TRUE as well. Both statements hold. \[ \boxed{\text{Both statements (I) and (II) are correct}} \]