Complex Eigenvalues & Real Solutions - Linear Algebra

A $3 \times 3$ real matrix $A$ must have at least one real eigenvalue. Which of the following sets could NOT be the complete set of eigenvalues for such a matrix $A$ ?
Answer: $\{5, 1+2i, -1-2i\}$ The characteristic polynomial of a real $n \times n$ matrix has real coefficients. As a result, any complex roots (eigenvalues) must occur in conjugate pairs. That is, if $a+bi$ is an eigenvalue, then $a-bi$ must also be an eigenvalue. In choice D, the set contains the eigenvalue $1+2i$ . Its conjugate is $1-2i$ , which is not in the set. The value $-1-2i$ is the negative of $1+2i$ , not its conjugate. Therefore, this set cannot be the complete set of eigenvalues for a real matrix. The other sets are all valid: A has one real root and a conjugate pair, B has three real roots, and C has one real root (0) and a pure imaginary conjugate pair.
Let $A$ be a real $n \times n$ matrix. If $\lambda = 5 - 2i$ is an eigenvalue of $A$ with a corresponding eigenvector $\mathbf{v}$ , which of the following statements is necessarily true?
Answer: The matrix $A$ has an eigenvector $\bar{\mathbf{v}}$ corresponding to the eigenvalue $5 + 2i$ . A fundamental theorem of linear algebra states that if a real matrix $A$ has a complex eigenvalue $\lambda = a+bi$ , then its complex conjugate $\bar{\lambda} = a-bi$ is also an eigenvalue. Furthermore, if $\mathbf{v}$ is an eigenvector for $\lambda$ , then its complex conjugate $\bar{\mathbf{v}}$ is an eigenvector for $\bar{\lambda}$ . Given $\lambda = 5 - 2i$ , its conjugate is $\bar{\lambda} = 5 + 2i$ . Therefore, $\bar{\mathbf{v}}$ must be an eigenvector corresponding to $5 + 2i$ . Choice A is incorrect; neither the real nor imaginary part of a complex eigenvector is generally an eigenvector itself. Choice C is incorrect; the real part of the eigenvalue is $+5$ , which leads to solutions growing, not decaying. Choice D is incorrect; we only know of one eigenvector, so the geometric multiplicity is at least 1.
Consider the linear system $\mathbf{x}' = A\mathbf{x}$ where $A$ has eigenvalues $\lambda_1 = 1 + 2i$ and $\lambda_2 = 1 - 2i$ with corresponding eigenvector $\mathbf{v}_1 = \begin{pmatrix} 1 \\ i \end{pmatrix}$ . What is the general real solution to this system?
Answer: $\mathbf{x}(t) = c_1 e^t \begin{pmatrix} \cos(2t) \\ -\sin(2t) \end{pmatrix} + c_2 e^t \begin{pmatrix} \sin(2t) \\ \cos(2t) \end{pmatrix}$ For complex eigenvalue $\lambda = 1 + 2i$ with eigenvector $\begin{pmatrix} 1 \\ i \end{pmatrix}$ , we have $e^{\lambda t}\mathbf{v}_1 = e^{(1+2i)t}\begin{pmatrix} 1 \\ i \end{pmatrix} = e^t(\cos(2t) + i\sin(2t))\begin{pmatrix} 1 \\ i \end{pmatrix} = e^t\begin{pmatrix} \cos(2t) + i\sin(2t) \\ i\cos(2t) - \sin(2t) \end{pmatrix}$ . The real and imaginary parts give us the fundamental solutions: $e^t\begin{pmatrix} \cos(2t) \\ -\sin(2t) \end{pmatrix}$ and $e^t\begin{pmatrix} \sin(2t) \\ \cos(2t) \end{pmatrix}$ . Choice B has wrong signs. Choice C has wrong exponential sign. Choice D misapplies the trigonometric decomposition.
A real $2 \times 2$ matrix $A$ has trace $\text{tr}(A) = -4$ . For what range of values of the determinant, $\det(A)$ , will solutions to the system $\mathbf{x}'=A\mathbf{x}$ spiral into the origin (a spiral sink)?
Answer: $\det(A) > 4$ For a system to be a spiral sink, two conditions must be met. First, the eigenvalues must be complex conjugates, which means the discriminant of the characteristic equation must be negative: $(\text{tr}(A))^2 - 4\det(A) < 0$ . Second, the real part of the eigenvalues must be negative. The real part of the eigenvalues is given by $\text{Re}(\lambda) = \frac{\text{tr}(A)}{2}$ . Given $\text{tr}(A) = -4$ , the real part is $-4/2 = -2$ , which is negative. So the second condition is automatically satisfied. We only need to satisfy the first condition: $(-4)^2 - 4\det(A) < 0$ , which simplifies to $16 - 4\det(A) < 0$ . Adding $4\det(A)$ to both sides gives $16 < 4\det(A)$ . Dividing by 4 yields $\det(A) > 4$ .
Given that $\begin{pmatrix} 1 + 2i \\ 3 - i \end{pmatrix}$ is an eigenvector of real matrix $C$ corresponding to eigenvalue $\mu = 2 - i$ , what is the trace of $C$ ?
Answer: $4$ Since $C$ is real and has complex eigenvalue $\mu_1 = 2 - i$ , its complex conjugate $\mu_2 = 2 + i$ must also be an eigenvalue. For a $2 \times 2$ real matrix with complex eigenvalues, the trace equals the sum of the eigenvalues: $\text{tr}(C) = \mu_1 + \mu_2 = (2 - i) + (2 + i) = 4$ . This follows from the fact that complex eigenvalues of real matrices come in conjugate pairs, and the trace is always real (sum of diagonal entries). Choices B, C, and D represent common errors in computing eigenvalue sums or incorrectly handling the complex parts.
A real $2 \times 2$ system $\mathbf{x}'=A\mathbf{x}$ has a complex eigenvalue $\lambda=4i$ with eigenvector $\mathbf{v}=\begin{pmatrix} 1 \\ 2i \end{pmatrix}$ . One real solution is $\mathbf{x}_1(t) = \begin{pmatrix} \cos(4t) \\ -2\sin(4t) \end{pmatrix}$ . Which of the following is a second, linearly independent real solution $\mathbf{x}_2(t)$ ?
Answer: $\begin{pmatrix} \sin(4t) \\ 2\cos(4t) \end{pmatrix}$ The complex solution is $\mathbf{z}(t) = e^{4it}\mathbf{v} = (\cos(4t)+i\sin(4t))(\begin{pmatrix} 1 \\ 0 \end{pmatrix} + i\begin{pmatrix} 0 \\ 2 \end{pmatrix})$ . Expanding this gives $\mathbf{z}(t) = [\cos(4t)\begin{pmatrix} 1 \\ 0 \end{pmatrix} - \sin(4t)\begin{pmatrix} 0 \\ 2 \end{pmatrix}] + i[\sin(4t)\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \cos(4t)\begin{pmatrix} 0 \\ 2 \end{pmatrix}]$ . The real part is $\text{Re}(\mathbf{z}(t)) = \begin{pmatrix} \cos(4t) \\ -2\sin(4t) \end{pmatrix}$ , which is the given solution $\mathbf{x}_1(t)$ . The imaginary part is $\text{Im}(\mathbf{z}(t)) = \begin{pmatrix} \sin(4t) \\ 2\cos(4t) \end{pmatrix}$ , which is a second, linearly independent real solution $\mathbf{x}_2(t)$ . Choice A is just $-\mathbf{x}_1(t)$ and is not linearly independent. Choices C and D have sign errors.
A real $2 \times 2$ matrix $A$ has eigenvalues $\lambda_1 = e^{i\pi/3}$ and $\lambda_2 = e^{-i\pi/3}$ . Which matrix is equal to $A^6$ ?
Answer: The identity matrix, $I$ . If a matrix $A$ has eigenvalues $\lambda_1, \dots, \lambda_n$ , then the matrix $A^k$ has eigenvalues $\lambda_1^k, \dots, \lambda_n^k$ . The eigenvalues of $A^6$ are $(\lambda_1)^6$ and $(\lambda_2)^6$ . Using de Moivre's formula, $(\lambda_1)^6 = (e^{i\pi/3})^6 = e^{i(6\pi/3)} = e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1$ . Similarly, $(\lambda_2)^6 = (e^{-i\pi/3})^6 = e^{-i2\pi} = \cos(-2\pi) + i\sin(-2\pi) = 1$ . So, both eigenvalues of $A^6$ are 1. If $A$ is diagonalizable, $A = PDP^{-1}$ , then $A^6 = PD^6P^{-1}$ . Since $D^6 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$ , it follows that $A^6 = PIP^{-1} = I$ . This holds even if $A$ is not diagonalizable. The only matrix with all eigenvalues equal to 1 that could result from this process is the identity matrix.
The characteristic polynomial of a real $2 \times 2$ matrix $A$ is $p(\lambda) = \lambda^2 - 2\lambda + 10$ . What is the long-term behavior of trajectories of the system $\mathbf{x}' = A\mathbf{x}$ near the origin?
Answer: Trajectories move away from the origin in a spiral pattern (spiral source). The eigenvalues of matrix $A$ are the roots of its characteristic polynomial. Using the quadratic formula, $\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)} = \frac{2 \pm \sqrt{4 - 40}}{2} = \frac{2 \pm \sqrt{-36}}{2} = \frac{2 \pm 6i}{2} = 1 \pm 3i$ . The eigenvalues are a complex conjugate pair $\lambda = a \pm bi$ with $a=1$ and $b=3$ . Since $b \neq 0$ , the trajectories involve rotation. Since the real part $a=1$ is positive, the trajectories spiral away from the origin. This behavior is known as a spiral source.
The phase portrait for a $2 \times 2$ system $\mathbf{x}' = A\mathbf{x}$ shows trajectories that are concentric ellipses centered at the origin. Which of the following conditions on the trace and determinant of $A$ must be true?
Answer: $\text{tr}(A) = 0$ and $\det(A) > 0$ Trajectories that are concentric ellipses correspond to a center. This behavior occurs when the eigenvalues of the matrix $A$ are purely imaginary, i.e., $\lambda = \pm bi$ for some real $b \neq 0$ . The sum of the eigenvalues is the trace, so $\text{tr}(A) = bi + (-bi) = 0$ . The product of the eigenvalues is the determinant, so $\det(A) = (bi)(-bi) = -b^2i^2 = b^2$ . Since $b \neq 0$ , the determinant must be positive. Thus, the conditions are $\text{tr}(A) = 0$ and $\det(A) > 0$ .
The general real solution to a $2 \times 2$ linear system $\mathbf{x}' = A\mathbf{x}$ is given by $\mathbf{x}(t) = c_1 e^{3t} \begin{pmatrix} \cos(4t) \\ -2\sin(4t) \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} \sin(4t) \\ 2\cos(4t) \end{pmatrix}$ . What are the eigenvalues of the matrix $A$ ?
Answer: $3 \pm 4i$ The form of the real solutions to a system with complex eigenvalues $\lambda = a \pm bi$ is $\mathbf{x}(t) = e^{at} (c_1 (\dots \cos(bt) \dots) + c_2 (\dots \sin(bt) \dots))$ . By inspecting the given solution, we can identify the exponential term $e^{3t}$ , which implies the real part of the eigenvalue is $a=3$ . The trigonometric terms $\cos(4t)$ and $\sin(4t)$ imply the imaginary part of the eigenvalue is $b=4$ . Therefore, the eigenvalues of matrix $A$ are $3 \pm 4i$ .
For a real $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ , the system $\mathbf{x}' = A\mathbf{x}$ has solutions that exhibit rotational behavior (either as a center or a spiral). This occurs if and only if the eigenvalues of $A$ are complex. Which of the following inequalities ensures this?
Answer: $(a+d)^2 - 4(ad-bc) < 0$ The eigenvalues of $A$ are the roots of the characteristic equation $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$ . For a $2 \times 2$ matrix, $\text{tr}(A) = a+d$ and $\det(A) = ad-bc$ . The eigenvalues are complex if and only if the discriminant of this quadratic equation is negative. The discriminant is $B^2 - 4AC = (-\text{tr}(A))^2 - 4(1)(\det(A)) = (a+d)^2 - 4(ad-bc)$ . Therefore, the condition for complex eigenvalues is $(a+d)^2 - 4(ad-bc) < 0$ .
A $3 \times 3$ real matrix has eigenvalues $\lambda_1 = 2$ , $\lambda_2 = -1 + 3i$ , and $\lambda_3 = -1 - 3i$ . If the matrix represents a dynamical system $\mathbf{x}' = A\mathbf{x}$ , which statement best describes the long-term behavior of solutions?
Answer: Solutions grow exponentially and oscillate due to the positive real eigenvalue dominating the complex eigenvalues The real eigenvalue $\lambda_1 = 2 > 0$ dominates the complex eigenvalues $\lambda_{2,3} = -1 \pm 3i$ which have negative real part $-1$ . Since $2 > |-1|$ , the positive real eigenvalue determines the long-term behavior. Solutions will grow exponentially in the direction of the eigenvector corresponding to $\lambda_1 = 2$ , while also exhibiting oscillatory behavior from the complex eigenvalues. Choice A incorrectly focuses on complex eigenvalues. Choice C is wrong because real parts are not zero. Choice D describes saddle points which occur when eigenvalues have mixed signs in purely real cases.
The solution to a system $\mathbf{x}' = A\mathbf{x}$ can be written as $\mathbf{x}(t) = e^{At}\mathbf{x}(0)$ . If one of the entries in the matrix $e^{At}$ contains the term $e^{5t}\sin(2t)$ , what can be concluded about the eigenvalues of the real matrix $A$ ?
Answer: The eigenvalues of $A$ are a complex conjugate pair $5 \pm 2i$ . When you encounter a differential equation system $\mathbf{x}' = A\mathbf{x}$ with solution $\mathbf{x}(t) = e^{At}\mathbf{x}(0)$ , the key insight is that the entries of the matrix exponential $e^{At}$ directly reflect the eigenvalues of matrix $A$ . The term $e^{5t}\sin(2t)$ tells us everything we need to know about the eigenvalues. This expression has the characteristic form that arises from complex eigenvalues: $e^{\alpha t}\sin(\beta t)$ , where $\alpha$ is the real part and $\beta$ is the imaginary part of the eigenvalue. Since we see $e^{5t}\sin(2t)$ , we can identify $\alpha = 5$ and $\beta = 2$ , meaning the eigenvalues are $5 \pm 2i$ . Since $A$ is real, complex eigenvalues must occur in conjugate pairs. Looking at the wrong answers: Choice A incorrectly assumes the eigenvalues are real—but real eigenvalues produce exponential terms like $e^{5t}$ and $e^{2t}$ , never trigonometric functions. Choice B reverses the roles of the real and imaginary parts; $2 \pm 5i$ would generate $e^{2t}\sin(5t)$ or $e^{2t}\cos(5t)$ terms instead. Choice C suggests purely imaginary eigenvalues $\pm 2i$ , which would produce terms like $\sin(2t)$ and $\cos(2t)$ without the exponential growth factor $e^{5t}$ . The answer is D: the eigenvalues are $5 \pm 2i$ . Remember this pattern: in $e^{At}$ , terms like $e^{\alpha t}\sin(\beta t)$ always signal complex eigenvalues $\alpha \pm \beta i$ .
A real $2 \times 2$ matrix $A$ has an eigenvalue $\lambda = 1 - 3i$ with eigenvector $\mathbf{v} = \begin{pmatrix} 2+i \\ 5 \end{pmatrix}$ . Let $\mathbf{u} = \text{Re}(\mathbf{v})$ and $\mathbf{w} = \text{Im}(\mathbf{v})$ . The set $S = \text{span}\{\mathbf{u}, \mathbf{w}\}$ forms a plane in $\mathbb{R}^2$ . What is the primary significance of this plane with respect to the matrix $A$ ?
Answer: The plane $S$ is an invariant subspace under the transformation $A$ . When you encounter complex eigenvalues of real matrices, you're working with a fundamental concept: real matrices with complex eigenvalues create rotational transformations, and certain real subspaces remain invariant under these transformations. Since $A$ is real but has complex eigenvalue $\lambda = 1 - 3i$ , its complex conjugate $\bar{\lambda} = 1 + 3i$ is also an eigenvalue. The key insight is that while the eigenvectors are complex, we can extract real vectors that span an invariant subspace. Given $\mathbf{v} = \begin{pmatrix} 2+i \\ 5 \end{pmatrix}$ , we have $\mathbf{u} = \text{Re}(\mathbf{v}) = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and $\mathbf{w} = \text{Im}(\mathbf{v}) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ . The plane $S = \text{span}\{\mathbf{u}, \mathbf{w}\}$ is invariant under $A$ , meaning $A$ maps any vector in $S$ to another vector in $S$ . This is answer D. Answer A is wrong because $S$ contains real vectors, but the actual eigenvectors are complex. Answer B is incorrect because $A - I$ would have eigenvalue $-3i \neq 0$ , so its null space is trivial. Answer C misunderstands the geometry—both eigenspaces correspond to the same invariant plane when projected to real space. Study tip: For real matrices with complex eigenvalues, always remember that the real and imaginary parts of any complex eigenvector span a 2D invariant subspace. This pattern appears frequently in applications involving rotations and oscillations.
Let $A$ be a real $2 \times 2$ matrix. A complex eigenvalue and corresponding eigenvector for the system $\mathbf{x}' = A\mathbf{x}$ are $\lambda = -2 + i$ and $\mathbf{v} = \begin{pmatrix} 1 \\ 1-i \end{pmatrix}$ . Which of the following is a valid real-valued solution $\mathbf{x}(t)$ to the system?
Answer: $e^{-2t} \left( \cos(t)\begin{pmatrix} 1 \\ 1 \end{pmatrix} - \sin(t)\begin{pmatrix} 0 \\ -1 \end{pmatrix} \right)$ A complex solution is $\mathbf{z}(t) = e^{\lambda t}\mathbf{v}$ . The real and imaginary parts of $\mathbf{z}(t)$ form two linearly independent real solutions. Let $\mathbf{v} = \text{Re}(\mathbf{v}) + i\text{Im}(\mathbf{v})$ , where $\text{Re}(\mathbf{v}) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\text{Im}(\mathbf{v}) = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$ . Using Euler's formula, $e^{(-2+i)t} = e^{-2t}(\cos(t) + i\sin(t))$ . Then $\mathbf{z}(t) = e^{-2t}(\cos(t) + i\sin(t))(\text{Re}(\mathbf{v}) + i\text{Im}(\mathbf{v}))$ . The real part of this expression is $\text{Re}(\mathbf{z}(t)) = e^{-2t}(\cos(t)\text{Re}(\mathbf{v}) - \sin(t)\text{Im}(\mathbf{v}))$ . Substituting the vectors gives the solution in choice A. Choice B has a sign error. Choice C incorrectly swaps the real and imaginary parts of $\lambda$ . Choice D swaps the real and imaginary parts of $\mathbf{v}$ .

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