Linear Algebra Quiz: Complex Eigenvalues And Real Solutions

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Complex Eigenvalues And Real SolutionsQuestion 1 of 15

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?

(a+d)^2 - 4(ad-bc) < 0

(a+d)^2 - 4(ad-bc) > 0

ad-bc = 0

(a-d)^2 + 4bc > 0

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Linear Algebra Quiz

Linear Algebra Quiz: Complex Eigenvalues And Real Solutions

Practice Complex Eigenvalues And Real Solutions in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Complex Eigenvalues And Real Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$(a+d)^2 - 4(ad-bc) < 0$ (correct answer)
$(a+d)^2 - 4(ad-bc) > 0$
$ad-bc = 0$
$(a-d)^2 + 4bc > 0$

Explanation: 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

Question 2

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?

$e^{-2t} \left( \cos(t)\begin{pmatrix} 1 \\ 1 \end{pmatrix} - \sin(t)\begin{pmatrix} 0 \\ -1 \end{pmatrix} \right)$ (correct answer)
$e^{-2t} \left( \cos(t)\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \sin(t)\begin{pmatrix} 0 \\ -1 \end{pmatrix} \right)$
$e^{t} \left( \cos(-2t)\begin{pmatrix} 1 \\ 1 \end{pmatrix} - \sin(-2t)\begin{pmatrix} 0 \\ -1 \end{pmatrix} \right)$
$e^{-2t} \left( \cos(t)\begin{pmatrix} 0 \\ -1 \end{pmatrix} - \sin(t)\begin{pmatrix} 1 \\ 1 \end{pmatrix} \right)$

Explanation: 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}

Question 3

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 $A^6$ ?

The zero matrix, $O$ .
The identity matrix, $I$ . (correct answer)
The negative identity matrix, $-I$ .
The original matrix, $A$ .

Explanation: 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.

Question 4

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)?

$\det(A) < 4$
$\det(A) = 4$
$\det(A) > 4$ (correct answer)
$\det(A) < 0$

Explanation: 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

Question 5

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?

$\text{tr}(A) = 0$ and $\det(A) > 0$ (correct answer)
$\text{tr}(A) \neq 0$ and $\det(A) > 0$
$\text{tr}(A) < 0$ and $\det(A) = 0$
$\text{tr}(A) = 0$ and $\det(A) < 0$

Explanation: 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

Question 6

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 $A$ ?

$3$ and $4$
$4 \pm 3i$
$3 \pm 4i$ (correct answer)
$-3 \pm 4i$

Explanation: The form of the real solutions to a system with complex eigenvalues

\lambda = a \pm bi

\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

Question 7

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?

The vector $\text{Re}(\mathbf{v})$ is an eigenvector of $A$ with eigenvalue $5$ .
The matrix $A$ has an eigenvector $\bar{\mathbf{v}}$ corresponding to the eigenvalue $5 + 2i$ . (correct answer)
The real part of any solution to $\mathbf{x}'=A\mathbf{x}$ must decay to zero as $t \to \infty$ .
The geometric multiplicity of the eigenvalue $\lambda = 5 - 2i$ must be at least 2.

Explanation: 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.

Question 8

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 $A$ ?

The eigenvalues of $A$ are real and are $5$ and $2$ .
The eigenvalues of $A$ are a complex conjugate pair $2 \pm 5i$ .
The eigenvalues of $A$ are purely imaginary, $\pm 2i$ .
The eigenvalues of $A$ are a complex conjugate pair $5 \pm 2i$ . (correct answer)

Explanation: 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)

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

Question 9

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?

Trajectories form a stable center, orbiting the origin in closed ellipses.
Trajectories approach the origin in a spiral pattern (spiral sink).
Trajectories move away from the origin in a spiral pattern (spiral source). (correct answer)
Trajectories approach the origin along one direction and move away along another (saddle point).

Explanation: 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.

Question 10

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 $\mathbf{x}_2(t)$ ?

$\begin{pmatrix} -\cos(4t) \\ 2\sin(4t) \end{pmatrix}$
$\begin{pmatrix} \sin(4t) \\ 2\cos(4t) \end{pmatrix}$ (correct answer)
$\begin{pmatrix} \cos(4t) \\ 2\sin(4t) \end{pmatrix}$
$\begin{pmatrix} \sin(4t) \\ -2\cos(4t) \end{pmatrix}$

Explanation: 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.

Question 11

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 $A$ ?

The plane $S$ contains only eigenvectors of $A$ .
The plane $S$ is the null space of the matrix $A - I$ .
The plane $S$ is orthogonal to the eigenspace of the eigenvalue $\bar{\lambda} = 1 + 3i$ .
The plane $S$ is an invariant subspace under the transformation $A$ . (correct answer)

Explanation: 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.

Question 12

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 $C$ ?

$4$ (correct answer)
$2$
$6$
$8$

Explanation: 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.

Question 13

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}

(correct answer)

Explanation: 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}

Question 14

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?

All solutions approach the origin along spiraling trajectories due to the dominant complex eigenvalues with negative real parts
Solutions grow exponentially and oscillate due to the positive real eigenvalue dominating the complex eigenvalues (correct answer)
Solutions approach a limit cycle because the complex eigenvalues have zero real parts and create periodic motion
The system exhibits saddle behavior with some solutions growing and others decaying depending on initial conditions

Explanation: 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.

Question 15

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 $A$ ?

$\{2, 3+i, 3-i\}$
$\{4, 4, 1\}$
$\{0, i, -i\}$
$\{5, 1+2i, -1-2i\}$ (correct answer)

Explanation: 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.