Linear Algebra Quiz: Algebraic Vs Geometric Multiplicity

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Algebraic Vs Geometric MultiplicityQuestion 1 of 20

A $5 \times 5$ matrix $A$ has the characteristic polynomial $p(\lambda) = (\lambda - 4)^3 (\lambda + 2)^2$ . Let $m_a(\lambda)$ denote the algebraic multiplicity and $m_g(\lambda)$ denote the geometric multiplicity of an eigenvalue $\lambda$ . Which of the following is a possible set of geometric multiplicities for the eigenvalues of $A$ ?

m_g(4) = 3

and

m_g(-2) = 3

m_g(4) = 4

and

m_g(-2) = 2

m_g(4) = 2

and

m_g(-2) = 1

m_g(4) = 3

and

m_g(-2) = 0

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Linear Algebra Quiz

Linear Algebra Quiz: Algebraic Vs Geometric Multiplicity

Practice Algebraic Vs Geometric Multiplicity 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 Algebraic Vs Geometric Multiplicity, 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

$m_g(4) = 3$ and $m_g(-2) = 3$
$m_g(4) = 4$ and $m_g(-2) = 2$
$m_g(4) = 2$ and $m_g(-2) = 1$ (correct answer)
$m_g(4) = 3$ and $m_g(-2) = 0$

Explanation: The algebraic multiplicity (

m_a

) of an eigenvalue is its power in the characteristic polynomial. Here,

m_a(4) = 3

and

m_a(-2) = 2

. The geometric multiplicity (

m_g

) must satisfy the inequality

1 \le m_g(\lambda) \le m_a(\lambda)

for every eigenvalue

\lambda

. We check the options: (A)

m_g(-2)=3

violates

m_g(-2) \le m_a(-2)=2

. (B)

m_g(4)=4

violates

m_g(4) \le m_a(4)=3

. (D)

m_g(-2)=0

violates

m_g(-2) \ge 1

. (C) is the only option where

1 \le m_g(4) \le 3

and

1 \le m_g(-2) \le 2

are both satisfied. This scenario corresponds to a non-diagonalizable matrix.

Question 2

Let $A$ be a $5 \times 5$ matrix with an eigenvalue $\lambda = -1$ whose algebraic multiplicity is 3. The reduced row echelon form of the matrix $A + I$ is given by: $\mathrm{rref}(A+I) = \begin{pmatrix} 1 & 0 & 3 & 0 & 5 \\ 0 & 1 & -2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$ What can be concluded about the eigenvalue $\lambda = -1$ and the diagonalizability of $A$ ?

The geometric multiplicity is 2, and the matrix $A$ is not diagonalizable. (correct answer)
The geometric multiplicity is 3, and the matrix $A$ is diagonalizable.
The geometric multiplicity is 2, but more information is needed to determine if $A$ is diagonalizable.
The geometric multiplicity is 3, and the matrix $A$ is not diagonalizable.

Explanation: The geometric multiplicity of

\lambda = -1

is the dimension of the null space of

A - (-1)I = A + I

. The dimension of the null space is equal to the number of free variables in the system

(A+I)\vec{x} = \vec{0}

. The RREF of

A+I

has 3 pivot columns (columns 1, 2, and 4). Since the matrix is

5 \times 5

, the number of free variables is

5 - (\text{number of pivots}) = 5 - 3 = 2

. Thus, the geometric multiplicity is 2. We are given that the algebraic multiplicity is 3. Since the geometric multiplicity (2) is less than the algebraic multiplicity (3) for this eigenvalue, the matrix

A

is not diagonalizable.

Question 3

A $4 \times 4$ matrix $A$ is known to be diagonalizable. It has two distinct eigenvalues, $\lambda_1 = 2$ and $\lambda_2 = 5$ . The eigenspace for $\lambda_1=2$ is spanned by the vectors $\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$ . What is the algebraic multiplicity of the eigenvalue $\lambda_2=5$ ?

1
2 (correct answer)
3
4

Explanation: The eigenspace for

\lambda_1=2

is spanned by two linearly independent vectors, so its dimension, the geometric multiplicity of

\lambda_1=2

, is

m_g(2)=2

. Since the matrix

A

is diagonalizable, the algebraic multiplicity must equal the geometric multiplicity for every eigenvalue. Thus, the algebraic multiplicity of

\lambda_1=2

m_a(2)=2

. The sum of the algebraic multiplicities of all eigenvalues must equal the size of the matrix, which is 4. Therefore,

m_a(2) + m_a(5) = 4

. Substituting the value we found, we get

2 + m_a(5) = 4

, which implies

m_a(5) = 2

Question 4

A $6 \times 6$ matrix has the characteristic polynomial $p(\lambda) = (\lambda-1)^4(\lambda-3)^2$ . Which of the following values CANNOT be the sum of the geometric multiplicities of its eigenvalues?

2
5
6
7 (correct answer)

Explanation: The eigenvalues are

\lambda_1=1

with algebraic multiplicity

m_a(1)=4

, and

\lambda_2=3

with algebraic multiplicity

m_a(3)=2

. The geometric multiplicity

m_g(\lambda)

must satisfy

1 \le m_g(\lambda) \le m_a(\lambda)

. Thus,

1 \le m_g(1) \le 4

and

1 \le m_g(3) \le 2

. The sum of the geometric multiplicities is

S = m_g(1) + m_g(3)

. The minimum possible sum is

1+1=2

. The maximum possible sum is

4+2=6

. Any integer value between 2 and 6 is a possible sum. For example, a sum of 5 is possible if

m_g(1)=3

and

m_g(3)=2

. A sum of 2 is possible if

m_g(1)=1

and

m_g(3)=1

. A sum of 6 is possible if the matrix is diagonalizable. However, a sum of 7 is impossible because the sum cannot exceed the maximum possible value of 6.

Question 5

A $3 \times 3$ matrix $A$ has the characteristic polynomial $p(\lambda) = -(\lambda - 5)^2(\lambda + 1)$ . The eigenspace corresponding to the eigenvalue $\lambda=5$ is the plane in $\mathbb{R}^3$ defined by the equation $x_1 + 2x_2 - 3x_3 = 0$ . Which conclusion can be drawn about matrix $A$ ?

The matrix $A$ is not diagonalizable.
The geometric multiplicity of $\lambda=5$ is 1.
The matrix $A$ is diagonalizable. (correct answer)
More information is needed to determine if $A$ is diagonalizable.

Explanation: From the characteristic polynomial, the eigenvalues are

\lambda_1=5

with algebraic multiplicity

m_a(5)=2

, and

\lambda_2=-1

with algebraic multiplicity

m_a(-1)=1

. The eigenspace for

\lambda=5

is a plane in

\mathbb{R}^3

. A plane is a two-dimensional subspace, so the dimension of this eigenspace, which is the geometric multiplicity, is

m_g(5)=2

. For the eigenvalue

\lambda=-1

, its algebraic multiplicity is 1, which forces its geometric multiplicity to also be 1 (since

1 \le m_g(-1) \le 1

). Since the geometric multiplicity equals the algebraic multiplicity for all eigenvalues (

m_g(5)=m_a(5)=2

and

m_g(-1)=m_a(-1)=1

), the matrix

A

is diagonalizable.

Question 6

A matrix $A \in \mathbb{R}^{4 \times 4}$ has exactly one distinct eigenvalue, $\lambda = 7$ , which has an algebraic multiplicity of 4. Let $m_g(7)$ be its geometric multiplicity. Which statement about $A$ must be true?

The matrix $A$ is not diagonalizable.
The geometric multiplicity $m_g(7)$ must be 1.
The matrix $A$ is diagonalizable if and only if $m_g(7) = 4$ . (correct answer)
$A$ must be the matrix $7I_4$ , where $I_4$ is the $4 \times 4$ identity matrix.

Explanation: A matrix is diagonalizable if and only if the geometric multiplicity equals the algebraic multiplicity for all eigenvalues. In this case, there is only one eigenvalue,

\lambda = 7

, with algebraic multiplicity 4. Therefore, the matrix

A

is diagonalizable if and only if the geometric multiplicity of this eigenvalue,

m_g(7)

, is also 4. It is possible that

m_g(7) < 4

, in which case

A

would not be diagonalizable, so (A) is not necessarily true.

m_g(7)

can be any integer from 1 to 4, so (B) is not necessarily true. While

7I_4

is an example of such a diagonalizable matrix, other matrices (like a diagonal matrix with three 7s and one 7) also fit the description, so (D) is not a necessary condition.

Question 7

Consider the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ representing a horizontal shear, with standard matrix $A = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ for some nonzero constant $k$ . Which statement accurately describes the algebraic and geometric multiplicities for this matrix?

The matrix has one eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. (correct answer)
The matrix has one eigenvalue with algebraic multiplicity 2 and geometric multiplicity 2.
The matrix has two distinct eigenvalues, each with algebraic and geometric multiplicity 1.
The matrix has one eigenvalue with algebraic multiplicity 1 and geometric multiplicity 1.

Explanation: The characteristic equation is

\det(A - \lambda I) = \det \begin{pmatrix} 1-\lambda & k \\ 0 & 1-\lambda \end{pmatrix} = (1-\lambda)^2 = 0

. This gives a single eigenvalue

\lambda=1

with algebraic multiplicity 2. To find the geometric multiplicity, we find the nullity of

A - 1I = \begin{pmatrix} 0 & k \\ 0 & 0 \end{pmatrix}

. Since

k

is nonzero, this matrix has rank 1. The nullity is the number of columns minus the rank, which is

2 - 1 = 1

. Therefore, the geometric multiplicity is 1. Since the algebraic multiplicity (2) is greater than the geometric multiplicity (1), this matrix is not diagonalizable.

Question 8

Let $A$ be a $4 \times 4$ matrix with characteristic polynomial $p(\lambda) = (\lambda - 2)^3(\lambda + 1)$ . If the eigenspace corresponding to eigenvalue $\lambda = 2$ has dimension 1, what can be concluded about the diagonalizability of $A$ ?

$A$ is diagonalizable because the sum of geometric multiplicities equals 4
$A$ is diagonalizable because all eigenvalues are real and the matrix is $4 \times 4$
$A$ may or may not be diagonalizable depending on the geometric multiplicity of $\lambda = -1$
$A$ is not diagonalizable because the geometric multiplicity of $\lambda = 2$ is less than its algebraic multiplicity (correct answer)

Explanation: When you encounter questions about diagonalizability, the key concept to remember is that a matrix is diagonalizable if and only if the geometric multiplicity equals the algebraic multiplicity for every eigenvalue. From the characteristic polynomial

p(\lambda) = (\lambda - 2)^3(\lambda + 1)

, you can identify the eigenvalues and their algebraic multiplicities:

\lambda = 2

has algebraic multiplicity 3, and

\lambda = -1

has algebraic multiplicity 1. The problem states that the eigenspace for

\lambda = 2

has dimension 1, meaning its geometric multiplicity is 1. Since the geometric multiplicity (1) is less than the algebraic multiplicity (3) for eigenvalue

\lambda = 2

, the matrix

A

cannot be diagonalizable. This immediately confirms that answer D is correct. Answer A is wrong because the geometric multiplicities don't sum to 4. We have geometric multiplicity 1 for

\lambda = 2

, and at most 1 for

\lambda = -1

, giving us at most 2 total linearly independent eigenvectors. Answer B reflects a common misconception. While having real eigenvalues is necessary for real diagonalizability, it's not sufficient. The geometric multiplicity condition must also be satisfied. Answer C is incorrect because regardless of the geometric multiplicity of

\lambda = -1

, we already know

A

is not diagonalizable due to the deficiency at

\lambda = 2

. Study tip: Always check that geometric multiplicity equals algebraic multiplicity for every eigenvalue. If even one eigenvalue fails this test, the matrix is not diagonalizable.

Question 9

Consider the matrix $B = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}$ . What is the relationship between the algebraic and geometric multiplicities of the eigenvalue $\lambda = 3$ ?

Algebraic multiplicity is 3, geometric multiplicity is 3, so the matrix is diagonalizable
Algebraic multiplicity is 1, geometric multiplicity is 1, so the matrix is diagonalizable
Algebraic multiplicity is 3, geometric multiplicity is 1, so the matrix is not diagonalizable (correct answer)
Algebraic multiplicity is 2, geometric multiplicity is 1, so the matrix is not diagonalizable

Explanation: For the upper triangular matrix

B

, the eigenvalues are the diagonal entries, so

\lambda = 3

is the only eigenvalue with algebraic multiplicity 3. To find the geometric multiplicity, we solve

(B - 3I)\mathbf{v} = \mathbf{0}

. We have

B - 3I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

, which has rank 2, so the nullspace has dimension

3 - 2 = 1

. Thus geometric multiplicity is 1 while algebraic multiplicity is 3, making the matrix non-diagonalizable. Choice A incorrectly states the geometric multiplicity. Choices B and D incorrectly state the algebraic multiplicity.

Question 10

Consider a $3 \times 3$ matrix $E$ with eigenvalues $\lambda = 2$ (algebraic multiplicity 2) and $\lambda = -1$ (algebraic multiplicity 1). The matrix $(E - 2I)$ has rank 2. Which statement about the Jordan canonical form of $E$ is correct?

$E$ has Jordan canonical form with one $2 \times 2$ Jordan block for $\lambda = 2$ and one $1 \times 1$ block for $\lambda = -1$ (correct answer)
$E$ has Jordan canonical form with two $1 \times 1$ Jordan blocks for $\lambda = 2$ and one $1 \times 1$ block for $\lambda = -1$
$E$ has Jordan canonical form with one $1 \times 1$ Jordan block for $\lambda = 2$ and one $2 \times 2$ block for $\lambda = -1$
The Jordan canonical form cannot be determined without additional information about the rank of $(E + I)$

Explanation: When you encounter Jordan canonical form problems, the key insight is that the geometric multiplicity (number of linearly independent eigenvectors) determines how eigenvalues split into Jordan blocks. The geometric multiplicity equals the nullity of

(A - \lambda I)

, which is the matrix size minus the rank. For eigenvalue

\lambda = 2

with algebraic multiplicity 2, we're told that

(E - 2I)

has rank 2. Since

E

3 \times 3

, the nullity of

(E - 2I)

3 - 2 = 1

. This means the geometric multiplicity of

\lambda = 2

is 1. When geometric multiplicity is less than algebraic multiplicity, the eigenvalue forms a single Jordan block of size equal to its algebraic multiplicity. So

\lambda = 2

creates one

2 \times 2

Jordan block. For

\lambda = -1

with algebraic multiplicity 1, the geometric multiplicity must also be 1 (it cannot exceed algebraic multiplicity), creating one

1 \times 1

Jordan block. Therefore, answer A is correct: one

2 \times 2

Jordan block for

\lambda = 2

and one

1 \times 1

block for

\lambda = -1

. Answer B is wrong because two

1 \times 1

blocks would require geometric multiplicity 2 for

\lambda = 2

, but we found it's only 1. Answer C incorrectly assigns the

2 \times 2

block to

\lambda = -1

, which has algebraic multiplicity 1. Answer D is wrong because the rank of

(E - 2I)

already provides sufficient information to determine the Jordan form. Study tip: Remember that geometric multiplicity = nullity = matrix size - rank. When geometric multiplicity < algebraic multiplicity, you get larger Jordan blocks, not more blocks.

Question 11

Consider the family of matrices $H_t = \begin{pmatrix} t & 1 & 0 \\ 0 & t & 1 \\ 0 & 0 & t \end{pmatrix}$ where $t$ is a real parameter. For which values of $t$ do the algebraic and geometric multiplicities of all eigenvalues coincide?

For all real values of $t$ , since the matrix is always upper triangular
Never, since the matrix always has a superdiagonal of ones (correct answer)
Only when $t = 0$ , since this makes the matrix nilpotent
Only when $t = 1$ , since this makes all eigenvalues equal to 1

Explanation: For any value of

t

, the matrix

H_t

is upper triangular with diagonal entries all equal to

t

, so

\lambda = t

is the only eigenvalue with algebraic multiplicity 3. To find the geometric multiplicity, we compute

H_t - tI = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

, which is independent of

t

and has rank 2. Therefore,

\text{gm}(t) = 3 - 2 = 1

for all values of

t

. Since

\text{am}(t) = 3

and

\text{gm}(t) = 1

for all

t

, the algebraic and geometric multiplicities never coincide. The presence of the superdiagonal of ones creates Jordan blocks regardless of the value of

t

. Choice A is wrong because being upper triangular doesn't guarantee diagonalizability. Choice C is wrong because even when

t = 0

, the geometric multiplicity is still 1. Choice D is wrong because the value of

t

doesn't affect the geometric multiplicity.

Question 12

Let $M$ be a $4 \times 4$ matrix with characteristic polynomial $p(\lambda) = (\lambda - 1)^2(\lambda - 3)^2$ . Suppose that $\dim(\text{null}(M - I)) = 1$ and $\dim(\text{null}((M - I)^2)) = 2$ . What can be concluded about the Jordan canonical form of $M$ ?

$M$ has two $2 \times 2$ Jordan blocks, one for each eigenvalue
$M$ has one $2 \times 2$ Jordan block for $\lambda = 1$ and the Jordan structure for $\lambda = 3$ cannot be determined
$M$ has one $2 \times 2$ Jordan block for $\lambda = 1$ and two $1 \times 1$ Jordan blocks for $\lambda = 3$ (correct answer)
$M$ has two $1 \times 1$ Jordan blocks for $\lambda = 1$ and one $2 \times 2$ Jordan block for $\lambda = 3$

Explanation: For eigenvalue

\lambda = 1

with algebraic multiplicity 2:

\text{gm}(1) = \dim(\text{null}(M - I)) = 1

. Since

\text{gm}(1) = 1 < 2 = \text{am}(1)

, there is exactly one Jordan block of size

2 \times 2

for

\lambda = 1

. The condition

\dim(\text{null}((M - I)^2)) = 2

confirms this structure. For eigenvalue

\lambda = 3

with algebraic multiplicity 2: Since the matrix is

4 \times 4

and the Jordan blocks for

\lambda = 1

occupy a

2 \times 2

space, the remaining

2 \times 2

space corresponds to

\lambda = 3

. Since the sum of geometric multiplicities must equal the number of Jordan blocks, and we know

\text{gm}(1) = 1

, we have

\text{gm}(3) = 4 - 2 = 2

(where 2 is the size of the Jordan block for

\lambda = 1

). Therefore,

\lambda = 3

has geometric multiplicity 2, giving two

1 \times 1

Jordan blocks.

Question 13

A $5 \times 5$ matrix $C$ has eigenvalues $\lambda_1 = 1$ with algebraic multiplicity 2, $\lambda_2 = -2$ with algebraic multiplicity 2, and $\lambda_3 = 0$ with algebraic multiplicity 1. If $C$ is diagonalizable, which of the following must be true about the geometric multiplicities?

$\text{gm}(1) = 1, \text{gm}(-2) = 2, \text{gm}(0) = 2$ where gm denotes geometric multiplicity
$\text{gm}(1) + \text{gm}(-2) + \text{gm}(0) = 5$ but individual values can vary within algebraic constraints
$\text{gm}(1) \leq 2, \text{gm}(-2) \leq 2, \text{gm}(0) \leq 1$ where gm denotes geometric multiplicity
$\text{gm}(1) = 2, \text{gm}(-2) = 2, \text{gm}(0) = 1$ where gm denotes geometric multiplicity (correct answer)

Explanation: When you encounter diagonalizability questions, remember that a matrix is diagonalizable if and only if the geometric multiplicity equals the algebraic multiplicity for every eigenvalue. This is the fundamental connection between eigenspaces and diagonalization. For matrix

C

to be diagonalizable, each eigenvalue's geometric multiplicity (the dimension of its eigenspace) must exactly match its algebraic multiplicity (how many times it appears as a root of the characteristic polynomial). Since we're told

C

is diagonalizable, we can determine the geometric multiplicities directly. Given the algebraic multiplicities:

\lambda_1 = 1

(multiplicity 2),

\lambda_2 = -2

(multiplicity 2), and

\lambda_3 = 0

(multiplicity 1), the geometric multiplicities must be identical to these values. Therefore:

\text{gm}(1) = 2

\text{gm}(-2) = 2

, and

\text{gm}(0) = 1

. Option A incorrectly states

\text{gm}(1) = 1

and

\text{gm}(0) = 2

, which would make the matrix non-diagonalizable since geometric multiplicities wouldn't match algebraic multiplicities. Option B suggests the geometric multiplicities can vary as long as they sum to 5, but this ignores the constraint that each geometric multiplicity must equal its corresponding algebraic multiplicity for diagonalizability. Option C only gives upper bounds (which are always true), but doesn't use the crucial information that

C

is diagonalizable. Option D correctly recognizes that diagonalizability forces geometric multiplicities to equal algebraic multiplicities. Key takeaway: Diagonalizable means geometric multiplicity equals algebraic multiplicity for every eigenvalue—no exceptions, no flexibility.

Question 14

A $6 \times 6$ matrix $K$ has exactly two distinct eigenvalues: $\lambda_1 = 4$ with algebraic multiplicity 4, and $\lambda_2 = -3$ with algebraic multiplicity 2. Given that $\text{rank}(K - 4I) = 2$ and $K$ is diagonalizable, what must be true about $\text{rank}(K + 3I)$ ?

$\text{rank}(K + 3I) = 4$ because the geometric multiplicity of $\lambda_2 = -3$ must equal 2 (correct answer)
$\text{rank}(K + 3I) = 0$ because the matrix must be nilpotent for this eigenvalue
$\text{rank}(K + 3I) = 2$ because this matches the algebraic multiplicity pattern
$\text{rank}(K + 3I)$ cannot be determined from the given information about diagonalizability

Explanation: When you encounter eigenvalue problems involving diagonalizability, focus on the relationship between algebraic multiplicity, geometric multiplicity, and matrix rank. For a matrix to be diagonalizable, the geometric multiplicity of each eigenvalue must equal its algebraic multiplicity. Since

K

is diagonalizable, each eigenvalue's geometric multiplicity equals its algebraic multiplicity. The geometric multiplicity of eigenvalue

\lambda

equals

\dim(\text{null}(K - \lambda I))

, which relates to rank through the equation:

\text{rank}(K - \lambda I) + \dim(\text{null}(K - \lambda I)) = n

(where

n = 6

). For

\lambda_1 = 4

with algebraic multiplicity 4: Since

K

is diagonalizable, the geometric multiplicity is also 4, so

\dim(\text{null}(K - 4I)) = 4

. This gives us

\text{rank}(K - 4I) = 6 - 4 = 2

, which matches the given information. For

\lambda_2 = -3

with algebraic multiplicity 2: Since

K

is diagonalizable, the geometric multiplicity must also be 2, so

\dim(\text{null}(K + 3I)) = 2

. Therefore,

\text{rank}(K + 3I) = 6 - 2 = 4

. Answer A correctly identifies this relationship. Answer B incorrectly suggests nilpotency, which doesn't apply here. Answer C mistakes the rank as equaling the algebraic multiplicity rather than

n

minus the algebraic multiplicity. Answer D incorrectly claims insufficient information when diagonalizability completely determines the geometric multiplicities. Study tip: Remember that for diagonalizable matrices,

\text{rank}(A - \lambda I) = n - \text{(algebraic multiplicity of } \lambda\text{)}

Question 15

Let $G$ be a $5 \times 5$ matrix with characteristic polynomial $p(\lambda) = (\lambda + 1)^3(\lambda - 2)^2$ . If $\text{rank}(G + I) = 2$ and $\text{rank}(G - 2I) = 4$ , determine whether $G$ is diagonalizable.

$G$ is diagonalizable because the sum of geometric multiplicities equals 5
$G$ is not diagonalizable because the geometric multiplicity of $\lambda = -1$ exceeds its algebraic multiplicity
$G$ is not diagonalizable because the geometric multiplicity of $\lambda = 2$ is less than its algebraic multiplicity (correct answer)
$G$ is diagonalizable because both eigenvalues have positive geometric multiplicities

Explanation: From the characteristic polynomial,

\text{am}(-1) = 3

and

\text{am}(2) = 2

. The geometric multiplicity of

\lambda = -1

equals

\dim(\text{null}(G + I)) = 5 - \text{rank}(G + I) = 5 - 2 = 3

. The geometric multiplicity of

\lambda = 2

equals

\dim(\text{null}(G - 2I)) = 5 - \text{rank}(G - 2I) = 5 - 4 = 1

. For

\lambda = -1

\text{gm}(-1) = 3 = \text{am}(-1)

✓. For

\lambda = 2

\text{gm}(2) = 1 < 2 = \text{am}(2)

✗. Since the geometric multiplicity of

\lambda = 2

is less than its algebraic multiplicity,

G

is not diagonalizable. Choice A is wrong because diagonalizability requires geometric multiplicity to equal algebraic multiplicity for each eigenvalue. Choice B is wrong because

\text{gm}(-1) = \text{am}(-1)

. Choice D is wrong because having positive geometric multiplicities is not sufficient.

Question 16

Consider the matrix $A = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix}$ . What are the algebraic and geometric multiplicities of the eigenvalue $\lambda = 3$ ?

Algebraic multiplicity is 2, geometric multiplicity is 2.
Algebraic multiplicity is 2, geometric multiplicity is 1. (correct answer)
Algebraic multiplicity is 1, geometric multiplicity is 2.
Algebraic multiplicity is 1, geometric multiplicity is 1.

Explanation: Since

A

is an upper triangular matrix, the eigenvalues are the diagonal entries:

\lambda = 3, 3, 5

. The algebraic multiplicity of

\lambda=3

is 2, as it appears twice. The geometric multiplicity is the dimension of the null space of

A - 3I

A - 3I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix}

. Row reducing gives

\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

. This matrix has two pivot columns and one free variable (

x_1

). The dimension of the null space (the nullity) is 1. Therefore, the geometric multiplicity of

\lambda=3

is 1.

Question 17

Let $A$ be an $n \times n$ matrix for which the characteristic polynomial is known. To determine conclusively whether $A$ is diagonalizable, what is the minimum additional information required?

The determinant of the matrix $A$ .
The trace of the matrix $A$ .
The geometric multiplicity of every eigenvalue of $A$ . (correct answer)
The complete set of eigenvectors for every eigenvalue of $A$ .

Explanation: A matrix is diagonalizable if and only if for every eigenvalue, its geometric multiplicity equals its algebraic multiplicity. The characteristic polynomial gives all eigenvalues and their algebraic multiplicities. Therefore, the minimum additional information needed is the geometric multiplicity for each of these eigenvalues to compare with the known algebraic multiplicities. The determinant and trace are products and sums of the eigenvalues, which are already known from the characteristic polynomial. Knowing the complete set of eigenvectors is sufficient but more information than necessary; only the dimension of each eigenspace (the geometric multiplicity) is required.

Question 18

The matrix $A = \begin{pmatrix} 3 & a \\ 0 & b \end{pmatrix}$ is diagonalizable under which of the following conditions on $a$ and $b$ ?

$a=0$ only
$b=3$ only
$a=0$ and $b=3$
$b \neq 3$ or $a=0$ (correct answer)

Explanation: There are two cases. Case 1: The eigenvalues are distinct. The eigenvalues are the diagonal entries, 3 and

b

. If

b \neq 3

, the

2 \times 2

matrix has two distinct eigenvalues, which guarantees it is diagonalizable, regardless of the value of

a

. Case 2: The eigenvalues are repeated. This occurs if

b=3

. The matrix is

A = \begin{pmatrix} 3 & a \\ 0 & 3 \end{pmatrix}

. The eigenvalue

\lambda=3

has algebraic multiplicity 2. For

A

to be diagonalizable, the geometric multiplicity must also be 2. The geometric multiplicity is the nullity of

A - 3I = \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}

. If

a=0

, this is the zero matrix, which has nullity 2. If

a \neq 0

, the matrix has rank 1 and nullity 1. So, if

b=3

, we need

a=0

for it to be diagonalizable. Combining both cases, the matrix is diagonalizable if (

b \neq 3

) OR (

b=3

and

a=0

). This is equivalent to the condition '

b \neq 3

a=0

Question 19

Let $A$ be an $n \times n$ matrix with real entries. Which of the following statements is always true?

If $A$ has $n$ distinct real eigenvalues, then the geometric multiplicity of each eigenvalue is 1. (correct answer)
If an eigenvalue $\lambda$ has algebraic multiplicity $k$ , its eigenspace $E_\lambda$ contains exactly $k$ linearly independent eigenvectors.
The sum of the geometric multiplicities of all eigenvalues is always equal to $n$ .
If a matrix is not diagonalizable, it must have at least one eigenvalue with algebraic multiplicity 1.

Explanation: (A) If an eigenvalue

\lambda

is distinct, its algebraic multiplicity is 1. Since the geometric multiplicity

m_g(\lambda)

must satisfy

1 \le m_g(\lambda) \le m_a(\lambda)

, if

m_a(\lambda)=1

, then

m_g(\lambda)

must be 1. This is true for all

n

distinct eigenvalues. (B) is false; the number of linearly independent eigenvectors is the geometric multiplicity, which can be less than the algebraic multiplicity

k

. (C) is false; this is the condition for a matrix to be diagonalizable, which is not always true. (D) is false; for a matrix to be non-diagonalizable, there must be an eigenvalue where

m_g < m_a

, which implies

m_a

must be at least 2 for that eigenvalue.

Question 20

A square matrix $A$ is guaranteed to be non-diagonalizable if which one of the following conditions is met?

The characteristic polynomial has at least one repeated root.
The matrix is singular, meaning $\det(A)=0$ .
For at least one eigenvalue $\lambda$ , the dimension of its eigenspace is strictly less than its algebraic multiplicity. (correct answer)
The sum of the algebraic multiplicities is equal to the dimension of the matrix.

Explanation: The condition for a matrix to be diagonalizable is that for every eigenvalue, its geometric multiplicity (the dimension of its eigenspace) is equal to its algebraic multiplicity. Therefore, a matrix is guaranteed to be non-diagonalizable if this condition fails for even one eigenvalue. (A) is not sufficient; the identity matrix

I_n

has repeated eigenvalues but is diagonal (and thus diagonalizable). (B) being singular only means

\lambda=0

is an eigenvalue; it doesn't guarantee non-diagonalizability. (D) is true for any matrix whose characteristic polynomial splits over the field, including all diagonalizable matrices.