Algebraic vs. Geometric Multiplicity - Linear Algebra

1. 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?
   Answer: $E$ has Jordan canonical form with one $2 \times 2$ Jordan block for $\lambda = 2$ and one $1 \times 1$ block for $\lambda = -1$ 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 $E$ is $3 \times 3$, the nullity of $(E - 2I)$ is $3 - 2 = 1$. 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.
2. 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 $\text{rank}(K + 3I)$?
   Answer: $\text{rank}(K + 3I) = 4$ because the geometric multiplicity of $\lambda_2 = -3$ must equal 2 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, $\text{rank}(A - \lambda I) = n - \text{(algebraic multiplicity of } \lambda\text{)}$.
3. A $4 \times 4$ matrix $F$ has the property that $(F - 3I)^2 = 0$ but $(F - 3I) \neq 0$. What are the algebraic and geometric multiplicities of the eigenvalue $\lambda = 3$?
   Answer: Algebraic multiplicity is 4, geometric multiplicity is 2, and $F$ is not diagonalizable Since $(F - 3I)^2 = 0$, the minimal polynomial of $F$ divides $(\lambda - 3)^2$. Since $(F - 3I) \neq 0$, the minimal polynomial is exactly $(\lambda - 3)^2$. For a $4 \times 4$ matrix, if 3 is the only eigenvalue, then the algebraic multiplicity of $\lambda = 3$ must be 4. The condition $(F - 3I)^2 = 0$ means that the minimal polynomial has degree 2, which implies the largest Jordan block has size $2 \times 2$. For a $4 \times 4$ matrix with eigenvalue 3 of algebraic multiplicity 4, if the largest Jordan block is $2 \times 2$, then there are two $2 \times 2$ Jordan blocks, giving geometric multiplicity 2. Since geometric multiplicity $<$ algebraic multiplicity, $F$ is not diagonalizable.
4. 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 $M$?
   Answer: $M$ has one $2 \times 2$ Jordan block for $\lambda = 1$ and two $1 \times 1$ Jordan blocks for $\lambda = 3$ 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.
5. 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 $\lambda = 3$?
   Answer: Algebraic multiplicity is 2, geometric multiplicity is 1. 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.
6. 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.
   Answer: $G$ is not diagonalizable because the geometric multiplicity of $\lambda = 2$ is less than its algebraic multiplicity 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.
7. 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 $\lambda = 3$?
   Answer: Algebraic multiplicity is 3, geometric multiplicity is 1, so the matrix is not diagonalizable 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.
8. 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 $\lambda_2=5$?
   Answer: 2 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 $\lambda_1=2$ is $m_a(2)=2$. 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 $m_a(5) = 2$.
9. 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?
   Answer: The matrix has one eigenvalue with algebraic multiplicity 2 and geometric multiplicity 1. 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.
10. 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?
    Answer: The matrix $A$ is diagonalizable if and only if $m_g(7) = 4$. 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.
11. 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 $A$?
    Answer: The geometric multiplicity is 2, and the matrix $A$ is not diagonalizable. 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.
12. 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 $A$?
    Answer: $m_g(4) = 2$ and $m_g(-2) = 1$ 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.
13. 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 $A$?
    Answer: The matrix $A$ is diagonalizable. 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.
14. The matrix $A = \begin{pmatrix} 3 & a \\ 0 & b \end{pmatrix}$ is diagonalizable under which of the following conditions on $a$ and $b$?
    Answer: $b \neq 3$ or $a=0$ 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 '$b \neq 3$ or $a=0$'.
15. 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 $A$?
    Answer: $A$ is not diagonalizable because the geometric multiplicity of $\lambda = 2$ is less than its algebraic multiplicity 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.
16. 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?
    Answer: The geometric multiplicity of every eigenvalue of $A$. 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.
17. A square matrix $A$ is guaranteed to be non-diagonalizable if which one of the following conditions is met?
    Answer: For at least one eigenvalue $\lambda$, the dimension of its eigenspace is strictly less than its algebraic multiplicity. 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.
18. 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?
    Answer: $\text{gm}(1) = 2, \text{gm}(-2) = 2, \text{gm}(0) = 1$ where gm denotes geometric multiplicity 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.
19. Let $A$ be an $n \times n$ matrix with real entries. Which of the following statements is always true?
    Answer: If $A$ has $n$ distinct real eigenvalues, then the geometric multiplicity of each eigenvalue is 1. (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.
20. 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?
    Answer: 7 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.

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