Rank & Nullity - Linear Algebra

Let $V$ be the vector space of all $3 \times 3$ matrices, and let $T: V \to V$ be defined by $T(X) = AX - XA$ for some fixed $3 \times 3$ matrix $A$ . If $A$ has eigenvalues $1, 2, 3$ , what is the nullity of $T$ ?
Answer: 3 The nullspace of $T$ consists of matrices $X$ such that $AX = XA$ , i.e., matrices that commute with $A$ . Since $A$ has distinct eigenvalues $1, 2, 3$ , it is diagonalizable with $A = PDP^{-1}$ where $D = \text{diag}(1,2,3)$ . A matrix $X$ commutes with $A$ if and only if $P^{-1}XP$ commutes with $D$ . Matrices that commute with a diagonal matrix with distinct entries are themselves diagonal. So $P^{-1}XP$ must be diagonal, meaning $X = P \cdot \text{diag}(d_1, d_2, d_3) \cdot P^{-1}$ for arbitrary scalars $d_1, d_2, d_3$ . This gives a 3-dimensional space of matrices $X$ , so nullity $T = 3$ . Choice A undercounts the diagonal degrees of freedom. Choice C would apply if $A$ had repeated eigenvalues. Choice D would mean $T$ is the zero transformation.
A $4 \times 6$ matrix $A$ has the property that for every $2 \times 4$ matrix $B$ , the matrix $BA$ has rank at most 2. What can be concluded about the rank of $A$ ?
Answer: rank $A \leq 2$ This question tests your understanding of how matrix multiplication affects rank, specifically the fundamental property that rank( $BA$ ) ≤ min(rank( $B$ ), rank( $A$ )). Since $A$ is $4 \times 6$ and $B$ is $2 \times 4$ , their product $BA$ is $2 \times 6$ . The key insight is that we're told this holds for every $2 \times 4$ matrix $B$ . Consider what happens when $B$ has rank 2 (the maximum possible for a $2 \times 4$ matrix). Since rank( $BA$ ) ≤ min(rank( $B$ ), rank( $A$ )) = min(2, rank( $A$ )), and we know rank( $BA$ ) ≤ 2, this constraint is automatically satisfied if rank( $A$ ) ≥ 2. However, the universal quantifier "for every $2 \times 4$ matrix $B$ " is crucial. If rank( $A$ ) were 3 or 4, we could find specific matrices $B$ of rank 2 such that $BA$ would have rank greater than 2, violating our condition. The only way to guarantee that rank( $BA$ ) ≤ 2 for all possible $B$ is if rank( $A$ ) ≤ 2. Option A is wrong because ranks 3 and 4 would allow violations of the given condition. Option B is too restrictive—rank( $A$ ) could be 1 or 2 and still satisfy the condition. Option C contradicts our conclusion entirely, as ranks 3 or higher would create problems. Study tip: When you see "for every" or "for all" in matrix problems, think about the strongest constraints this imposes. Universal conditions often force tighter bounds than you might initially expect.
For a non-zero $3 \times 3$ skew-symmetric matrix $A$ (where $A^T = -A$ ), which of the following is a possible value for the rank of $A$ ?
Answer: 2 For any skew-symmetric matrix $A$ of odd dimension $n$ , its determinant is always zero. Here $n=3$ . $\det(A) = \det(A^T) = \det(-A) = (-1)^3 \det(A) = -\det(A)$ . This implies $2\det(A) = 0$ , so $\det(A) = 0$ . Since the determinant is zero, the matrix is singular, and its rank must be less than 3. The problem states $A$ is non-zero, so its rank cannot be 0. The rank of any non-zero skew-symmetric matrix is always an even number. For a $3 \times 3$ matrix, the only possible even rank greater than 0 and less than 3 is 2.
If $A$ is an $m \times n$ matrix and $B$ is an $n \times p$ matrix, which inequality concerning the rank of the product $AB$ must be true?
Answer: $\mathrm{rank}(AB) \le \mathrm{rank}(A)$ When you encounter questions about the rank of matrix products, think about how matrix multiplication affects the column space and row space of the matrices involved. The key insight is that multiplying by matrix $B$ can only decrease or maintain the rank of $A$ , never increase it. Here's why: the rank of $A$ equals the dimension of its column space. When you multiply $A$ by $B$ to get $AB$ , each column of $AB$ is a linear combination of the columns of $A$ . Since linear combinations cannot create dimensions beyond those already present in the column space of $A$ , we have $\text{rank}(AB) \leq \text{rank}(A)$ . This makes option A correct—the rank of the product is always less than or equal to the rank of the left matrix. Option B suggests the opposite inequality, which contradicts our reasoning above. Matrix multiplication cannot magically create new dimensions in the column space. Option C claims the rank equals the product of individual ranks, but this is rarely true. For instance, if $A$ and $B$ both have rank 2, their product typically won't have rank 4. Option D states the rank equals the minimum of the two ranks. While $\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B))$ is actually true, equality doesn't always hold. Consider when $A$ and $B$ have the same rank but $AB$ has smaller rank due to linear dependencies. Remember this pattern: matrix multiplication on the right can only maintain or reduce rank, never increase it. This principle applies broadly in linear algebra problems involving matrix products.
Let $A$ be a $5 \times 7$ matrix. If the dimension of the solution space for the homogeneous system $A\mathbf{x} = \mathbf{0}$ is 3, what is the rank of $A$ ?
Answer: 4 The Rank-Nullity Theorem states that for an $m \times n$ matrix $A$ , $\mathrm{rank}(A) + \mathrm{nullity}(A) = n$ . The dimension of the solution space for $A\mathbf{x} = \mathbf{0}$ is the nullity of $A$ . Here, $A$ is a $5 \times 7$ matrix, so $n=7$ . We are given that $\mathrm{nullity}(A) = 3$ . Plugging these values into the theorem gives $\mathrm{rank}(A) + 3 = 7$ . Solving for the rank, we get $\mathrm{rank}(A) = 7 - 3 = 4$ .
Let $\mathbf{u}$ be a non-zero vector in $\mathbb{R}^4$ and $\mathbf{v}$ be a non-zero vector in $\mathbb{R}^3$ . What is the rank of the $4 \times 3$ matrix $A = \mathbf{u}\mathbf{v}^T$ ?
Answer: 1 When you encounter a matrix formed as an outer product $\mathbf{u}\mathbf{v}^T$ , you're dealing with a fundamental concept in linear algebra: the rank of such matrices is always determined by the linear independence of the vectors involved. The matrix $A = \mathbf{u}\mathbf{v}^T$ is a $4 \times 3$ matrix where each column is a scalar multiple of vector $\mathbf{u}$ . Specifically, if $\mathbf{v} = [v_1, v_2, v_3]^T$ , then the columns of $A$ are $v_1\mathbf{u}$ , $v_2\mathbf{u}$ , and $v_3\mathbf{u}$ . Since all three columns are scalar multiples of the same vector $\mathbf{u}$ , they all lie in the one-dimensional subspace spanned by $\mathbf{u}$ . Because $\mathbf{u}$ is non-zero, this subspace has dimension 1, making the rank of $A$ equal to 1. Choice B (rank 0) would only occur if $A$ were the zero matrix, which requires either $\mathbf{u} = \mathbf{0}$ or $\mathbf{v} = \mathbf{0}$ . Since both vectors are explicitly non-zero, this is impossible. Choice C (rank 3) would require three linearly independent columns, but since all columns are multiples of $\mathbf{u}$ , they're linearly dependent. Choice D suggests the rank cannot be determined, but the outer product structure completely determines the rank regardless of the specific vector components. Study tip: Remember that any outer product $\mathbf{u}\mathbf{v}^T$ of non-zero vectors always has rank 1. This is a powerful pattern that appears frequently in linear algebra applications like matrix factorization and data compression.
Let $A$ be a $3 \times 5$ matrix with $\mathrm{rank}(A) = 2$ . What is the nullity of the transpose matrix, $A^T$ ?
Answer: 1 The matrix $A$ is $3 \times 5$ , so its transpose $A^T$ is $5 \times 3$ . A key property of rank is that $\mathrm{rank}(A) = \mathrm{rank}(A^T)$ . Given $\mathrm{rank}(A) = 2$ , it follows that $\mathrm{rank}(A^T) = 2$ . To find the nullity of $A^T$ , we apply the Rank-Nullity Theorem to $A^T$ . For $A^T$ , the number of columns is $n=3$ . The theorem states $\mathrm{rank}(A^T) + \mathrm{nullity}(A^T) = 3$ . Substituting the known rank: $2 + \mathrm{nullity}(A^T) = 3$ . Therefore, $\mathrm{nullity}(A^T) = 1$ .
Let $A$ be a $3 \times 3$ matrix such that $A \neq \mathbf{0}$ but $A^2 = \mathbf{0}$ . What is the rank of $A$ ?
Answer: 1 The condition $A^2 = \mathbf{0}$ means that for any vector $\mathbf{x}$ , $A(A\mathbf{x}) = \mathbf{0}$ . This implies that every vector in the column space of $A$ , $\mathrm{Col}(A)$ , is in the null space of $A$ , $\mathrm{Nul}(A)$ . Therefore, $\mathrm{Col}(A) \subseteq \mathrm{Nul}(A)$ , which means $\dim(\mathrm{Col}(A)) \le \dim(\mathrm{Nul}(A))$ . In other words, $\mathrm{rank}(A) \le \mathrm{nullity}(A)$ . By the Rank-Nullity Theorem for a $3 \times 3$ matrix, $\mathrm{rank}(A) + \mathrm{nullity}(A) = 3$ . Substituting the inequality, we get $\mathrm{rank}(A) + \mathrm{rank}(A) \le \mathrm{rank}(A) + \mathrm{nullity}(A) = 3$ , so $2 \cdot \mathrm{rank}(A) \le 3$ . This implies $\mathrm{rank}(A) \le 1.5$ . Since rank must be an integer, the rank is either 0 or 1. As $A \neq \mathbf{0}$ , its rank cannot be 0. Thus, the only possibility is $\mathrm{rank}(A) = 1$ .
A linear transformation $T: \mathbb{R}^5 \to \mathbb{R}^4$ has a range that is a 2-dimensional subspace of $\mathbb{R}^4$ . What is the dimension of the kernel of $T$ ?
Answer: 3 The dimension of the range of a linear transformation is its rank, and the dimension of its kernel is its nullity. The Rank-Nullity Theorem states that $\mathrm{rank}(T) + \mathrm{nullity}(T) = \dim(\text{domain})$ . We are given that the range is a 2-dimensional subspace, so $\mathrm{rank}(T) = 2$ . The domain is $\mathbb{R}^5$ , so its dimension is 5. Applying the theorem: $2 + \mathrm{nullity}(T) = 5$ . Solving for the nullity gives $\mathrm{nullity}(T) = 3$ .
Let $A$ be a $5 \times 7$ matrix with rank $A = 3$ , and let $B$ be a $7 \times 4$ matrix such that the nullspace of $A$ is contained in the nullspace of $B$ . What is the maximum possible value of rank $(AB)$ ?
Answer: 3 Since rank $A = 3$ , we have nullity $A = 7 - 3 = 4$ . The condition that null $A \subseteq$ null $B$ means $B$ maps every vector in null $A$ to zero. By the rank inequality, rank $(AB) \leq \min(\text{rank}A, \text{rank}B) = \min(3, \text{rank}B)$ . Since the column space of $AB$ is contained in the column space of $B$ , and $A$ can map at most 3 linearly independent vectors to linearly independent outputs, rank $(AB) \leq 3$ . This maximum is achievable when the 3-dimensional column space of $A$ maps injectively under $B$ . Choice B ignores the rank constraint from $A$ . Choice C exceeds the number of rows in $AB$ . Choice D exceeds the dimensions of both matrices.
The row-reduced echelon form of a matrix $A$ is given by $R$ below. What is the nullity of $A$ ? $R = \begin{pmatrix} 1 & 2 & 0 & -1 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$
Answer: 2 The nullity of a matrix is the dimension of its null space, which is equal to the number of free variables in the system $A\mathbf{x} = \mathbf{0}$ . In the row-reduced echelon form, the number of free variables is the number of columns that do not contain a leading 1 (a pivot). The matrix $R$ has 5 columns. The pivot columns are columns 1, 3, and 5. The columns without pivots are columns 2 and 4. Therefore, there are 2 free variables, and the nullity of $A$ is 2. Alternatively, the rank is the number of pivots, which is 3. By the Rank-Nullity Theorem, $\mathrm{rank}(A) + \mathrm{nullity}(A) = n$ , so $3 + \mathrm{nullity}(A) = 5$ , which gives $\mathrm{nullity}(A) = 2$ .
Let $A$ be a $4 \times 4$ matrix with $\mathrm{rank}(A)=3$ . What is the rank of its adjugate matrix, $\mathrm{adj}(A)$ ?
Answer: 1 There is a standard property for the rank of the adjugate of an $n \times n$ matrix $A$ : - If $\mathrm{rank}(A) = n$ , then $\mathrm{adj}(A)$ is invertible and $\mathrm{rank}(\mathrm{adj}(A)) = n$ . - If $\mathrm{rank}(A) = n-1$ , then $\mathrm{rank}(\mathrm{adj}(A)) = 1$ . - If $\mathrm{rank}(A) < n-1$ , then $\mathrm{adj}(A) = \mathbf{0}$ and $\mathrm{rank}(\mathrm{adj}(A)) = 0$ . In this problem, $n=4$ and we are given $\mathrm{rank}(A)=3$ . This corresponds to the case $\mathrm{rank}(A) = n-1$ . Therefore, the rank of $\mathrm{adj}(A)$ must be 1.
Let $T: \mathbb{R}^6 \to \mathbb{R}^4$ be a linear transformation with nullity 3. If $S: \mathbb{R}^4 \to \mathbb{R}^5$ is a linear transformation such that $S \circ T$ has rank 2, what is the minimum possible nullity of $S$ ?
Answer: 1 Since nullity $T = 3$ and $T: \mathbb{R}^6 \to \mathbb{R}^4$ , we have rank $T = 6 - 3 = 3$ . The composition $S \circ T$ has rank 2, so rank $(S \circ T) \leq \min(\text{rank}T, \text{rank}S) = \min(3, \text{rank}S)$ . Since rank $(S \circ T) = 2$ , we need rank $S \geq 2$ . Also, rank $(S \circ T) \leq$ rank $S$ , and the image of $T$ (which has dimension 3) must be mapped by $S$ to a 2-dimensional space. For this to happen with minimum nullity of $S$ , we want maximum rank $S$ , which is $\min(4,5) = 4$ . But we need $S$ to compress the 3-dimensional image of $T$ to dimension 2, so rank $S$ can be 3, giving nullity $S = 4 - 3 = 1$ . Choices B, C, D represent insufficient compression of the image of $T$ .
Let $A$ be a $4 \times 6$ matrix and $B$ be a $6 \times 4$ matrix such that $AB = I_4$ . If the $2 \times 6$ matrix $C$ is formed by taking the first two rows of $A$ , what is the maximum possible rank of $C$ ?
Answer: 2 Since $AB = I_4$ , we know $A$ is a left inverse of $B$ , so rank $A = 4$ and rank $B = 4$ . The matrix $A$ has 4 linearly independent rows. When we take the first two rows to form $C$ , we get at most 2 linearly independent rows, so rank $C \leq 2$ . This maximum is achievable: since $A$ has rank 4, we can choose $A$ such that its first two rows are linearly independent (while maintaining the constraint $AB = I_4$ ). For example, if $A$ is constructed to have its first two rows forming part of a basis for $\mathbb{R}^6$ , then rank $C = 2$ . Choice B would require $C$ to have 3 linearly independent rows, but $C$ has only 2 rows. Choice C would require the 2 rows of $C$ to span a 4-dimensional space, which is impossible. Choice D exceeds the dimension of the ambient space that the rows of $C$ live in.
Consider matrices $A$ ( $m \times n$ ), $B$ ( $n \times p$ ), and $C$ ( $p \times q$ ) such that rank $(AB) =$ rank $(BC) =$ rank $(ABC) = r$ . If $B$ is square and invertible, which of the following must be true?
Answer: rank $A =$ rank $C = r$ When you encounter problems involving matrix products and an invertible matrix in the chain, the key insight is that invertible matrices preserve rank. Since matrix multiplication can only decrease or maintain rank, but never increase it, an invertible matrix acts as a "rank bridge" that allows you to connect the ranks of matrices on either side. Since $B$ is invertible, we can use the fundamental property that multiplying by an invertible matrix preserves rank. Given that rank $(AB) = r$ , we know rank $A \geq r$ because matrix multiplication can only decrease rank. However, since $B$ is invertible, we can multiply $(AB)$ by $B^{-1}$ to get $A = (AB)B^{-1}$ . This means rank $A =$ rank $(AB) = r$ . Similarly, from rank $(BC) = r$ , we get rank $C \geq r$ . Since $B$ is invertible, $C = B^{-1}(BC)$ , so rank $C =$ rank $(BC) = r$ . The condition rank $(ABC) = r$ is consistent with our findings and doesn't provide additional constraints beyond what we already determined. Option A incorrectly suggests rank $A \geq r$ (which would be true without $B$ 's invertibility) rather than the exact equality. Option B reverses the relationships, suggesting rank $A = r$ but only rank $C \geq r$ . Option D incorrectly claims rank $B = r$ , but since $B$ is invertible and $n \times n$ , we have rank $B = n$ , which may not equal $r$ . Remember: invertible matrices in matrix products act as rank preservers, allowing you to "cancel" them when analyzing rank relationships.
Let $A$ be a $4 \times 6$ matrix. Which of the following statements about the nullity of $A$ must be true?
Answer: $\mathrm{nullity}(A) \geq 2$ When you encounter questions about nullity and matrix dimensions, always think about the rank-nullity theorem, which provides fundamental constraints on any matrix's properties. The rank-nullity theorem states that for any $m \times n$ matrix $A$ : $\text{rank}(A) + \text{nullity}(A) = n$ . For your $4 \times 6$ matrix, this means $\text{rank}(A) + \text{nullity}(A) = 6$ . Since $A$ is a $4 \times 6$ matrix, its rank cannot exceed the number of rows, so $\text{rank}(A) \leq 4$ . Substituting into the rank-nullity theorem: $\text{nullity}(A) = 6 - \text{rank}(A) \geq 6 - 4 = 2$ . Therefore, the nullity must be at least 2, making answer D correct. Looking at the wrong answers: Option A claims $\text{nullity}(A) \leq 2$ , which contradicts our finding that nullity must be at least 2. While nullity could equal 2, it could also be larger. Option B states $\text{nullity}(A) = 2$ , but this only occurs when the matrix has full row rank (rank = 4). The matrix could have lower rank, making nullity larger than 2. Option C claims $\text{nullity}(A) = 0$ , which would require rank = 6, impossible since the matrix only has 4 rows. Remember this pattern: for any $m \times n$ matrix where $n > m$ , the nullity is always at least $n - m$ . This guarantees a non-trivial null space when you have more columns than rows, which frequently appears in linear algebra problems.
For what value of $k$ will the matrix $M$ have a nullity of 1? $M = \begin{pmatrix} 1 & 0 & -2 \\ 2 & 1 & 1 \\ 0 & 1 & k \end{pmatrix}$
Answer: 5 For a $3 \times 3$ matrix, having a nullity of 1 is equivalent to having a rank of $3 - 1 = 2$ . A square matrix has rank less than its dimension if and only if its determinant is zero. We calculate the determinant of $M$ : $\det(M) = 1(1 \cdot k - 1 \cdot 1) - 0(2k - 0) + (-2)(2 \cdot 1 - 1 \cdot 0)$ $= 1(k - 1) - 0 - 2(2)$ $= k - 1 - 4 = k - 5$ . For the rank to be less than 3, we set the determinant to zero: $k - 5 = 0$ , which gives $k=5$ . When $k=5$ , the rank is not 0 (since it's not the zero matrix), so it must be 1 or 2. Row reduction for $k=5$ would show 2 pivots, confirming the rank is 2 and nullity is 1.
Let $A$ be an $n \times n$ matrix such that $A^3 = 0$ but $A^2 \neq 0$ . If the nullspace of $A^2$ has dimension $n-1$ , what is the dimension of the nullspace of $A$ ?
Answer: $n-2$ When you encounter a nilpotent matrix (where some power equals zero), the key is understanding how the dimensions of successive nullspaces relate through the rank-nullity theorem and properties of nilpotent transformations. Since $A^3 = 0$ but $A^2 \neq 0$ , we know $A$ is nilpotent with index 3. This creates a chain of subspaces: $\text{null}(A) \subseteq \text{null}(A^2) \subseteq \text{null}(A^3) = \mathbb{R}^n$ . The dimensions must increase strictly at each step since $A^2 \neq 0$ and $A \neq 0$ . Given that $\dim(\text{null}(A^2)) = n-1$ , we can find $\text{rank}(A^2) = n - (n-1) = 1$ by rank-nullity. Since $A^3 = 0$ , we have $A \cdot A^2 = 0$ , meaning the column space of $A^2$ lies entirely within the nullspace of $A$ . Since $\text{rank}(A^2) = 1$ , the column space of $A^2$ is one-dimensional and must be contained in $\text{null}(A)$ . For nilpotent matrices, there's a key relationship: the dimension increases by exactly 1 at each step in the chain. Since $\dim(\text{null}(A^2)) = n-1$ and we need $\dim(\text{null}(A)) < \dim(\text{null}(A^2))$ , we have $\dim(\text{null}(A)) = n-2$ . Choice A) $\frac{n+1}{2}$ doesn't follow from any nilpotent matrix property. Choice B) $n-1$ would mean $\text{null}(A) = \text{null}(A^2)$ , contradicting $A^2 \neq 0$ . Choice D) $n$ would mean $A = 0$ , contradicting our conditions. Study tip: For nilpotent matrices, nullspace dimensions increase by exactly 1 at each step until reaching the full space.
Let $A$ be a $4 \times 4$ matrix whose columns are $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ . If it is known that $3\mathbf{v}_1 - \mathbf{v}_3 = \mathbf{0}$ , which of the following statements must be true?
Answer: $\mathrm{nullity}(A) \ge 1$ This question tests your understanding of linear dependence and the rank-nullity theorem. When you see a linear relationship between columns of a matrix, immediately think about what this tells you about the matrix's rank and nullity. The given condition $3\mathbf{v}_1 - \mathbf{v}_3 = \mathbf{0}$ can be rewritten as $3\mathbf{v}_1 + 0\mathbf{v}_2 + (-1)\mathbf{v}_3 + 0\mathbf{v}_4 = \mathbf{0}$ . This is a nontrivial linear combination of the columns that equals zero (the coefficients aren't all zero), which means the columns are linearly dependent. This linear dependence relation gives us a vector in the null space of $A$ : the vector $\mathbf{x} = \begin{bmatrix} 3 \\ 0 \\ -1 \\ 0 \end{bmatrix}$ satisfies $A\mathbf{x} = \mathbf{0}$ . Since the null space contains at least this nonzero vector, we know $\mathrm{nullity}(A) \geq 1$ . This makes choice A correct. Choice B is wrong because we only know there's at least one linearly dependent relationship—there could be more, making the nullity greater than 1. Choice C is incorrect for the same reason: while the rank is at most 3 (since the columns are dependent), it could be less if there are additional dependencies. Choice D directly contradicts our finding that the columns are linearly dependent. Remember: whenever you find a nontrivial linear combination of matrix columns equaling zero, you've discovered that the nullity is at least 1. Look for the weakest statement that must be true—it's often the correct answer in "must be true" questions.

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