Rank and Nullity Questions and Answers

Q: Let the linear transformation \(T : F^{2} \to F^{3}\) be defined by \(T(x_{1}, x_{2}) = (x_{1}, x_{1} + x_{2}, x_{2})\). Then the nullity of \(T\) is

The nullity of \(T\) is 0.The kernel consists of the vectors with \(T(x_{1}, x_{2}) = (0, 0, 0)\): the first and third coordinates force \(x_{1} = 0\) and \(x_{2} = 0\). So the kernel is \(\{0\}\) and the nullity is 0.By rank-nullity, \(\dim F^{2} = 2 = \text{rank} + \text{nullity}\), and the images \(T(1, 0) = (1, 1, 0)\) and \(T(0, 1) = (0, 1, 1)\) are independent, so the rank is 2, confirming nullity 0. The other options confuse nullity with the dimension of the codomain (3), of the domain (2), or with the codimension of the image (1).

Q: Let \(T : \mathbb{R}^{4} \to \mathbb{R}^{4}\) be a linear transformation and the null space of T be the subspace of \(\mathbb{R}^{4}\) given by\[ \begin{aligned} \{(x_{1},x_{2},x_{3},x_{4}) &\in \mathbb{R}^{4} : \\ 4x_{1}+3x_{2}+2x_{3}+x_{4} &= 0\}. \end{aligned} \]If \(Rank(T - 3I) = 3\), where I is the identity map on \(\mathbb{R}^{4}\), then the minimal polynomial of T is

The minimal polynomial is x(x - 3).The null space of T is a hyperplane in \(\mathbb{R}^{4}\), of dimension 3, so 0 is an eigenvalue with geometric multiplicity 3. Rank(T - 3I) = 3 means T - 3I has nullity 1, so 3 is an eigenvalue with geometric multiplicity 1.The algebraic multiplicities are at least the geometric ones and add up to 4, so they are exactly 3 and 1. Each eigenvalue then has as many independent eigenvectors as its multiplicity, so T is diagonalizable with eigenvalues 0, 0, 0, 3, and its minimal polynomial has only simple factors: x(x - 3).The options with higher powers would describe non-diagonalizable maps, which would need fewer eigenvectors than the null space supplies.

Q: If\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 3 & -2 \\ 0 & -6 & 4 \end{pmatrix}, \]then the rank of the matrix \(AA^{t}\) is :

The rank of \(AA^{t}\) is 2.For a real matrix, \(\text{rank}(AA^{t}) = \text{rank}(A)\), because \(AA^{t}x = 0\) gives \(x^{t}AA^{t}x = |A^{t}x|^{2} = 0\), so \(AA^{t}\) and \(A^{t}\) have the same null space.So find the rank of A. The third row \((0, -6, 4)\) is \(-2\) times the second row \((0, 3, -2)\), so the rows are dependent and the rank is at most 2. The first two rows are clearly independent (only the first has a nonzero first entry). Hence rank A = 2, and \(AA^{t}\) has rank 2 as well.Rank 3 would need \(\det A \ne 0\), but \(\det A = 1 \cdot (3 \cdot 4 - (-2)(-6)) = 0\). Rank 1 or 0 would need the first two rows to be dependent, which they are not.

Q: If P is a \(3 \times 3\) matrix of rank three and :\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \]then the rank of PA is :

The rank of PA is 2.A \(3 \times 3\) matrix of rank three is invertible, and multiplying by an invertible matrix does not change rank. PA and A have the same null space, since \(PAx = 0\) gives \(Ax = P^{-1}0 = 0\). So rank PA = rank A.For A, the rows are in arithmetic progression: row 3 minus row 2 equals row 2 minus row 1, both \((3, 3, 3)\). So row 1 - 2 row 2 + row 3 = 0 and \(\det A = 0\), while the first two rows are independent. Hence rank A = 2 and rank PA = 2.3 would need A invertible. 1 would need all rows proportional. 'Depends upon matrix P' is the trap: it would be right if P could be singular, but rank three rules that out.

Q: Let \(A = \begin{pmatrix} 1 & 2 \\ 3 & 6 \\ 1 & 2 \end{pmatrix}\) and \(f : \mathbf{R}^{2} \to \mathbf{R}^{3}\) be defined as \(f(x) = Ax\). Then which of the following statements is false ?

The false statement is \(f\) maps an open neighbourhood of the origin to an open neighbourhood.The columns of A are proportional, (2, 6, 2) = 2(1, 3, 1), so A has rank 1. The image of \(\mathbf{R}^{2}\) is the line spanned by (1, 3, 1) in \(\mathbf{R}^{3}\), and a line contains no open set of \(\mathbf{R}^{3}\). So the image of a neighbourhood of the origin cannot be open.The other statements are true. A linear map between finite dimensional spaces is continuous and differentiable, with derivative A at every point. The directional derivative in the direction (1, 0) is \(A(1, 0)^{t}\), the first column (1, 3, 1).

Q: Let A be a 4 × 3 matrix and B a 3 × 4 matrix. Which of the following is always true ?

The statement always true is det (AB) = 0, det (BA) need not be zero.AB is 4 × 4, but its rank is at most the rank of A, which is at most 3. A 4 × 4 matrix of rank at most 3 is singular, so \(\det(AB) = 0\) always.BA is 3 × 3 and can be invertible. For example, if A is the 4 × 3 matrix with \(I_{3}\) on top and a zero row below, and B is \([I_{3}\ 0]\), then \(BA = I_{3}\) with determinant 1.So 'both need not be zero' and 'both are zero' are wrong, and det(AB) = 0 holds even when neither AB nor BA is the zero matrix.

Q: Let A be a 6 × 7 matrix of rank 6 and B be a 7 × 5 matrix of rank 4, then the rank of AB is :

The rank of AB cannot be determined exactly, but is at most 4.Upper bound: \(\text{rank}(AB) \le \min(\text{rank } A, \text{rank } B) = \min(6, 4) = 4\).Lower bound, by Sylvester's inequality with inner dimension 7:\[ \begin{aligned} \text{rank}(AB) &\ge 6 + 4 - 7 \\ &= 3 \end{aligned} \]Both values occur. A has a one-dimensional null space; if that null space lies inside the column space of B, AB loses one dimension and has rank 3, otherwise it keeps rank 4. Since the data do not decide this, the rank is 3 or 4, so the exact values 4, 5 and 6 are all wrong as certainties (5 and 6 are impossible).

Q: The matrix\[ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 2 \\ 1 & 3 & 3 & 3 \\ 4 & 4 & 4 & 4 \end{pmatrix} \]has rank :

The rank is 2.Row 2 is twice row 1 and row 4 is four times row 1, so they add nothing new. Row 3, \((1, 3, 3, 3)\), is not a multiple of row 1, because its first entry differs from the rest. So the row space is spanned by two independent rows:\[ R_{3} - R_{1} = (0, 2, 2, 2) \neq 0 \]Every \(3 \times 3\) minor vanishes, since any three rows include two proportional ones, while the minor from rows 1 and 3, columns 1 and 2, is \(1 \cdot 3 - 1 \cdot 1 = 2 \ne 0\). Rank 1 misses row 3; ranks 3 and 4 treat the proportional rows as independent.

Q: Let A be a 5 × 5 real matrix and let B be the adjoint of A. Then :

A is invertible if and only if B is invertible.The adjoint satisfies \(A\,\mathrm{adj}(A) = |A|\, I\), and for an \(n \times n\) matrix \(|\mathrm{adj}(A)| = |A|^{n-1}\). With \(n = 5\), \(|B| = |A|^{4}\), which is non-zero exactly when \(|A|\) is non-zero.The other options fail in general. The determinants \(|A|\) and \(|A|^{4}\) differ unless \(|A|\) is 0 or 1; ranks differ too (if A has rank 3 or less, adj A is the zero matrix). And \(AB = |A|\, I_{5}\), not \(5I_{5}\); the 5 in option D is the size of the matrix, not its determinant.

Let the linear transformation \(T : F^{2} \to F^{3}\) be defined by \(T(x_{1}, x_{2}) = (x_{1}, x_{1} + x_{2}, x_{2})\). Then the nullity of \(T\) is

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The nullity of \(T\) is 0.

The kernel consists of the vectors with \(T(x_{1}, x_{2}) = (0, 0, 0)\): the first and third coordinates force \(x_{1} = 0\) and \(x_{2} = 0\). So the kernel is \(\{0\}\) and the nullity is 0.

By rank-nullity, \(\dim F^{2} = 2 = \text{rank} + \text{nullity}\), and the images \(T(1, 0) = (1, 1, 0)\) and \(T(0, 1) = (0, 1, 1)\) are independent, so the rank is 2, confirming nullity 0. The other options confuse nullity with the dimension of the codomain (3), of the domain (2), or with the codimension of the image (1).

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Let \(T : \mathbb{R}^{4} \to \mathbb{R}^{4}\) be a linear transformation and the null space of T be the subspace of \(\mathbb{R}^{4}\) given by

\[ \begin{aligned} \{(x_{1},x_{2},x_{3},x_{4}) &\in \mathbb{R}^{4} : \\ 4x_{1}+3x_{2}+2x_{3}+x_{4} &= 0\}. \end{aligned} \]

If \(Rank(T - 3I) = 3\), where I is the identity map on \(\mathbb{R}^{4}\), then the minimal polynomial of T is

A.\(x(x-3)\)
B.\(x(x-3)^{3}\)
C.\(x^{3}(x-3)\)
D.\(x^{2}(x-3)^{2}\)

Exam Year(s)

GATE Mathematics - 2023,

Answer: A

The minimal polynomial is x(x - 3).

The null space of T is a hyperplane in \(\mathbb{R}^{4}\), of dimension 3, so 0 is an eigenvalue with geometric multiplicity 3. Rank(T - 3I) = 3 means T - 3I has nullity 1, so 3 is an eigenvalue with geometric multiplicity 1.

The algebraic multiplicities are at least the geometric ones and add up to 4, so they are exactly 3 and 1. Each eigenvalue then has as many independent eigenvectors as its multiplicity, so T is diagonalizable with eigenvalues 0, 0, 0, 3, and its minimal polynomial has only simple factors: x(x - 3).

The options with higher powers would describe non-diagonalizable maps, which would need fewer eigenvectors than the null space supplies.

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\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 3 & -2 \\ 0 & -6 & 4 \end{pmatrix}, \]

then the rank of the matrix \(AA^{t}\) is :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The rank of \(AA^{t}\) is 2.

For a real matrix, \(\text{rank}(AA^{t}) = \text{rank}(A)\), because \(AA^{t}x = 0\) gives \(x^{t}AA^{t}x = |A^{t}x|^{2} = 0\), so \(AA^{t}\) and \(A^{t}\) have the same null space.

So find the rank of A. The third row \((0, -6, 4)\) is \(-2\) times the second row \((0, 3, -2)\), so the rows are dependent and the rank is at most 2. The first two rows are clearly independent (only the first has a nonzero first entry). Hence rank A = 2, and \(AA^{t}\) has rank 2 as well.

Rank 3 would need \(\det A \ne 0\), but \(\det A = 1 \cdot (3 \cdot 4 - (-2)(-6)) = 0\). Rank 1 or 0 would need the first two rows to be dependent, which they are not.

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If P is a \(3 \times 3\) matrix of rank three and :

\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}, \]

then the rank of PA is :

A.1
B.2
C.3
D.Depends upon matrix P

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The rank of PA is 2.

A \(3 \times 3\) matrix of rank three is invertible, and multiplying by an invertible matrix does not change rank. PA and A have the same null space, since \(PAx = 0\) gives \(Ax = P^{-1}0 = 0\). So rank PA = rank A.

For A, the rows are in arithmetic progression: row 3 minus row 2 equals row 2 minus row 1, both \((3, 3, 3)\). So row 1 - 2 row 2 + row 3 = 0 and \(\det A = 0\), while the first two rows are independent. Hence rank A = 2 and rank PA = 2.

3 would need A invertible. 1 would need all rows proportional. 'Depends upon matrix P' is the trap: it would be right if P could be singular, but rank three rules that out.

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Let \(A = \begin{pmatrix} 1 & 2 \\ 3 & 6 \\ 1 & 2 \end{pmatrix}\) and \(f : \mathbf{R}^{2} \to \mathbf{R}^{3}\) be defined as \(f(x) = Ax\). Then which of the following statements is false ?

A.\(f\) is continuous function
B.\(f\) is differentiable function
C.\(f\) maps open neighbourhood of origin to an open neighbourhood
D.the directional derivative of \(f\) in direction of (1, 0) is \(\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The false statement is \(f\) maps an open neighbourhood of the origin to an open neighbourhood.

The columns of A are proportional, (2, 6, 2) = 2(1, 3, 1), so A has rank 1. The image of \(\mathbf{R}^{2}\) is the line spanned by (1, 3, 1) in \(\mathbf{R}^{3}\), and a line contains no open set of \(\mathbf{R}^{3}\). So the image of a neighbourhood of the origin cannot be open.

The other statements are true. A linear map between finite dimensional spaces is continuous and differentiable, with derivative A at every point. The directional derivative in the direction (1, 0) is \(A(1, 0)^{t}\), the first column (1, 3, 1).

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Let A be a 4 × 3 matrix and B a 3 × 4 matrix. Which of the following is always true ?

A.det (AB) = 0, det (BA) need not be zero
B.det (AB) and det (BA) need not be zero
C.det (AB) = det (BA) = 0
D.det (AB) = 0 only if either AB = 0 or BA = 0

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The statement always true is det (AB) = 0, det (BA) need not be zero.

AB is 4 × 4, but its rank is at most the rank of A, which is at most 3. A 4 × 4 matrix of rank at most 3 is singular, so \(\det(AB) = 0\) always.

BA is 3 × 3 and can be invertible. For example, if A is the 4 × 3 matrix with \(I_{3}\) on top and a zero row below, and B is \([I_{3}\ 0]\), then \(BA = I_{3}\) with determinant 1.

So 'both need not be zero' and 'both are zero' are wrong, and det(AB) = 0 holds even when neither AB nor BA is the zero matrix.

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Let A be a 6 × 7 matrix of rank 6 and B be a 7 × 5 matrix of rank 4, then the rank of AB is :

A.5
B.6
C.4
D.Cannot be determined exactly, but less than or equal to 4

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The rank of AB cannot be determined exactly, but is at most 4.

Upper bound: \(\text{rank}(AB) \le \min(\text{rank } A, \text{rank } B) = \min(6, 4) = 4\).

Lower bound, by Sylvester's inequality with inner dimension 7:

\[ \begin{aligned} \text{rank}(AB) &\ge 6 + 4 - 7 \\ &= 3 \end{aligned} \]

Both values occur. A has a one-dimensional null space; if that null space lies inside the column space of B, AB loses one dimension and has rank 3, otherwise it keeps rank 4. Since the data do not decide this, the rank is 3 or 4, so the exact values 4, 5 and 6 are all wrong as certainties (5 and 6 are impossible).

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The matrix

\[ \begin{pmatrix} 1 & 1 & 1 & 1 \\ 2 & 2 & 2 & 2 \\ 1 & 3 & 3 & 3 \\ 4 & 4 & 4 & 4 \end{pmatrix} \]

has rank :

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The rank is 2.

Row 2 is twice row 1 and row 4 is four times row 1, so they add nothing new. Row 3, \((1, 3, 3, 3)\), is not a multiple of row 1, because its first entry differs from the rest. So the row space is spanned by two independent rows:

\[ R_{3} - R_{1} = (0, 2, 2, 2) \neq 0 \]

Every \(3 \times 3\) minor vanishes, since any three rows include two proportional ones, while the minor from rows 1 and 3, columns 1 and 2, is \(1 \cdot 3 - 1 \cdot 1 = 2 \ne 0\). Rank 1 misses row 3; ranks 3 and 4 treat the proportional rows as independent.

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Let A be a 5 × 5 real matrix and let B be the adjoint of A. Then :

A.A and B have same rank
B.A and B have same determinant
C.A is invertible if and only if B is invertible
D.\(AB = 5 I_{5}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

A is invertible if and only if B is invertible.

The adjoint satisfies \(A\,\mathrm{adj}(A) = |A|\, I\), and for an \(n \times n\) matrix \(|\mathrm{adj}(A)| = |A|^{n-1}\). With \(n = 5\), \(|B| = |A|^{4}\), which is non-zero exactly when \(|A|\) is non-zero.

The other options fail in general. The determinants \(|A|\) and \(|A|^{4}\) differ unless \(|A|\) is 0 or 1; ranks differ too (if A has rank 3 or less, adj A is the zero matrix). And \(AB = |A|\, I_{5}\), not \(5I_{5}\); the 5 in option D is the size of the matrix, not its determinant.

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