Vector Spaces and Subspaces Questions and Answers

Q: Let\[ \begin{aligned} S = \Big\{ w &= \begin{bmatrix} w_{1} \\ w_{2} \\ w_{3} \end{bmatrix} \in \mathbb{R}^{3} : \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix} \begin{bmatrix} w_{1} & w_{2} & w_{3} \end{bmatrix} \\ &\text{is diagonalizable and } \|w\| = 1 \Big\}, \end{aligned} \]where \(\|w\| = (w_{1}^{2} + w_{2}^{2} + w_{3}^{2})^{1/2}\). Then, which one of the following is TRUE?

S is neither compact nor connected.Write \(v = (1, -3, 2)^{T}\). The matrix \(vw^{T}\) has rank 1, and its only possibly non-zero eigenvalue is its trace, \(w^{T}v = w_{1} - 3w_{2} + 2w_{3}\). A rank-1 matrix is diagonalizable exactly when this trace is non-zero; if it is zero the matrix is nilpotent and not zero, so not diagonalizable.So S is the unit sphere with the great circle \(w_{1} - 3w_{2} + 2w_{3} = 0\) removed. That leaves two open hemispheres. S is not closed, since points on the removed circle are limits of points of S, so it is not compact; and it falls into two disjoint open pieces, so it is not connected.

Q: Consider the following subsets of the Euclidean space \(\mathbb{R}^{4}\):\[ \begin{aligned} S &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = 0\}, \\ T &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = 1\}, \text{ and} \\ U &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = -1\}. \end{aligned} \]Then, which one of the following is TRUE?

S and T are connected, but U is not connected.Write \(r^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}\).S: \(x_{4} = \pm r\), a double cone whose two halves meet at the origin, so it is connected (every point joins the origin along a straight segment in S).T: \(r^{2} = 1 + x_{4}^{2}\). For each value of \(x_{4}\) the slice is a 2-sphere of radius \(\sqrt{1 + x_{4}^{2}}\), so T is homeomorphic to \(S^{2} \times \mathbb{R}\), which is connected.U: \(x_{4}^{2} = 1 + r^{2} \ge 1\), so \(x_{4} \ge 1\) or \(x_{4} \le -1\). The two sheets are separated by the slab \(|x_{4}| \lt 1\), so U is not connected.Every other option misjudges one of these three.

Q: Let \(M_3(\mathbb{R})\) be the vector space of all \(3 \times 3\) real matrices over \(\mathbb{R}\) under usual matrix addition and scalar multiplication. Let \(T_3(\mathbb{R})\) be the set of all \(3 \times 3\) real upper triangular matrices. Which one of the following is TRUE?

The quotient is isomorphic to the space of 3 by 3 real skew symmetric matrices.Finite-dimensional real vector spaces are isomorphic exactly when their dimensions are equal. \(M_3(\mathbb{R})\) has dimension 9 and the upper triangular matrices have dimension 6 (the diagonal and the three entries above it), so\[ \dim M_3(\mathbb{R})/T_3(\mathbb{R}) = 9 - 6 = 3. \]Skew symmetric 3 by 3 matrices are fixed by the three entries above a zero diagonal, so they also have dimension 3. Symmetric and lower triangular matrices each have dimension 6, and trace-zero matrices have dimension 8, so none of those is isomorphic to the quotient.

Q: Which one of the following statements is TRUE?

The true statement is that the polynomials with real coefficients form a Unique Factorization Domain.\(\mathbb{R}[x]\) is a polynomial ring over a field, hence a Euclidean domain (division with remainder by degree), hence a principal ideal domain and so a UFD. Every nonzero non-unit factors uniquely into irreducibles, which over \(\mathbb{R}\) are linear and irreducible quadratic factors.The other rings are not even integral domains. \(2 \times 2\) real matrices are not commutative and have zero divisors. Real valued functions on a set with two or more points have zero divisors: a function vanishing on one half times one vanishing on the other is 0. The same construction works for Riemann integrable functions on [0,1].

Let

\[ \begin{aligned} S = \Big\{ w &= \begin{bmatrix} w_{1} \\ w_{2} \\ w_{3} \end{bmatrix} \in \mathbb{R}^{3} : \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix} \begin{bmatrix} w_{1} & w_{2} & w_{3} \end{bmatrix} \\ &\text{is diagonalizable and } \|w\| = 1 \Big\}, \end{aligned} \]

where \(\|w\| = (w_{1}^{2} + w_{2}^{2} + w_{3}^{2})^{1/2}\). Then, which one of the following is TRUE?

A.S is compact and connected
B.S is neither compact nor connected
C.S is compact but not connected
D.S is connected but not compact

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

S is neither compact nor connected.

Write \(v = (1, -3, 2)^{T}\). The matrix \(vw^{T}\) has rank 1, and its only possibly non-zero eigenvalue is its trace, \(w^{T}v = w_{1} - 3w_{2} + 2w_{3}\). A rank-1 matrix is diagonalizable exactly when this trace is non-zero; if it is zero the matrix is nilpotent and not zero, so not diagonalizable.

So S is the unit sphere with the great circle \(w_{1} - 3w_{2} + 2w_{3} = 0\) removed. That leaves two open hemispheres. S is not closed, since points on the removed circle are limits of points of S, so it is not compact; and it falls into two disjoint open pieces, so it is not connected.

Discuss in Forum

Consider the following subsets of the Euclidean space \(\mathbb{R}^{4}\):

\[ \begin{aligned} S &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = 0\}, \\ T &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = 1\}, \text{ and} \\ U &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = -1\}. \end{aligned} \]

Then, which one of the following is TRUE?

A.S is connected, but T and U are not connected
B.T and U are connected, but S is not connected
C.S and U are connected, but T is not connected
D.S and T are connected, but U is not connected

Exam Year(s)

GATE Mathematics - 2025,

Answer: D

S and T are connected, but U is not connected.

Write \(r^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}\).

S: \(x_{4} = \pm r\), a double cone whose two halves meet at the origin, so it is connected (every point joins the origin along a straight segment in S).
T: \(r^{2} = 1 + x_{4}^{2}\). For each value of \(x_{4}\) the slice is a 2-sphere of radius \(\sqrt{1 + x_{4}^{2}}\), so T is homeomorphic to \(S^{2} \times \mathbb{R}\), which is connected.
U: \(x_{4}^{2} = 1 + r^{2} \ge 1\), so \(x_{4} \ge 1\) or \(x_{4} \le -1\). The two sheets are separated by the slab \(|x_{4}| \lt 1\), so U is not connected.

Every other option misjudges one of these three.

Discuss in Forum

Let \(M_3(\mathbb{R})\) be the vector space of all \(3 \times 3\) real matrices over \(\mathbb{R}\) under usual matrix addition and scalar multiplication. Let \(T_3(\mathbb{R})\) be the set of all \(3 \times 3\) real upper triangular matrices. Which one of the following is TRUE?

A.The quotient space \(M_3(\mathbb{R})/T_3(\mathbb{R})\) is isomorphic to the vector space of all \(3 \times 3\) real skew symmetric matrices over \(\mathbb{R}\) under usual matrix addition and scalar multiplication.
B.The quotient space \(M_3(\mathbb{R})/T_3(\mathbb{R})\) is isomorphic to the vector space of all \(3 \times 3\) real symmetric matrices over \(\mathbb{R}\) under usual matrix addition and scalar multiplication.
C.The quotient space \(M_3(\mathbb{R})/T_3(\mathbb{R})\) is isomorphic to the vector space of all \(3 \times 3\) real lower triangular matrices over \(\mathbb{R}\) under usual matrix addition and scalar multiplication.
D.The quotient space \(M_3(\mathbb{R})/T_3(\mathbb{R})\) is isomorphic to the vector space of all \(3 \times 3\) real matrices with trace zero over \(\mathbb{R}\) under usual matrix addition and scalar multiplication.

Exam Year(s)

GATE Mathematics - 2026,

Answer: A

The quotient is isomorphic to the space of 3 by 3 real skew symmetric matrices.

Finite-dimensional real vector spaces are isomorphic exactly when their dimensions are equal. \(M_3(\mathbb{R})\) has dimension 9 and the upper triangular matrices have dimension 6 (the diagonal and the three entries above it), so

\[ \dim M_3(\mathbb{R})/T_3(\mathbb{R}) = 9 - 6 = 3. \]

Skew symmetric 3 by 3 matrices are fixed by the three entries above a zero diagonal, so they also have dimension 3. Symmetric and lower triangular matrices each have dimension 6, and trace-zero matrices have dimension 8, so none of those is isomorphic to the quotient.

Discuss in Forum

Which one of the following statements is TRUE?

A.The set of all \(2 \times 2\) real matrices under usual matrix addition and multiplication is a Unique Factorization Domain.
B.The set of all real valued functions under usual function addition and multiplication is a Unique Factorization Domain.
C.The set of all polynomials with coefficients in \(\mathbb{R}\) under usual polynomial addition and multiplication is a Unique Factorization Domain.
D.The set of Riemann integrable functions in [0,1] under usual function addition and multiplication is a Unique Factorization Domain.

Exam Year(s)

GATE Mathematics - 2026,

Answer: C

The true statement is that the polynomials with real coefficients form a Unique Factorization Domain.

\(\mathbb{R}[x]\) is a polynomial ring over a field, hence a Euclidean domain (division with remainder by degree), hence a principal ideal domain and so a UFD. Every nonzero non-unit factors uniquely into irreducibles, which over \(\mathbb{R}\) are linear and irreducible quadratic factors.

The other rings are not even integral domains. \(2 \times 2\) real matrices are not commutative and have zero divisors. Real valued functions on a set with two or more points have zero divisors: a function vanishing on one half times one vanishing on the other is 0. The same construction works for Riemann integrable functions on [0,1].

Discuss in Forum

Let \(e_1, e_2, e_3, e_4, e_5\) denote the standard basis vectors of the real vector space \(\mathbb{R}^{5}\). Let \(V\) be the subspace of \(\mathbb{R}^{5}\) spanned by the vectors \(e_1 + e_2, e_2 + e_3, e_3 + e_4 + e_5\). Consider the real vector space \(M_{2,5}(\mathbb{R})\). Let

\(S = \{P \in M_{2,5}(\mathbb{R}) \mid \text{nullspace}(P) = V\}\).

Which ONE of the following statements is TRUE?

A.\(S\) is the empty set.
B.\(S = \{0\}\).
C.\(S\) is non-empty and \(S\) is not a subspace of \(M_{2,5}(\mathbb{R})\).
D.\(S\) is a nonzero subspace of \(M_{2,5}(\mathbb{R})\).

Exam Year(s)

IIT JAM - 2026,

Answer: C

The TRUE statement is: \(S\) is non-empty and \(S\) is not a subspace of \(M_{2,5}(\mathbb{R})\).

The three spanning vectors are linearly independent, so \(\dim V = 3\). A matrix \(P \in M_{2,5}(\mathbb{R})\) has rank at most 2, so by rank-nullity its null space has dimension \(5 - \text{rank}(P) \geq 3\). It equals \(V\) exactly when \(P\) has rank 2 and its two rows span the orthogonal complement \(V^{\perp}\), which has dimension 2.

Such matrices exist: the rows \((1,-1,1,-1,0)\) and \((1,-1,1,0,-1)\) are independent and orthogonal to all three spanning vectors. So \(S\) is not empty.

The zero matrix is not in \(S\), because its null space is all of \(\mathbb{R}^{5}\), not \(V\). A subspace must contain 0, so \(S\) is not a subspace, and \(S = \{0\}\) is wrong for the same reason. Closure fails too: if \(P \in S\) then \(-P \in S\), but \(P + (-P) = 0 \notin S\).

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Vector Spaces and Subspaces Questions and Answers || Vector Spaces and Subspaces MCQ Questions and Answers || Vector Spaces and Subspaces GK Questions and Answers || Vector Spaces and Subspaces GK MCQ Questions || Vector Spaces and Subspaces Multiple Choice Questions and Answers || Vector Spaces and Subspaces GK || GK on Vector Spaces and Subspaces || Linear Algebra Questions and Answers || Linear Algebra MCQ Questions and Answers || Linear Algebra GK Questions and Answers || GK on Linear Algebra

Vector Spaces and Subspaces Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications