Linear Transformations Questions and Answers

Q: Let V, W be two complex vector spaces and \(T \in L(V, W)\). If T has matrix representation :\[ \begin{bmatrix} 2 & 1 + i & 3 \\ 4 + i & 1 - i & i \end{bmatrix} \]Which of the following matrices represents the adjoint map \(T^{*}\) ?

\(T^{*}\) is represented by the conjugate transpose, \(\begin{pmatrix} 2 & 4 - i \\ 1 - i & 1 + i \\ 3 & -i \end{pmatrix}\).With respect to orthonormal bases, the matrix of the adjoint is the conjugate transpose \(A^{*} = \bar{A}^{T}\). Transpose the \(2 \times 3\) matrix (rows become columns) and conjugate every entry: \(1 + i \to 1 - i\), \(4 + i \to 4 - i\), \(1 - i \to 1 + i\), \(i \to -i\).Option B is the transpose without conjugation, option A conjugates without transposing (and has the wrong shape for a map from W to V), and option D is the original matrix.

Q: If \(T : \mathbf{R}^{n} \to \mathbf{R}^{m}\) is a linear map, then it is represented by a matrix of order :

The matrix is of order \(m \times n\).The matrix of \(T\) has one column for each basis vector of the domain \(\mathbf{R}^{n}\), namely \(T(e_{j})\), and each column is a vector in \(\mathbf{R}^{m}\), so it has \(m\) entries. That gives \(m\) rows and \(n\) columns, and \(T(x) = Ax\) is defined because an \(m \times n\) matrix times an \(n \times 1\) vector is \(m \times 1\).\(n \times m\) is the transpose, which maps \(\mathbf{R}^{m}\) to \(\mathbf{R}^{n}\). Square orders \(m \times m\) or \(n \times n\) would fit only when \(m = n\).

Q: \(T_{1}\), \(T_{2}\) and \(T_{3}\) are three maps defined on \(\mathbf{R}^{3}\) as :\[ T_{1}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + 1 \\ y \\ z \end{bmatrix}, \]\[ T_{2}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} xy \\ z \end{bmatrix} \]\[ T_{3}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + y \\ y + z \\ z + x \end{bmatrix}. \]Which of these maps are linear ?

Only \(T_{3}\) is linear.\(T_{3}\) sends \((x, y, z)\) to \((x + y, y + z, z + x)\); each component is a linear combination of \(x, y, z\), so it is the linear map given by a 3 by 3 matrix.\(T_{1}\) is a translation: \(T_{1}(0) = (1, 0, 0) \ne 0\), and a linear map must send 0 to 0. \(T_{2}\) contains the product \(xy\): doubling the input multiplies the first component by 4, not 2, so it is not homogeneous. Any option listing \(T_{1}\) or \(T_{2}\) is wrong.

Q: Let T be a linear operator on \(\mathbf{R}^{3}\), given by :\[ T\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2x + 2y + z \\ 2y + z \\ 2x + 3y + z \end{bmatrix} \]Then the matrix representation of T with respect to standard basis of \(\mathbf{R}^{3}\) is :

The matrix is \(\begin{bmatrix} 2 & 2 & 1 \\ 0 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}\).The columns are the images of the standard basis vectors:\[ \begin{aligned} T(e_{1}) &= (2, 0, 2) \\ T(e_{2}) &= (2, 2, 3) \\ T(e_{3}) &= (1, 1, 1) \end{aligned} \]Placed as columns, they give rows (2, 2, 1), (0, 2, 1), (2, 3, 1), which are just the coefficients of each output component. Option C is the transpose, with the images written as rows, and the other two rearrange the entries.

Q: Let V and V' be finite dimensional vector spaces over K and let \(\{\sqrt{1}, \ldots, \sqrt{n}\}\) be a basis of V. If T \(\in\) L(V, V') and T is injective, then which of the following statements is FALSE ?

The false statement is dim V = dim V', option C.The basis is printed with square-root signs (\(\sqrt{1}, \ldots, \sqrt{n}\)), which look like a typesetting slip for basis vectors. The argument below uses only that it is a basis of V with n elements and that T is injective.Injective means \(\operatorname{Ker} T = \{0\}\), which is option D, so D is true. By rank-nullity:\[ \begin{aligned} \dim V &= \dim \operatorname{Ker} T + \dim \operatorname{Im} T \\ &= \dim \operatorname{Im} T \le \dim V' \end{aligned} \]so \(\dim V \le \dim V'\), option B, is true. Equality needs T to be onto as well, which is not given: the map \(x \mapsto (x, 0)\) from \(\mathbf{R}\) to \(\mathbf{R}^{2}\) is injective, yet 1 is not 2. So C can fail. An injective map takes a linearly independent set to a linearly independent set, so the images of the basis vectors stay independent and A is true.

Q: Let T be a linear transformation from R3 to R given by\[ T[x, y, z]^{T} = 2x - 5y + 9z, \]where [x, y, z]T donote the transpose of the vector [x, y, z]. Then, the dimension of the null space of T is :

The null space has dimension 2.T maps \(\mathbf{R}^{3}\) onto R: it is not the zero map, since \(T(1, 0, 0) = 2\), so its rank is 1. By the rank-nullity theorem\[ \dim N(T) = 3 - \text{rank}(T) = 3 - 1 = 2 \]Geometrically the null space is the plane \(2x - 5y + 9z = 0\) through the origin, which is two-dimensional. 1 would be a line and 3 would make T the zero map; 0 is impossible for a map from a 3-dimensional space to a 1-dimensional one.

Q: Let \(P : \mathbf{R}^{3} \to \mathbf{R}^{3}\) represent the projection of R3 onto xy-plane along z-axis. Then :

The projection satisfies P3 = P.P sends (x, y, z) to (x, y, 0). Applying it again changes nothing, so \(P^{2} = P\), and then \(P^{3} = P^{2}P = P \cdot P = P\).\(P^{2} = I\) is false, because \(P^{2} = P \neq I\). P is not invertible, since it sends (0, 0, 1) to 0. And P is diagonalizable: in the standard basis its matrix is already diagonal, diag(1, 1, 0), with eigenvalues 1, 1, 0.

Q: Let \(T : \mathbf{R}^{3} \to \mathbf{R}^{2}\) be a linear map such that :\[ T\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad T\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]and\[ T\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \]Then \(T\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} =\)

\(T(1, 1, 1)\) is \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\).Write the vector as a combination of the vectors whose images are given:\[ (1, 1, 1) = (1, 0, 0) + (0, 1, 1) \]By linearity,\[ T(1, 1, 1) = (1, 0) + (1, 1) = (2, 1) \]The information \(T(0, 1, 0) = (0, 1)\) is not needed. Adding all three given images gives (2, 2), which is the trap in option D; (1, 1) is the image of (0, 1, 1) alone.

11.

Let V, W be two complex vector spaces and \(T \in L(V, W)\). If T has matrix representation :

\[ \begin{bmatrix} 2 & 1 + i & 3 \\ 4 + i & 1 - i & i \end{bmatrix} \]

Which of the following matrices represents the adjoint map \(T^{*}\) ?

A.\(\begin{pmatrix} 2 & 1 - i & 3 \\ 4 - i & 1 + i & -i \end{pmatrix}\)
B.\(\begin{pmatrix} 2 & 4 + i \\ 1 + i & 1 - i \\ 3 & i \end{pmatrix}\)
C.\(\begin{pmatrix} 2 & 4 - i \\ 1 - i & 1 + i \\ 3 & -i \end{pmatrix}\)
D.\(\begin{pmatrix} 2 & 1 + i & 3 \\ 4 + i & 1 - i & i \end{pmatrix}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

\(T^{*}\) is represented by the conjugate transpose, \(\begin{pmatrix} 2 & 4 - i \\ 1 - i & 1 + i \\ 3 & -i \end{pmatrix}\).

With respect to orthonormal bases, the matrix of the adjoint is the conjugate transpose \(A^{*} = \bar{A}^{T}\). Transpose the \(2 \times 3\) matrix (rows become columns) and conjugate every entry: \(1 + i \to 1 - i\), \(4 + i \to 4 - i\), \(1 - i \to 1 + i\), \(i \to -i\).

Option B is the transpose without conjugation, option A conjugates without transposing (and has the wrong shape for a map from W to V), and option D is the original matrix.

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12.

If \(T : \mathbf{R}^{n} \to \mathbf{R}^{m}\) is a linear map, then it is represented by a matrix of order :

A.\(m \times n\)
B.\(m \times m\)
C.\(n \times n\)
D.\(n \times m\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The matrix is of order \(m \times n\).

The matrix of \(T\) has one column for each basis vector of the domain \(\mathbf{R}^{n}\), namely \(T(e_{j})\), and each column is a vector in \(\mathbf{R}^{m}\), so it has \(m\) entries. That gives \(m\) rows and \(n\) columns, and \(T(x) = Ax\) is defined because an \(m \times n\) matrix times an \(n \times 1\) vector is \(m \times 1\).

\(n \times m\) is the transpose, which maps \(\mathbf{R}^{m}\) to \(\mathbf{R}^{n}\). Square orders \(m \times m\) or \(n \times n\) would fit only when \(m = n\).

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13.

\(T_{1}\), \(T_{2}\) and \(T_{3}\) are three maps defined on \(\mathbf{R}^{3}\) as :

\[ T_{1}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + 1 \\ y \\ z \end{bmatrix}, \]

\[ T_{2}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} xy \\ z \end{bmatrix} \]

\[ T_{3}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + y \\ y + z \\ z + x \end{bmatrix}. \]

Which of these maps are linear ?

A.\(T_{1}, T_{2}, T_{3}\)
B.\(T_{1}, T_{2},\)
C.\(T_{3}\)
D.\(T_{1}, T_{3}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

Only \(T_{3}\) is linear.

\(T_{3}\) sends \((x, y, z)\) to \((x + y, y + z, z + x)\); each component is a linear combination of \(x, y, z\), so it is the linear map given by a 3 by 3 matrix.

\(T_{1}\) is a translation: \(T_{1}(0) = (1, 0, 0) \ne 0\), and a linear map must send 0 to 0. \(T_{2}\) contains the product \(xy\): doubling the input multiplies the first component by 4, not 2, so it is not homogeneous. Any option listing \(T_{1}\) or \(T_{2}\) is wrong.

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14.

Let T be a linear operator on \(\mathbf{R}^{3}\), given by :

\[ T\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2x + 2y + z \\ 2y + z \\ 2x + 3y + z \end{bmatrix} \]

Then the matrix representation of T with respect to standard basis of \(\mathbf{R}^{3}\) is :

A.\(\begin{bmatrix} 2 & 2 & 1 \\ 0 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}\)
B.\(\begin{bmatrix} 2 & 3 & 1 \\ 0 & 2 & 1 \\ 2 & 2 & 1 \end{bmatrix}\)
C.\(\begin{bmatrix} 2 & 0 & 2 \\ 2 & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix}\)
D.\(\begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 2 \\ 2 & 2 & 3 \end{bmatrix}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The matrix is \(\begin{bmatrix} 2 & 2 & 1 \\ 0 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}\).

The columns are the images of the standard basis vectors:

\[ \begin{aligned} T(e_{1}) &= (2, 0, 2) \\ T(e_{2}) &= (2, 2, 3) \\ T(e_{3}) &= (1, 1, 1) \end{aligned} \]

Placed as columns, they give rows (2, 2, 1), (0, 2, 1), (2, 3, 1), which are just the coefficients of each output component. Option C is the transpose, with the images written as rows, and the other two rearrange the entries.

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15.

Let V and V' be finite dimensional vector spaces over K and let \(\{\sqrt{1}, \ldots, \sqrt{n}\}\) be a basis of V. If T \(\in\) L(V, V') and T is injective, then which of the following statements is FALSE ?

A.T\(\sqrt{1}\), \(\ldots\), T\(\sqrt{n}\) are linearly independent
B.dim V \(\le\) dim V'
C.dim V = dim V'
D.Ker T = {0}

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The false statement is dim V = dim V', option C.

The basis is printed with square-root signs (\(\sqrt{1}, \ldots, \sqrt{n}\)), which look like a typesetting slip for basis vectors. The argument below uses only that it is a basis of V with n elements and that T is injective.

Injective means \(\operatorname{Ker} T = \{0\}\), which is option D, so D is true. By rank-nullity:

\[ \begin{aligned} \dim V &= \dim \operatorname{Ker} T + \dim \operatorname{Im} T \\ &= \dim \operatorname{Im} T \le \dim V' \end{aligned} \]

so \(\dim V \le \dim V'\), option B, is true. Equality needs T to be onto as well, which is not given: the map \(x \mapsto (x, 0)\) from \(\mathbf{R}\) to \(\mathbf{R}^{2}\) is injective, yet 1 is not 2. So C can fail. An injective map takes a linearly independent set to a linearly independent set, so the images of the basis vectors stay independent and A is true.

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16.

Let T be a linear transformation from R² to R² such that :

\[ T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\ T\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}. \]

Then, \(T\begin{bmatrix} 3 \\ 4 \end{bmatrix}\) :

A.is \(\begin{bmatrix} 3 \\ 2 \end{bmatrix}\)
B.is \(\begin{bmatrix} 4 \\ 5 \end{bmatrix}\)
C.is \(\begin{bmatrix} 5 \\ 3 \end{bmatrix}\)
D.Cannot be determined from the given information

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The values determine T completely, and \(T(3, 4)^{T} = (5, 3)^{T}\).

The vectors (1, 0) and (1, 2) form a basis of R², so a linear map is fixed by their images. Since \((0, 2) = (1, 2) - (1, 0)\):

\[ \begin{aligned} T(0, 2) &= (2, 1) - (1, 1) = (1, 0) \\ T(0, 1) &= (1/2, 0) \\ T(3, 4) &= 3T(1, 0) + 4T(0, 1) \\ &= (3, 3) + (2, 0) = (5, 3) \end{aligned} \]

As a check, \((3, 4) = 1 \cdot (1, 0) + 2 \cdot (1, 2)\), and \(1 \cdot (1, 1) + 2 \cdot (2, 1) = (5, 3)\). 'Cannot be determined' would be right only if the two given vectors were dependent. (3, 2) is \(T(1, 0) + T(1, 2)\), the image of (2, 2) rather than (3, 4), and (4, 5) is not the image of (3, 4) under any map satisfying both conditions.

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17.

Let T be a linear transformation from R³ to R given by

\[ T[x, y, z]^{T} = 2x - 5y + 9z, \]

where [x, y, z]^T donote the transpose of the vector [x, y, z]. Then, the dimension of the null space of T is :

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The null space has dimension 2.

T maps \(\mathbf{R}^{3}\) onto R: it is not the zero map, since \(T(1, 0, 0) = 2\), so its rank is 1. By the rank-nullity theorem

\[ \dim N(T) = 3 - \text{rank}(T) = 3 - 1 = 2 \]

Geometrically the null space is the plane \(2x - 5y + 9z = 0\) through the origin, which is two-dimensional. 1 would be a line and 3 would make T the zero map; 0 is impossible for a map from a 3-dimensional space to a 1-dimensional one.

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18.

Let \(P : \mathbf{R}^{3} \to \mathbf{R}^{3}\) represent the projection of R³ onto xy-plane along z-axis. Then :

A.\(P^{2} = I\)
B.\(P^{3} = P\)
C.P is invertible
D.P is not diagonalizable

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The projection satisfies P³ = P.

P sends (x, y, z) to (x, y, 0). Applying it again changes nothing, so \(P^{2} = P\), and then \(P^{3} = P^{2}P = P \cdot P = P\).

\(P^{2} = I\) is false, because \(P^{2} = P \neq I\). P is not invertible, since it sends (0, 0, 1) to 0. And P is diagonalizable: in the standard basis its matrix is already diagonal, diag(1, 1, 0), with eigenvalues 1, 1, 0.

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19.

Let \(T : \mathbf{R}^{3} \to \mathbf{R}^{2}\) be a linear map such that :

\[ T\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad T\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

and

\[ T\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \]

Then \(T\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} =\)

A.\(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\)
B.\(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
C.\(\begin{bmatrix} -1 \\ 2 \end{bmatrix}\)
D.\(\begin{bmatrix} 2 \\ 2 \end{bmatrix}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

\(T(1, 1, 1)\) is \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\).

Write the vector as a combination of the vectors whose images are given:

\[ (1, 1, 1) = (1, 0, 0) + (0, 1, 1) \]

By linearity,

\[ T(1, 1, 1) = (1, 0) + (1, 1) = (2, 1) \]

The information \(T(0, 1, 0) = (0, 1)\) is not needed. Adding all three given images gives (2, 2), which is the trap in option D; (1, 1) is the image of (0, 1, 1) alone.

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