Linear Transformations Questions and Answers

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.

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 \(m = n\).

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.

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.

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.

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.

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.

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.

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.