Your Answer : (Not Answered)

Correct Answer : D

A is invertible.

The determinant equals the product of the eigenvalues, the roots of \((x - 1)(x^{2} + 1)\): \(1 \cdot i \cdot (-i) = 1 \ne 0\). Equivalently, 0 is not a root of the characteristic polynomial, so A is invertible.

The others fail. A is not diagonalizable or triangulable over \(\mathbf{R}\), because both need the characteristic polynomial to split into real linear factors, and \(x^{2} + 1\) does not. A is not nilpotent, since a nilpotent matrix has characteristic polynomial \(x^{3}\).

Your Answer : (Not Answered)

Correct Answer : C

Only (ii) defines an inner product on \(\mathbf{C}^{2}\).

(ii) is the standard Hermitian inner product: it is linear in \(x\), conjugate symmetric, and \((x, x) = |x_{1}|^{2} + |x_{2}|^{2} \gt 0\) for \(x \ne 0\).

(i) fails positivity: \(x = (1, 0)\) gives \((x, x) = 1 \cdot \bar{0} = 0\) with \(x \ne 0\). (iii) fails over \(\mathbf{C}\): \(x = (i, 0)\) gives \((x, x) = i^{2} = -1\), and it is not conjugate symmetric. So options naming (i) or (iii) are wrong.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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 \(m = n\).

Your Answer : (Not Answered)

Correct Answer : D

T is nilpotent.

T is the backward shift: \(e_{n} \mapsto e_{n-1} \mapsto \cdots \mapsto e_{1} \mapsto 0\). Applying it \(n\) times sends every basis vector to 0, so \(T^{n} = 0\).

It is linear by construction, since it is defined on a basis and extended linearly. It is not idempotent: \(T^{2}e_{3} = e_{1}\) while \(Te_{3} = e_{2}\) (for \(n \ge 3\); for \(n = 2\), \(T^{2} = 0 \ne T\)). Its kernel contains \(e_{1}\), so Ker T is not {0}.

Your Answer : (Not Answered)

Correct Answer : B

\(W^{\perp}\) is generated by \((1, 2, 3, 4)^{t}\) and \((1, -1, 1, 0)^{t}\).

W is the set of vectors orthogonal to \(a = (1, 2, 3, 4)\) and \(b = (1, -1, 1, 0)\), since its two equations say \(a \cdot x = 0\) and \(b \cdot x = 0\). So \(W = \mathrm{span}\{a, b\}^{\perp}\), and in a finite dimensional space \(W^{\perp} = \mathrm{span}\{a, b\}\). The two vectors are independent, so \(\dim W^{\perp} = 2\) and \(\dim W = 2\).

Option A spans a 3-dimensional space, too large. Options C and D give vectors that are not in the span of \(a\) and \(b\); for instance \((0, 0, 0, 1)\) would need a combination of \(a, b\) with zero first three entries, which forces both coefficients to be 0.

Your Answer : (Not Answered)

Correct Answer : B

The dimension is \(n + 1\).

Every polynomial of degree at most \(n\) is uniquely \(a_{0} + a_{1}x + \cdots + a_{n}x^{n}\), so \(\{1, x, x^{2}, \ldots, x^{n}\}\) is a basis. It has \(n + 1\) elements, because the constant term is counted as well as the powers 1 to \(n\).

The answer \(n\) forgets the constant polynomial 1, \(n^{2}\) has no connection with this space, and \(\infty\) is the dimension of the space of all polynomials, with no bound on the degree.

Your Answer : (Not Answered)

Correct Answer : C

The eigenvalues of an anti-Hermitian matrix are purely imaginary (or zero).

An anti-Hermitian matrix satisfies \( A^{\dagger} = -A \). If \( Av = \lambda v \), then

\[ \begin{aligned} v^{\dagger}Av &= \lambda\, v^{\dagger}v \\ (v^{\dagger}Av)^{*} &= v^{\dagger}A^{\dagger}v = -\lambda\, v^{\dagger}v \end{aligned} \]

so \( \lambda^{*} = -\lambda \), which means the real part of \( \lambda \) is zero. Equivalently, \( iA \) is Hermitian and has real eigenvalues.

Real eigenvalues, positive or negative, belong to Hermitian matrices, and a non-zero real part is exactly what the condition rules out.

Your Answer : (Not Answered)

Correct Answer : B

The eigenvalues of a nilpotent matrix are all zero.

Let \(A\vec{v} = \lambda\vec{v}\) with \(\vec{v} \neq 0\). Applying A repeatedly gives \(A^{K}\vec{v} = \lambda^{K}\vec{v}\). Since \(A^{K} = 0\), this means \(\lambda^{K}\vec{v} = 0\), and because \(\vec{v}\) is not the zero vector, \(\lambda^{K} = 0\), so \(\lambda = 0\).

Thus every eigenvalue is 0; equivalently the characteristic polynomial is \(\lambda^{n}\). They cannot be distinct (all are equal), and they are neither purely imaginary nor negative. A simple example is the matrix with a single 1 above the diagonal, whose square is zero.

Your Answer : (Not Answered)

Correct Answer : B

A is similar to the diagonal matrix diag(1, 1, 2).

Expanding along the first column, the characteristic polynomial is \((1 - \lambda)\) times that of the block \(\begin{pmatrix} 4 & -2 \\ 3 & -1 \end{pmatrix}\), whose trace is 3 and determinant is \(-4 + 6 = 2\). So the block has eigenvalues 1 and 2, and A has eigenvalues 1, 1, 2.

To decide between the diagonal form and a Jordan block for 1, compute

\[ A - I = \begin{pmatrix} 0 & 3 & -2 \\ 0 & 3 & -2 \\ 0 & 3 & -2 \end{pmatrix} \]

This has rank 1, so the eigenspace for 1 has dimension \(3 - 1 = 2\), equal to the algebraic multiplicity. A is diagonalisable, so it is similar to diag(1, 1, 2).

The matrix with a 1 above the diagonal in the eigenvalue-1 block would need a one-dimensional eigenspace. The options with eigenvalues 1, 2, 2 have the wrong trace: trace A = 1 + 4 - 1 = 4, not 5.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : C

The general solution is \(c_{1}e^{\lambda t}v + c_{2}e^{\lambda t}(tv + u)\), with \((A - \lambda I_{2})u = v\).

A non-diagonalisable \(2 \times 2\) matrix has a single eigenvector direction \(v\), and a generalised eigenvector \(u\) with \((A - \lambda I)u = v\). Check the second solution \(y = e^{\lambda t}(tv + u)\):

\[ \begin{aligned} y' &= \lambda e^{\lambda t}(tv + u) + e^{\lambda t}v \\ Ay &= e^{\lambda t}(t\lambda v + \lambda u + v) \end{aligned} \]

These agree. \(te^{\lambda t}v\) alone is not a solution (its derivative has an extra \(e^{\lambda t}v\) that \(A\) cannot produce), which rules out the first option; \(t^{2}e^{\lambda t}v\) fails the same way; and \(te^{\lambda t}(v + u)\) misplaces the factor \(t\), so it does not satisfy the system either.

Your Answer : (Not Answered)

Correct Answer : D

The FALSE statement is the one about cross products.

Because \(F\) is linear, \(\mathit{vect}(F(P),F(Q)) = F(Q) - F(P) = F(Q - P)\). So the condition says \(\|F(u)\| = \|u\|\) for every vector \(u\): \(F\) is an orthogonal map, \(F^{T}F = I\).

An orthogonal map has trivial kernel, so it is injective, and an injective linear map from \(\mathbb{R}^{3}\) to itself is also surjective. By polarisation, \(u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})\), so preserving lengths preserves dot products. Those three statements are true.

Cross products are not preserved. In general \(F(u) \times F(v) = (\det F)\,F(u \times v)\), which is not \(u \times v\). Take \(F\) the rotation by \(\frac{\pi}{2}\) about the \(z\)-axis, so \(F(e_2) = -e_1\) and \(F(e_3) = e_3\). Then \(F(e_2) \times F(e_3) = -e_1 \times e_3 = e_2\), but \(e_2 \times e_3 = e_1\).

Your Answer : (Not Answered)

Correct Answer : A

The characteristic polynomial is \(X^{4} - 8X^{2} + 16\).

Left multiplication by \(A\) acts on each column of \(B\) separately: if \(B = [b_{1}\ b_{2}]\), then \(AB = [Ab_{1}\ Ab_{2}]\). So \(T\) is \(A\) acting on two independent copies of \(\mathbb{C}^{2}\), and its characteristic polynomial is the square of that of \(A\).

\[ \begin{aligned} \det(XI - A) &= X^{2} - 4 \\ \chi_{T}(X) &= (X^{2} - 4)^{2} = X^{4} - 8X^{2} + 16 \end{aligned} \]

\(X^{2} - 4\) is the polynomial of \(A\) itself, but \(T\) acts on a 4-dimensional space so its polynomial has degree 4. \(X^{4} - 16\) has eigenvalues \(\pm 2, \pm 2i\), but \(T\)'s eigenvalues are only \(\pm 2\), each twice.

Your Answer : (Not Answered)

Correct Answer : B

The inverse is x = p + q; y = 2p - q.

Read the given data backwards: \(T^{-1}(1,0) = (1,2)\) and \(T^{-1}(1,1) = (2,1)\). Subtracting, \(T^{-1}(0,1) = (2,1) - (1,2) = (1,-1)\). By linearity

\[ \begin{aligned} T^{-1}(p,q) &= p(1,2) + q(1,-1) \\ &= (p + q,\ 2p - q) \end{aligned} \]

Check with (p, q) = (1, 1): x = 2, y = 1, which is (2, 1) as required. The option x = p - q fails this check, since it gives x = 0, and y = 2p + q gives y = 3, so the other three options are wrong.

Your Answer : (Not Answered)

Correct Answer : C

The dual basis is \(f_{1} = \frac{1}{2}(z_{1} - iz_{2}),\ f_{2} = \frac{1}{2}(-iz_{1} + z_{2})\).

A dual basis must satisfy \(f_{i}(v_{j}) = 1\) if \(i = j\) and 0 otherwise, with \(v_{1} = (1, i)\) and \(v_{2} = (i, 1)\). Checking this pair:

\[ \begin{aligned} f_{1}(1, i) &= \tfrac{1}{2}(1 - i^{2}) = 1 \\ f_{1}(i, 1) &= \tfrac{1}{2}(i - i) = 0 \\ f_{2}(1, i) &= \tfrac{1}{2}(-i + i) = 0 \\ f_{2}(i, 1) &= \tfrac{1}{2}(-i^{2} + 1) = 1 \end{aligned} \]

In the other options at least one condition fails. For instance \(f_{1} = \frac{1}{2}(z_{1} + iz_{2})\) gives \(f_{1}(1, i) = 0\), not 1, ruling out two options; and \(f_{2} = \frac{1}{2}(z_{1} + iz_{2})\) gives \(f_{2}(1, i) = 0\) and \(f_{2}(i, 1) = i\), not 1.

Your Answer : (Not Answered)

Correct Answer : C

Both P and Q are correct.

P: \(X^{T}AX\) is a \(1 \times 1\) matrix, so it equals its own transpose. Using \(A^{T} = -A\):

\[ X^{T}AX = (X^{T}AX)^{T} = X^{T}A^{T}X = -X^{T}AX \]

so \(X^{T}AX = 0\) for every real \(X\).

Q: take determinants of \(A^{T} = -A\): \(\det A = \det(-A) = (-1)^{n}\det A\). An orthogonal matrix has \(\det A = \pm 1 \neq 0\), so \((-1)^{n} = 1\) and \(n\) is even. The matrix \(\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\) shows the case \(n = 2\) does occur. Every odd-order skew-symmetric matrix is singular, so it cannot be orthogonal.

Your Answer : (Not Answered)

Correct Answer : B

The correct statements are Q and R only.

\(M\) equals its conjugate transpose: the diagonal entries 2, 5, 3 are real, and the off-diagonal pairs \(3 + 2i\) and \(3 - 2i\), \(-4\) and \(-4\), \(6i\) and \(-6i\) are conjugates. So \(M\) is Hermitian, and then \((iM)^{*} = -iM^{*} = -iM\), so \(iM\) is skew-Hermitian (Q).

A Hermitian matrix has real eigenvalues (R). The eigenvalues of \(iM\) are \(i\) times those of \(M\), so they are purely imaginary, and real only if they are 0. So S could hold only if every eigenvalue of \(M\) were 0, which for a Hermitian matrix means \(M = 0\); \(M\) is not zero, so S fails. P is false because \(M\) is not skew-Hermitian: its diagonal is not purely imaginary.

Your Answer : (Not Answered)

Correct Answer : D

The other solution is \(\begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix}\).

\(A\) has 3 columns and rank 2, so its null space has dimension \(3 - 2 = 1\). The difference of the two given solutions lies in it:

\[ \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} \]

so the null space is spanned by \((1,1,1)^{T}\) and every solution has the form \((1,2,3)^{T} + s(1,1,1)^{T}\). Taking \(s = 6\) gives \((7,8,9)^{T}\).

\((3,3,3)^{T}\) solves \(Ax = 0\), not \(Ax = b\), since \(b \neq 0\). \((5,7,9)^{T}\) is the sum of the two solutions, so \(A\) sends it to \(2b \neq b\). \((5,6,9)^{T}\) differs from \((1,2,3)^{T}\) by \((4,4,6)^{T}\), which is not a multiple of \((1,1,1)^{T}\).