**1. Check conjugate symmetry:** A matrix is Hermitian when \(B_{jk}=\overline{B_{kj}}\). Here the diagonal is real, the entries \(-2i,2i\) and \(-i,i\) are conjugate pairs, and the remaining off-diagonal pair is \(-1,-1\). Thus \(B=B^H\).
**2. Use the spectral consequence:** Every eigenvalue of a Hermitian matrix is real. For an eigenpair \(Bv=\lambda v\), the quantity \(v^HBv\) is real and equals \(\lambda\|v\|^2\), forcing \(\lambda\in\mathbb R\).
**3. Check the other numerical claims:** The first diagonal entry of \(B^2\) is \(4^2+(-2i)(2i)+(-1)^2=21\), whereas that of \(-B^H\) is \(-4\). Also, direct determinant expansion gives \(\det B=-16\ne0\).
**Answer: B, C — B is Hermitian and all its eigenvalues are real.**
A: **Incorrect:** The (1,1) entries 21 and −4 already disprove this matrix equality.
B: **Correct:** Conjugating and transposing reproduces every entry of B.
C: **Correct:** The Hermitian spectral theorem guarantees real eigenvalues.
D: **Incorrect:** The determinant is −16, so the matrix is nonsingular.
**1. Separate the diagonal shift:** Write \(A=2I+H\), where \(H\) has zero diagonal. Direct expansion gives \(\det(tI-H)=t(t^2-3)\), so the eigenvalues of \(H\) are \(0,\sqrt3,-\sqrt3\).
**2. Shift the eigenvalues:** Adding \(2I\) shifts each eigenvalue by 2. Therefore \[\Lambda=\operatorname{diag}(2,2+\sqrt3,2-\sqrt3).\] Their sum is 6, agreeing with the trace of A.
**3. Use Hermitian diagonalization:** Conjugate symmetry ensures an orthonormal complex eigenbasis, giving \(A=Q\Lambda Q^H\).
**Answer: B — diagonal eigenvalues 2, 2+√3, 2−√3.**
**Convention note:** The printed decomposition uses Q transposed. For this complex Hermitian matrix the spectral decomposition requires the conjugate transpose Q^H; the answer follows that intended correction.
A: **Incorrect:** These eigenvalues sum to 9, whereas the trace is 6.
B: **Correct:** These are exactly the three shifted eigenvalues.
C: **Incorrect:** These eigenvalues sum to 7, contradicting the trace.
D: **Incorrect:** A is Hermitian and hence unitarily diagonalizable.
**1. Match the quadratic-form convention:** For symmetric \(M\), \[\begin{pmatrix}x&y&z\end{pmatrix}M\begin{pmatrix}x\\y\\z\end{pmatrix}=m_{11}x^2+m_{22}y^2+m_{33}z^2+2m_{12}xy+2m_{13}xz+2m_{23}yz.\]
**2. Read the coefficients:** The diagonal is \((3,2,1)\). Halving the cross-term coefficients gives \(m_{12}=2,m_{13}=1,m_{23}=2\), hence \[M=\begin{pmatrix}3&2&1\\2&2&2\\1&2&1\end{pmatrix}.\]
**Answer: A — the symmetric matrix with rows (3,2,1), (2,2,2), (1,2,1).**
A: **Correct:** Its diagonal and twice each off-diagonal entry reproduce every term.
B: **Incorrect:** Its xz coefficient is 4 rather than 2.
C: **Incorrect:** It gives x² coefficient 2, xy coefficient 6 and yz coefficient 2.
D: **Incorrect:** Option A provides the required symmetric matrix.
**1. Differentiate and specify the update:** \(\nabla f(x,y)=(6x-6y,-6x+2y)\), and \(z_{t+1}=z_t-\eta_t\nabla f(z_t)\), with the first learning rate \(\eta_1=1\). The constant 4 contributes no gradient.
**2. First update:** At \((1,2)\), the gradient is \((-6,-2)\). Thus the new point is \((1,2)-(-6,-2)=(7,4)\).
**3. Second update:** Evaluate the gradient at the new point, obtaining \((18,-34)\). Since \(\eta_2=1/4\), \[(7,4)-\tfrac14(18,-34)=(2.5,12.5).\]
**Answer: B — (2.5,12.5).**
**Convention note:** The Hessian is indefinite, so these valid two updates do not imply convergence to a global minimum.
A: **Incorrect:** This is the point after only the first update.
B: **Correct:** It uses the new gradient and the second learning rate 1/4.
C: **Incorrect:** It does not result from either stated gradient update.
D: **Incorrect:** It does not follow the stated learning-rate schedule.
**1. Parameterize the feasible points:** Every point on the line is \((t,2t+1)\). Minimize squared distance instead of distance because squaring preserves the minimizer: \(D(t)=(t-4)^2+(2t+1)^2\).
**2. Find the minimum:** \(D\prime(t)=10t-4=0\) gives \(t=0.4\). Since \(D\prime\prime(t)=10>0\), this is the unique minimum.
**3. Recover the coordinates:** The y coordinate is \(2(0.4)+1=1.8\), so the nearest point is \((0.4,1.8)\). Its displacement from (4,0) is perpendicular to the line direction (1,2).
**Answer: B — (0.4,1.8).**
A: **Incorrect:** It lies on the line but is farther from (4,0).
B: **Correct:** It satisfies the line equation and the perpendicularity condition.
C: **Incorrect:** It does not satisfy y=2x+1.
D: **Incorrect:** For x=−1 the line requires y=−1, not 0.
**1. Find the left singular directions:** For \(A=\begin{pmatrix}1&0&1\\-1&1&0\end{pmatrix}\), \[AA^T=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.\] Its eigenvalues are 3 and 1, with eigenvectors (1,−1) and (1,1).
**2. Normalize and order the columns:** Their norms are both \(\sqrt2\). In descending singular-value order, \[Q_1=\frac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}.\] Each column has unit norm and the columns are orthogonal.
**Answer: B — Q₁ = (1/√2) times the matrix with rows (1,1), (−1,1).**
A: **Incorrect:** The directions are correct but the columns have norm √2 rather than 1.
B: **Correct:** It contains normalized eigenvectors of AA transposed in the required order.
C: **Incorrect:** Its columns are identical, so they cannot be orthonormal.
D: **Incorrect:** Multiplication by √2 makes the column norms 2.
**1. Use the singular-vector relation:** The left singular vectors are \(u_1=(1,-1)^T/\sqrt2\) and \(u_2=(1,1)^T/\sqrt2\), with singular values \(\sqrt3\) and 1. Thus \[v_1=\frac{A^Tu_1}{\sqrt3}=\frac{(2,-1,1)^T}{\sqrt6},\qquad v_2=A^Tu_2=\frac{(0,1,1)^T}{\sqrt2}.\]
**2. Complete the orthonormal basis:** Solving \(Av_3=0\) gives a vector proportional to \((-1,-1,1)\); normalize it by \(\sqrt3\). This vector is orthogonal to the first two.
**3. Place these vectors as rows:** \[Q_2^T=\frac1{\sqrt6}\begin{pmatrix}2&-1&1\\0&\sqrt3&\sqrt3\\-\sqrt2&-\sqrt2&\sqrt2\end{pmatrix}.\]
**Answer: D — Q₂ transposed with rows (2,−1,1)/√6, (0,1,1)/√2, (−1,−1,1)/√3.**
A: **Incorrect:** Its third row has squared norm 5/6 and is not a unit vector.
B: **Incorrect:** Its third row is not in the null space of A.
C: **Incorrect:** Its third row is not orthogonal to the second row and is not a null vector.
D: **Correct:** All rows are orthonormal and the last row spans the null space.
**1. Compute the singular values:** The eigenvalues of \(AA^T=\begin{pmatrix}2&-1\\-1&2\end{pmatrix}\) are 3 and 1. Singular values are their nonnegative square roots, giving \(\sqrt3\) and 1.
**2. Preserve the rectangular shape:** A is 2 by 3, so its full SVD has \[\Sigma=\begin{pmatrix}\sqrt3&0&0\\0&1&0\end{pmatrix}.\] The extra zero column accounts for the one-dimensional right null space.
**Answer: A — Σ with diagonal singular values √3 and 1.**
A: **Correct:** It has the correct shape and both square-root eigenvalues.
B: **Incorrect:** It uses 3 instead of its square root.
C: **Incorrect:** It incorrectly replaces the first singular value by 1.
D: **Incorrect:** It incorrectly replaces the second singular value by √3.
**1. Center the observations:** The mean of (2,1), (4,3), (6,5) is \(\mu=(4,3)\). The centered observations are \((-2,-2),(0,0),(2,2)\).
**2. Sum their outer products:** \[\sum_{j=1}^3(x_j-\mu)(x_j-\mu)^T=\begin{pmatrix}8&8\\8&8\end{pmatrix}.\]
**3. Apply the convention used by the options:** With empirical PCA divisor \(n=3\), \[C=\frac13\begin{pmatrix}8&8\\8&8\end{pmatrix}.\]
**Answer: A — (1/3) times the matrix whose four entries are 8.**
**Convention note:** Although the question says sample covariance, its options use the empirical PCA divisor n. The unbiased sample covariance divides by n−1 and would instead have all entries 4; that matrix is absent.
A: **Correct:** It is the centered scatter matrix divided by the three observations.
B: **Incorrect:** Its entries are half the correct empirical covariance.
C: **Incorrect:** The observations vary together, so the covariance is not zero.
D: **Incorrect:** Its entries are one quarter of the correct empirical covariance.
**1. Form the covariance matrix:** Centering the three observations gives \((-2,-2),(0,0),(2,2)\). Under the divisor-3 PCA convention, \(C=\frac83\begin{pmatrix}1&1\\1&1\end{pmatrix}\).
**2. Identify the eigenvectors:** Multiplication by C sends \((1,1)^T\) to \(\frac{16}3(1,1)^T\), and sends \((1,-1)^T\) to zero. Therefore the eigenvalues are \(16/3\) and 0.
**3. Interpret the zero value:** All centered data lie on one line, so there is variance in only one principal direction. The eigenvalues also sum to the trace, 16/3.
**Answer: D — 16/3 and 0.**
**Convention note:** This follows the empirical covariance convention used by the preceding question; using divisor n−1 would give eigenvalues 8 and 0.
A: **Incorrect:** 16 is the nonzero eigenvalue of the unnormalized scatter matrix.
B: **Incorrect:** 4 is not the nonzero eigenvalue of the stated covariance.
C: **Incorrect:** 1 does not agree with the covariance trace.
D: **Correct:** These values agree with both direct multiplication and the trace.
**1. Find the first principal direction:** With mean \(\mu=(4,3)\), every centered point is a multiple of \(u=(1,1)^T/\sqrt2\). Their scalar PC scores are \(-2\sqrt2,0,2\sqrt2\).
**2. Project and reconstruct:** The two-dimensional reconstruction from one component is \(\hat x=\mu+uu^T(x-\mu)\). Since each centered point already lies in the span of u, \(uu^T(x-\mu)=x-\mu\).
**3. Read the requested point coordinates:** Reconstruction returns exactly \((2,1),(4,3),(6,5)\); no variation is lost by keeping the first component.
**Answer: A — (2,1), (4,3), (6,5).**
**Convention note:** The offered two-dimensional points are interpreted as PCA projections with the mean restored. The one-dimensional scores themselves are −2√2, 0 and 2√2; raw projection of uncentered vectors is a different operation.
A: **Correct:** All three original observations lie on the affine first-principal-component line.
B: **Incorrect:** These points do not reconstruct the centered observations along the principal direction.
C: **Incorrect:** These are neither the PC scores nor the mean-restored projected points.
D: **Incorrect:** This projects onto the horizontal axis instead of the first principal direction.
**1. Test invertibility:** \(\det M=2\cdot2-4\cdot1=0\). The second column is twice the first, so M has rank one and is singular.
**2. Test diagonalizability separately:** \(\det(\lambda I-M)=\lambda(\lambda-4)\). The distinct eigenvalues 0 and 4 have independent eigenvectors, for example \((-2,1)^T\) and \((2,1)^T\). Thus M is diagonalizable despite being singular.
**Answer: B, D — singular and diagonalizable.**
A: **Incorrect:** A zero determinant rules out an inverse.
B: **Correct:** Dependent columns make the determinant zero.
C: **Incorrect:** The determinant is 0, while 4 is the trace and the nonzero eigenvalue.
D: **Correct:** Two distinct eigenvalues give an eigenbasis in dimension two.
A is negative semi-definite.
**Java**
**Section Id :** 64065373860
**Section Number :** 6
**Section type :** Online
**Mandatory or Optional :** Mandatory
**Number of Questions :** 16
**Number of Questions to be attempted :** 16
**Section Marks :** 100
**Display Number Panel :** Yes
**Section Negative Marks :** 0
**Group All Questions :** No
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Published solution
**1. Factor the quadratic form:** \[\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&4\\4&16\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=x^2+8xy+16y^2=(x+4y)^2\ge0.\]
**2. Distinguish definite from semidefinite:** The nonzero vector \((-4,1)\) gives value zero, so strict positive definiteness fails. Equivalently, the matrix is \(vv^T\) for \(v=(1,4)^T\), with eigenvalues 17 and 0.
**Answer: B — positive semidefinite.**
A: **Incorrect:** A nonzero vector can give quadratic-form value zero.
B: **Correct:** The quadratic form is always nonnegative, with a nontrivial null direction.
C: **Incorrect:** For (1,0) the quadratic form is positive, contradicting negative definiteness.
D: **Incorrect:** The positive value at (1,0) rules out negative semidefiniteness.
