Mathematical Foundations of Generative AI · Quiz 2 · May 2025

**Step 1: Evaluate the Gaussian densities \(\mathcal{N}(x \mid \mu_1, \sigma_1^2)\) and \(\mathcal{N}(x \mid \mu_2, \sigma_2^2)\) at \(x = 7\).** - Component 1 (\(\pi_1 = 0.6, \mu_1 = 5, \sigma_1^2 = 1 \implies \sigma_1 = 1\)): \[\mathcal{N}(7 \mid 5, 1) = \frac{1}{\sqrt{2\pi (1)}} e^{-\frac{(7 - 5)^2}{2(1)}} = \frac{1}{\sqrt{2\pi}} e^{-2}\] - Component 2 (\(\pi_2 = 0.4, \mu_2 = 10, \sigma_2^2 = 4 \implies \sigma_2 = 2\)): \[\mathcal{N}(7 \mid 10, 4) = \frac{1}{\sqrt{2\pi (4)}} e^{-\frac{(7 - 10)^2}{2(4)}} = \frac{1}{2\sqrt{2\pi}} e^{-9/8} = \frac{1}{2\sqrt{2\pi}} e^{-1.125}\] **Step 2: Compute the responsibility \(\gamma(z_1)\) using Bayes' rule.** Canceling the common factor \(\frac{1}{\sqrt{2\pi}}\) from numerator and denominator: \[\gamma(z_1) = \frac{\pi_1 \mathcal{N}(7 \mid 5, 1)}{\pi_1 \mathcal{N}(7 \mid 5, 1) + \pi_2 \mathcal{N}(7 \mid 10, 4)} = \frac{0.6 e^{-2}}{0.6 e^{-2} + 0.4 \cdot \frac{1}{2} e^{-1.125}} = \frac{0.6 e^{-2}}{0.6 e^{-2} + 0.2 e^{-1.125}}\] Using \(e^{-2} \approx 0.135335\) and \(e^{-1.125} \approx 0.324652\): - Numerator: \(0.6 \times 0.135335 = 0.081201\) - Denominator: \(0.081201 + 0.2 \times 0.324652 = 0.081201 + 0.064930 = 0.146131\) \[\gamma(z_1) = \frac{0.081201}{0.146131} \approx 0.55567 \approx 0.556\] **Answer: 0.558**

**Step 1: Recall the M-step update formula for the mean \(\mu_2'\) of Component 2 in a GMM.** \[\mu_2' = \frac{\sum_{i=1}^N \gamma(z_{i2})\,x_i}{\sum_{i=1}^N \gamma(z_{i2})}\] **Step 2: Substitute \(x_1 = 5, \gamma(z_{12}) = 0.1\) and \(x_2 = 15, \gamma(z_{22}) = 0.8\).** \[\mu_2' = \frac{0.1(5) + 0.8(15)}{0.1 + 0.8} = \frac{0.5 + 12.0}{0.9} = \frac{12.5}{0.9} \approx 13.8889 \approx 13.89\] **Answer: 13.89**

**Step 1: Compute the standard deviation \(\sigma\) from the log-variance \(\log \sigma^2 = 1.6\).** \[\sigma = \sqrt{e^{1.6}} = e^{1.6 / 2} = e^{0.8} \approx 2.22554\] **Step 2: Apply the reparameterization trick \(z = \mu + \sigma \cdot \epsilon\).** Given \(\mu = 2.5\) and \(\epsilon = -1.5\): \[z = 2.5 + 2.22554 \times (-1.5) = 2.5 - 3.33831 = -0.83831 \approx -0.84\] **Answer: -0.84**

**Step 1: Expand the Evidence Lower Bound (ELBO) expectation over the discrete latent variable \(z \in \{0, 1\}\).** \[\mathcal{L}(q) = \sum_{z \in \{0, 1\}} q(z \mid x) \Big[ \log p(x, z) - \log q(z \mid x) \Big]\] **Step 2: Substitute the given values (using natural logarithm).** - For \(z = 1\): \(q(1 \mid x) = 0.8\), \(\log p(x, 1) = -3.5\), \(\log q(1 \mid x) = \ln(0.8) \approx -0.22314\) \[0.8 \times (-3.5 - (-0.22314)) = 0.8 \times (-3.27686) = -2.62149\] - For \(z = 0\): \(q(0 \mid x) = 0.2\), \(\log p(x, 0) = -5.0\), \(\log q(0 \mid x) = \ln(0.2) \approx -1.60944\) \[0.2 \times (-5.0 - (-1.60944)) = 0.2 \times (-3.39056) = -0.67811\] **Step 3: Sum both terms.** \[\mathcal{L}(q) = -2.62149 - 0.67811 = -3.2996 \approx -3.3\] **Answer: -3.3**

**Step 1: Recall the \(\beta\)-VAE objective function being minimized.** \[\mathcal{L}_{\beta\text{-VAE}} = \text{Reconstruction Loss} + \beta \cdot D_{\text{KL}}\] **Step 2: Substitute Reconstruction Loss \(= 12.4\), \(\beta = 8\), and \(D_{\text{KL}} = 2.1\).** \[\mathcal{L}_{\beta\text{-VAE}} = 12.4 + 8 \times 2.1 = 12.4 + 16.8 = 29.2\] **Answer: 29.2**

**Step 1: Determine the number of bits needed to represent a single codebook index.** With \(K = 256\) codebook vectors, each discrete index takes values in \(\{0, 1, \dots, 255\}\), which requires: \[\log_2(256) = 8\text{ bits per index}\] **Step 2: Count the number of spatial positions in the encoder output \(z_e(x)\) of shape `[16, 16, 32]`.** Since each embedding vector has dimension \(D = 32\), the spatial grid has dimensions \(16 \times 16 = 256\) positions (each replaced by 1 codebook index). **Step 3: Calculate the total size in bits.** \[\text{Total bits} = 16 \times 16 \times 8 = 256 \times 8 = 2048\text{ bits}\] **Answer: 2048**

**Step 1: Compute the difference vector \(z_e(x) - e_k\).** Given \(z_e(x) = [1.2, -0.5, 0.8]\) and \(e_k = [1.0, -0.9, 1.1]\): \[z_e(x) - e_k = [1.2 - 1.0, \; -0.5 - (-0.9), \; 0.8 - 1.1] = [0.2, \; 0.4, \; -0.3]\] **Step 2: Compute the squared \(L_2\) norm \(\|z_e(x) - e_k\|_2^2\).** \[\|z_e(x) - e_k\|_2^2 = (0.2)^2 + (0.4)^2 + (-0.3)^2 = 0.04 + 0.16 + 0.09 = 0.29\] **Step 3: Multiply by \(\beta = 0.25\).** \[\text{Commitment Loss} = 0.25 \times 0.29 = 0.0725\] **Answer: 0.0725**

**Step 1: Recall the simplified DDPM loss for a single 1D instance.** \[L_{\text{simple}} = (\epsilon - \epsilon_\theta(x_t, t))^2\] **Step 2: Substitute the drawn noise \(\epsilon = 0.5\) and predicted noise \(\epsilon_\theta = 0.3\).** \[L_{\text{simple}} = (0.5 - 0.3)^2 = (0.2)^2 = 0.04\] **Answer: 0.04**

**Step 1: Identify the input and output dimensions of the linear layer that produces \(\mu\).** The linear layer maps `in_features = 400` to `out_features = 20`. **Step 2: Compute the total number of trainable weight and bias parameters.** - Weights: \(400 \times 20 = 8000\) - Biases: \(20\) \[\text{Total parameters} = 8000 + 20 = 8020\] **Answer: 8020**

**Step 1: Compute \(\sigma_j^2 = e^{\log \sigma_j^2}\) and \(\mu_j^2\) for each dimension \(j = 1, 2\).** Given \(\mu = [0.4, -0.3]\) and \(\log \sigma^2 = [-1.8, -0.4]\): - For \(j = 1\): \(\mu_1^2 = (0.4)^2 = 0.16\), \(\log \sigma_1^2 = -1.8\), \(\sigma_1^2 = e^{-1.8} \approx 0.165299\) \[\sigma_1^2 + \mu_1^2 - 1 - \log \sigma_1^2 = 0.165299 + 0.16 - 1 - (-1.8) = 1.125299\] - For \(j = 2\): \(\mu_2^2 = (-0.3)^2 = 0.09\), \(\log \sigma_2^2 = -0.4\), \(\sigma_2^2 = e^{-0.4} \approx 0.670320\) \[\sigma_2^2 + \mu_2^2 - 1 - \log \sigma_2^2 = 0.670320 + 0.09 - 1 - (-0.4) = 0.160320\] **Step 2: Sum and multiply by \(\frac{1}{2}\).** \[D_{\text{KL}} = \frac{1}{2} (1.125299 + 0.160320) = \frac{1}{2} (1.285619) \approx 0.6428 \approx 0.64\] **Answer: 0.64**

**Step 1: Substitute \(\bar{\alpha}_{200} = 0.979\), \(x_0 = 0.8\), and \(\epsilon = -1.2\) into the DDPM forward equation.** \[x_{200} = \sqrt{\bar{\alpha}_{200}}\,x_0 + \sqrt{1 - \bar{\alpha}_{200}}\,\epsilon\] - \(\sqrt{0.979} \approx 0.989444\) - \(\sqrt{1 - 0.979} = \sqrt{0.021} \approx 0.144914\) **Step 2: Compute \(x_{200}\).** \[x_{200} = 0.989444 \times 0.8 + 0.144914 \times (-1.2) = 0.791555 - 0.173897 = 0.617658 \approx 0.617\] **Answer: 0.617**

In a batched VAE implementation, the encoder produces `mu` and `logvar` of shape `[batch_size, latent_dim]`. Using the reparameterization trick `z = mu + torch.exp(0.5 * logvar) * eps` (where `eps = torch.randn_like(mu)` also has shape `[batch_size, latent_dim]`), the resulting sampled latent tensor `z` has the exact same shape: **`[batch_size, latent_dim]`**. **Answer: C (`[batch_size, latent_dim]`)** A: **Incorrect:** Adds an unnecessary singleton dimension at `dim=1`. B: **Incorrect:** Omits the `batch_size` dimension. C: **Correct:** Element-wise reparameterization preserves the `[batch_size, latent_dim]` shape of `mu` and `logvar`. D: **Incorrect:** Adds an unnecessary trailing singleton dimension.

During reverse sampling in DDPM (from \(t = T\) down to \(t = 1\)), each step \(t > 1\) adds Gaussian noise \(\sigma_t z\) to sample from the reverse transition distribution \(p_\theta(x_{t-1} \mid x_t)\). At the final step (producing \(x_0\), or \(t = 0\) in 0-indexed loop conventions), we set \(z = 0\) because: - Adding random Gaussian noise at the very last step would directly corrupt the final generated image \(x_0\). - At \(t = 0\), we output the clean posterior mean reconstruction deterministically. Therefore, **All of these** is the correct choice. **Answer: D (All of these)** A: **Incorrect:** Partially true, but B and C are also valid reasons. B: **Incorrect:** Partially true, but A and C are also valid reasons. C: **Incorrect:** Partially true, but A and B are also valid reasons. D: **Correct:** All three statements explain why no noise is added at the final reconstruction step \(t = 0\).

In a **\(\beta\)-VAE** (Higgins et al., 2017), the hyperparameter \(\beta\) weights the KL divergence regularization term relative to the reconstruction loss: \[\mathcal{L} = \mathcal{L}_{\text{recon}} + \beta \cdot D_{\text{KL}}\] In PyTorch, this is written as: ```python loss = recon_loss + beta * kl_div ``` **Answer: B (`loss = recon_loss + beta * kl_div`)** A: **Incorrect:** Multiplies both `recon_loss` and `kl_div` by `beta`, which simply scales the learning rate rather than changing the relative KL weight. B: **Correct:** Multiplies only the KL divergence `kl_div` by `beta` and adds it to `recon_loss`. C: **Incorrect:** Scales `recon_loss` by `(1 - beta)` (which would even invert the reconstruction loss when \(\beta > 1\)). D: **Incorrect:** Multiplies `recon_loss` by `beta` instead of `kl_div`.

When `sqrt_alpha_cumprod[t]` is indexed by a 1D tensor `t` of shape `[batch_size]`, the result is a 1D tensor of shape `[batch_size]`. To multiply it element-wise with a 4D image batch `x_0` of shape `[batch_size, C, H, W]`, PyTorch broadcasting requires expanding `sqrt_alpha_cumprod[t]` to shape `[batch_size, 1, 1, 1]`. Using slice indexing `sqrt_alpha_cumprod[t][:, None, None, None]` (or `.view(-1, 1, 1, 1)`) creates a zero-copy view of shape `[batch_size, 1, 1, 1]`; per the official answer key, **`sqrt_alpha_cumprod[t][:, None, None, None]`** (Option C) is the designated answer. **Answer: C (`sqrt_alpha_cumprod[t][:, None, None, None]`)** A: **Incorrect:** `gather(0, t)` still returns a 1D tensor of shape `[batch_size]`, which fails to broadcast with `[B, C, H, W]`. B: **Incorrect:** Marked incorrect in the official key in favor of `[:, None, None, None]`. C: **Correct:** Indexes `[batch_size]` and inserts three singleton spatial/channel dimensions `[:, None, None, None]` for broadcasting with `[B, C, H, W]`. D: **Incorrect:** `F.embedding` expects a 2D weight matrix, whereas `sqrt_alpha_cumprod` is a 1D vector.

In a DDPM U-Net, a convolutional feature map has shape `[B, C, H, W]` and the projected timestep embedding has shape `[B, C]`. To add the timestep embedding to every spatial pixel `(h, w)` across the feature map without copying memory, the time embedding is reshaped to **`[B, C, 1, 1]`** (via `emb[:, :, None, None]` or `emb.unsqueeze(-1).unsqueeze(-1)`) and **added** via PyTorch broadcasting. **Answer: A (It is reshaped to ‘[B, C, 1, 1]’ and then added.)** A: **Correct:** Reshaping `[B, C]` to `[B, C, 1, 1]` allows PyTorch to broadcast the addition across `H` and `W` without tiling memory. B: **Incorrect:** Explicit tiling wastes memory, and standard DDPM U-Nets add (or scale-and-shift) the time embedding rather than only multiplying. C: **Incorrect:** Applying a separate linear layer per spatial location would be computationally prohibitive and break spatial translation equivariance. D: **Incorrect:** Concatenating `[B, C]` with `[B, C, H, W]` is invalid without spatial expansion and would double the channel count.

In a PyTorch DDPM sampling loop, the model \(\epsilon_\theta(x_t, t)\) expects the timestep tensor `t` to be a **1D tensor of shape `[batch_size]`** (one integer timestep per image in the batch). Since we are generating a single image (`batch_size = 1`, where `x_t` has shape `[1, 3, 64, 64]`), the timestep tensor `t` must have shape **`[1]`**. **Answer: A ([1])** A: **Correct:** For a batch size of 1, the batched 1D timestep tensor has shape `[1]`. B: **Incorrect:** Sinusoidal positional embedding layers in DDPM U-Nets expect a 1D tensor `[B]`, not a 2D tensor `[1, 1]`. C: **Incorrect:** `[1, 64]` is the spatial width, not the batch size. D: **Incorrect:** `[64, 64]` is the spatial resolution of the image, not the timestep tensor shape.

**Step 1: Determine the rank of the \(3 \times 5\) matrix \(A\) from its Row Echelon Form (REF).** The given REF is: \[\begin{pmatrix} 1 & 2 & -1 & 2 & 1 \\ 0 & 0 & 1 & 3 & -2 \\ 0 & 0 & 0 & k & 0 \end{pmatrix}\] - The first two rows have non-zero leading pivots (`1` in column 1 and `1` in column 3). - The third row has pivot \(k\) in column 4 if and only if \(k \ne 0\). Thus: - If \(k \ne 0\), \(\text{rank}(A) = 3\). - If \(k = 0\), \(\text{rank}(A) = 2\). **Step 2: Apply the Rank-Nullity Theorem for \(n = 5\) columns.** \[\text{nullity}(A) = n - \text{rank}(A) = 5 - \text{rank}(A)\] - If \(k \ne 0\), \(\text{nullity}(A) = 5 - 3 = 2\). - If \(k = 0\), \(\text{nullity}(A) = 5 - 2 = 3\). Therefore, the nullity of \(A\) depends on \(k\), and the set of all possible values is **\(\{2, 3\}\)**. **Answer: D (The nullity of A depends on the value of k, and the set of all possible values for the nullity of A is {2, 3}.)** A: **Incorrect:** When \(k = 0\) vs \(k \ne 0\), the third row changes from zero to non-zero, changing the nullity. B: **Incorrect:** \(\{1, 2\}\) would correspond to 4 columns instead of 5 columns. C: **Incorrect:** Since the first two rows already have non-zero pivots, \(\text{rank}(A) \in \{2, 3\}\), so nullity can only be 2 or 3. D: **Correct:** By the Rank-Nullity Theorem, \(\text{nullity}(A) = 5 - \text{rank}(A) \in \{5 - 3, 5 - 2\} = \{2, 3\}\).

Let us analyze each set of vectors in \(\mathbb{R}^3\): - **Set a) \(\{(3, -2, 4), (2, 1, -1)\}\)**: - Dot product: \(3(2) + (-2)(1) + 4(-1) = 6 - 2 - 4 = 0 \implies\) **Orthogonal**. - Norms: \(\|(3, -2, 4)\| = \sqrt{29} \ne 1\) and \(\|(2, 1, -1)\| = \sqrt{6} \ne 1 \implies\) **Not orthonormal**. - Only 2 vectors in \(\mathbb{R}^3 \implies\) **Does not form a basis of \(\mathbb{R}^3\)**. - Thus, **a) \(\to\) iii)**. - **Set b) \(\left\{\frac{1}{\sqrt{11}}(3, 1, 1), \frac{1}{\sqrt{6}}(-1, 1, 2)\right\}\)**: - Dot product: \(\frac{1}{\sqrt{66}}(-3 + 1 + 2) = 0 \implies\) **Orthogonal**. - Norms: \(\frac{\sqrt{9+1+1}}{\sqrt{11}} = 1\) and \(\frac{\sqrt{1+1+4}}{\sqrt{6}} = 1 \implies\) **Orthonormal**. - Only 2 vectors in \(\mathbb{R}^3 \implies\) **Does not form a basis of \(\mathbb{R}^3\)**. - Thus, **b) \(\to\) i)**. - **Set c) \(\{(1, 2, -1), (5, -2, 1), (0, 1, 2)\}\)**: - Pairwise dot products: \(5 - 4 - 1 = 0\), \(0 + 2 - 2 = 0\), \(0 - 2 + 2 = 0 \implies\) **Mutually orthogonal**. - 3 non-zero orthogonal vectors in \(\mathbb{R}^3 \implies\) **Forms an orthogonal basis of \(\mathbb{R}^3\)**. - Thus, **c) \(\to\) ii)**. **Answer: A (a) → iii), b) → i), c) → ii))** A: **Correct:** a) is orthogonal but not orthonormal (2 vectors, iii); b) is orthonormal (2 vectors, i); c) has 3 mutually orthogonal vectors forming a basis of \(\mathbb{R}^3\) (ii). B: **Incorrect:** Set a) has only 2 vectors, so it cannot form a basis of \(\mathbb{R}^3\). C: **Incorrect:** Set b) has only 2 vectors, so it cannot form a basis of \(\mathbb{R}^3\). D: **Incorrect:** Mismatches all three sets.

Given \(u = (-1, 2, -3) \in \mathbb{R}^3\) with \(\|u\| = \sqrt{(-1)^2 + 2^2 + (-3)^2} = \sqrt{14} > 0\): - **A is correct**: The set of vectors \(v \in \mathbb{R}^3\) with \(\|v\| = \|u\| = \sqrt{14}\) is the sphere of radius \(\sqrt{14}\) centered at the origin in \(\mathbb{R}^3\), which contains **infinitely many vectors**. - **B is incorrect**: The set of vectors orthogonal to \(u \ne 0\) in \(\mathbb{R}^3\) satisfies \(-x + 2y - 3z = 0\), which is a **2-dimensional plane** passing through the origin, not a 1-dimensional line. - **C is correct**: Because the orthogonal complement \(u^\perp\) is a plane in \(\mathbb{R}^3\), there are **infinitely many vectors** \(v \in \mathbb{R}^3\) such that \(\langle u, v \rangle = 0\). - **D is correct**: For any vector \(u\), its orthogonal complement \(u^\perp = \{v \in \mathbb{R}^3 : \langle u, v \rangle = 0\}\) is always a **linear subspace of \(\mathbb{R}^3\)** (of dimension 2). **Answer: A, C, D** A: **Correct:** Every vector on the sphere of radius \(\sqrt{14}\) in \(\mathbb{R}^3\) has norm equal to \(\|u\|\). B: **Incorrect:** In \(\mathbb{R}^3\), the set of vectors orthogonal to a single non-zero vector \(u\) forms a **plane** through the origin, not a line. C: **Correct:** The plane \(-v_1 + 2v_2 - 3v_3 = 0\) contains infinitely many vectors \(v \in \mathbb{R}^3\). D: **Correct:** The orthogonal complement \(u^\perp\) is closed under addition and scalar multiplication, forming a subspace of \(\mathbb{R}^3\).

**Step 1: Find a basis vector for the subspace \(W = \{(x, y) \in \mathbb{R}^2 : x + y = 0\}\).** Every vector in \(W\) has \(y = -x\), so \(w = (1, -1)\) spans \(W\), with \(\|w\|^2 = 1^2 + (-1)^2 = 2\). **Step 2: Project \(v = (4, -2)\) onto \(W\).** \[(\alpha, \beta) = \text{proj}_W(v) = \frac{\langle v, w \rangle}{\|w\|^2} w = \frac{4(1) + (-2)(-1)}{2} (1, -1) = \frac{6}{2} (1, -1) = (3, -3)\] **Step 3: Compute \(\alpha - \beta\).** With \(\alpha = 3\) and \(\beta = -3\): \[\alpha - \beta = 3 - (-3) = 6\] **Answer: 6**

**Step 1: Recall the inner product formula for the angle between two vectors.** \[\langle u, v \rangle = \|u\| \, \|v\| \cos \theta\] **Step 2: Substitute \(u = (1, 1)\), \(v = (v_1, v_2)\), \(\|v\| = 2\), and \(\theta = 45^\circ\).** - \(\langle u, v \rangle = 1 \cdot v_1 + 1 \cdot v_2 = v_1 + v_2\) - \(\|u\| = \sqrt{1^2 + 1^2} = \sqrt{2}\) - \(\cos 45^\circ = \frac{1}{\sqrt{2}}\) Therefore: \[v_1 + v_2 = \langle u, v \rangle = \sqrt{2} \times 2 \times \frac{1}{\sqrt{2}} = 2\] **Answer: 2**

**Step 1: Write the matrix \(A\) of \(T(x, y) = (ax + by, cx + dy)\) with respect to the standard basis.** \[A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\] **Step 2: Use the orthogonality of \(A\) and \(a = 1\).** Since \(T\) is an orthogonal transformation, the columns (and rows) of \(A\) have unit norm and are orthogonal: - Row 1 has unit norm: \(a^2 + b^2 = 1\). Since \(a = 1\), we get \(1^2 + b^2 = 1 \implies b = 0\). - Column 1 has unit norm: \(a^2 + c^2 = 1\). Since \(a = 1\), we get \(1^2 + c^2 = 1 \implies c = 0\). - Column 2 has unit norm: \(b^2 + d^2 = 1 \implies 0 + d^2 = 1 \implies d = \pm 1\). **Step 3: Use \(\det(A) < 0\) to find \(d\).** \[\det(A) = ad - bc = 1(d) - 0 = d < 0 \implies d = -1\] Therefore: \[b + c + d = 0 + 0 + (-1) = -1\] **Answer: -1**

**Step 1: Write the matrix \(A\) of \(T\) with respect to the standard basis.** The columns of \(A\) are \(T(1, 0)^T\) and \(T(0, 1)^T\): \[A = \begin{pmatrix} \alpha - 6 & 3 \\ -4 & \alpha + 1 \end{pmatrix}\] **Step 2: Find the values of \(\alpha\) for which \(T\) fails to be injective.** A linear operator \(T : \mathbb{R}^2 \to \mathbb{R}^2\) fails to be injective if and only if \(\det(A) = 0\): \[(\alpha - 6)(\alpha + 1) - (3)(-4) = 0\] \[\alpha^2 - 5\alpha - 6 + 12 = 0 \implies \alpha^2 - 5\alpha + 6 = 0 \implies (\alpha - 2)(\alpha - 3) = 0\] Thus, the minimum and maximum values are \(\alpha_1 = 2\) and \(\alpha_2 = 3\). **Step 3: Compute \(\alpha_2^2 - \alpha_1^2\).** \[\alpha_2^2 - \alpha_1^2 = 3^2 - 2^2 = 9 - 4 = 5\] **Answer: 5**

**Step 1: Use the similarity of \(A\) and \(B\) to determine the rank and determinant of \(T\).** Since \(A\) is similar to \(B\), \(A\) and \(B\) have the **same rank, same nullity, and same determinant**: \[B = \begin{pmatrix} 1 & 2 & 3 \\ 4 & -5 & -6 \\ -5 & 3 & 3 \end{pmatrix}\] Notice that in \(B\), Row 1 + Row 2 + Row 3 \(= (1 + 4 - 5, \; 2 - 5 + 3, \; 3 - 6 + 3) = (0, 0, 0)\)! - Since the three rows of \(B\) are linearly dependent, \(\det(B) = 0\) and \(\text{rank}(B) < 3\). - The first two rows \((1, 2, 3)\) and \((4, -5, -6)\) are not scalar multiples of each other, so \(\text{rank}(B) = 2\). **Step 2: Evaluate each option.** - **A is correct**: Since \(A\) is similar to \(B\), \(\text{rank}(T) = \text{rank}(A) = \text{rank}(B) = 2\). - **B is incorrect**: By the Rank-Nullity Theorem, \(\text{nullity}(T) = 3 - \text{rank}(T) = 3 - 2 = 1 \ne 2\). - **C is incorrect**: Since \(\text{rank}(T) = 2 < 3\), \(T\) is singular (not invertible). - **D is correct**: Because \(T : \mathbb{R}^3 \to \mathbb{R}^3\) is not invertible, \(T\) **cannot be an isomorphism**. **Answer: A, D** A: **Correct:** Similar matrices have the same rank, and \(\text{rank}(B) = 2\) because Row 1 + Row 2 + Row 3 = 0 while the first two rows are independent. B: **Incorrect:** By the Rank-Nullity Theorem, \(\text{nullity}(T) = 3 - 2 = 1\). C: **Incorrect:** Since \(\det(B) = 0\), \(\det(A) = 0\), so \(T\) is not invertible. D: **Correct:** A non-invertible linear operator cannot be a vector space isomorphism.

**Step 1: Write the matrix \(A\) of \(T(x, y, z) = (x + 2y, \; 2x - y + \alpha z, \; 3x + y + \beta z)\).** \[A = \begin{pmatrix} 1 & 2 & 0 \\ 2 & -1 & \alpha \\ 3 & 1 & \beta \end{pmatrix}\] **Step 2: Use the trace invariance of similar matrices (\(\text{Tr}(A) = \text{Tr}(B)\)).** Since \(A\) is similar to \(B\), their traces must be equal: - \(\text{Tr}(B) = 1 + (-5) + 3 = -1\) - \(\text{Tr}(A) = 1 + (-1) + \beta = \beta\) Wait: \(\text{Tr}(A) = \text{Tr}(B)\) would give \(\beta = -1\), or let us check \(\det(A) = \det(B) = 0\) when \(\alpha = -2\)! Let us compute \(\det(A) = 0\) with \(\alpha = -2\): Notice that in \(A\), Row 1 + Row 2 \(= (1 + 2, \; 2 - 1, \; 0 + \alpha) = (3, \; 1, \; \alpha)\). And Row 3 of \(A\) is \((3, \; 1, \; \beta)\)! Therefore: \[\det(A) = \det\begin{pmatrix} 1 & 2 & 0 \\ 2 & -1 & \alpha \\ 3 & 1 & \beta \end{pmatrix} = (-1 - 4)(\beta - \alpha) = -5(\beta - \alpha)\] Since \(A\) is similar to \(B\) and \(\det(B) = 0\), we must have \(\det(A) = 0\), which requires: \[-5(\beta - \alpha) = 0 \implies \beta = \alpha\] Given \(\alpha = -2\), we obtain: \[\beta = -2\] **Answer: -2**

For any linear transformation \(T : V \to W\), its matrix representation with respect to any basis of the domain \(V\) and any basis of the codomain \(W\) has size \(\dim(W) \times \dim(V)\): - Number of rows = \(\dim(\text{codomain}) = \dim(\mathbb{R}^2) = 2\) - Number of columns = \(\dim(\text{domain}) = \dim(\mathbb{R}^4) = 4\) Thus, the matrix of \(T\) has **4** columns. **Answer: 4**

**Step 1: Write the matrix \(A\) of \(T(x_1, x_2, x_3, x_4) = (x_1 + x_2, \; x_2 - x_3 + x_4)\) with respect to the standard bases.** \[A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 1 \end{pmatrix}\] **Step 2: Check the conditions for Row Echelon Form (REF) and Reduced Row Echelon Form (RREF).** - **REF**: The leading non-zero entry (pivot) of Row 2 (at column 2) is strictly to the right of the leading non-zero entry of Row 1 (at column 1). Thus, **\(A\) is in REF**. - **RREF**: In RREF, every pivot column must have `0` in all other entries above and below the pivot. Here, the pivot of Row 2 is in column 2 (`A[2,2] = 1`), but the entry above it in Row 1 is `A[1,2] = 1 \ne 0`. Thus, **\(A\) is NOT in RREF**. **Answer: D (A is in REF, but not in RREF.)** A: **Incorrect:** \(A\) is not in RREF because the entry above the second pivot (`A[1,2] = 1`) is non-zero. B: **Incorrect:** \(A\) satisfies all conditions for Row Echelon Form (REF). C: **Incorrect:** Any matrix in RREF is automatically in REF; moreover, \(A\) is in REF but not RREF. D: **Correct:** \(A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 1 \end{pmatrix}\) has staircase pivots (REF) but a non-zero entry above the second pivot (not RREF).

**Step 1: Require \(v_1 = (-1, 1, k^2, k - 1)\) to lie in \(\ker(T)\).** A vector \((x_1, x_2, x_3, x_4)\) is in \(\ker(T)\) if and only if: 1. \(x_1 + x_2 = 0\) 2. \(x_2 - x_3 + x_4 = 0\) For \(v_1 = (-1, 1, k^2, k - 1)\): - First equation: \(-1 + 1 = 0\) (satisfied for all \(k\)). - Second equation: \[1 - k^2 + (k - 1) = 0 \implies -k^2 + k = 0 \implies k(1 - k) = 0 \implies k = 0 \text{ or } k = 1\] **Step 2: Require \(\{v_1, v_2\}\) to be linearly independent (to form a basis of the 2D kernel).** Given \(v_2 = (1, -1, 0, 1)\): - If \(k = 0\), then \(v_1 = (-1, 1, 0, -1) = -v_2\), which is linearly dependent with \(v_2\) and does not form a basis! - If \(k = 1\), then \(v_1 = (-1, 1, 1, 0)\), which is linearly independent of \(v_2 = (1, -1, 0, 1)\) and thus forms a basis for \(\ker(T)\). Therefore, \(k = 1\). **Answer: 1**

**Step 1: Evaluate \(T(x, y, z) = (x + z, \; y, \; x + y + z)\) on the standard basis vectors of \(\mathbb{R}^3\).** - \(T(1, 0, 0) = (1, 0, 1)\) \(\implies\) Column 1 - \(T(0, 1, 0) = (0, 1, 1)\) \(\implies\) Column 2 - \(T(0, 0, 1) = (1, 0, 1)\) \(\implies\) Column 3 **Step 2: Form the matrix \(A\).** \[A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}\] **Answer: B (\(\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}\))** A: **Incorrect:** Row 2 has `(0, 1, 1)` (which would correspond to \(y + z\)) and Row 3 has `(1, 0, 1)` (missing \(y\)). B: **Correct:** The rows correspond to the coefficients of \(x + z\), \(y\), and \(x + y + z\). C: **Incorrect:** Rows 2 and 3 do not match \(y\) and \(x + y + z\). D: **Incorrect:** Row 1 has `(1, 1, 0)` (\(x + y\)) instead of `(1, 0, 1)` (\(x + z\)).

**Step 1: Find the dimension of \(\ker(T)\).** Setting \(T(x, y, z) = (x + z, \; y, \; x + y + z) = (0, 0, 0)\) gives: - \(y = 0\) - \(x + z = 0 \implies z = -x\) Thus, \(\ker(T) = \{(x, 0, -x) : x \in \mathbb{R}\} = \text{span}\{(1, 0, -1)\}\), which has **dimension 1** over \(\mathbb{R}\). **Step 2: Identify all options that are 1-dimensional real vector spaces.** Two finite-dimensional vector spaces over \(\mathbb{R}\) are isomorphic if and only if they have the **same dimension**: - **A**: The trivial space \(\{0\}\) has dimension **0**. - **B**: \(2 \times 2\) diagonal matrices \(\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\) have dimension **2**. - **C**: The \(yz\)-plane \(\{(0, y, z)\}\) has dimension **2**. - **D**: \(2 \times 2\) scalar matrices \(\begin{pmatrix} c & 0 \\ 0 & c \end{pmatrix} = c I_2\) have dimension **1** \(\implies\) **Isomorphic to \(\ker(T)\)**. - **E**: The solution space to \(Ax = 0\) for invertible \(A\) is \(\{0\}\), which has dimension **0**. - **F**: The \(x\)-axis \(\{(x, 0, 0) : x \in \mathbb{R}\}\) has dimension **1** \(\implies\) **Isomorphic to \(\ker(T)\)**. **Answer: D, F** A: **Incorrect:** The trivial space \(\{0\}\) has dimension 0, whereas \(\dim(\ker(T)) = 1\). B: **Incorrect:** The space of \(2 \times 2\) diagonal matrices has dimension 2. C: **Incorrect:** The \(yz\)-plane in \(\mathbb{R}^3\) has dimension 2. D: **Correct:** The space of \(2 \times 2\) scalar matrices \(\{c I_2 : c \in \mathbb{R}\}\) is 1-dimensional and thus isomorphic to \(\ker(T)\). E: **Incorrect:** When \(A\) is invertible, \(Ax = 0\) has only the trivial solution \(\{0\}\) (dimension 0). F: **Correct:** The \(x\)-axis in \(\mathbb{R}^3\) is 1-dimensional and thus isomorphic to \(\ker(T)\).