update notations

content/CSE510/CSE510_L11.md

@@ -198,20 +198,20 @@ $$
 
 Take the softmax policy as example:
 
-Weight actions using the linear combination of features $\phi(s,a)^T\theta$:
+Weight actions using the linear combination of features $\phi(s,a)^\top\theta$:
 
 Probability of action is proportional to the exponentiated weights:
 
 $$
-\pi_\theta(s,a) \propto \exp(\phi(s,a)^T\theta)
+\pi_\theta(s,a) \propto \exp(\phi(s,a)^\top\theta)
 $$
 
 The score function is
 
 $$
 \begin{aligned}
-\nabla_\theta \ln\left[\frac{\exp(\phi(s,a)^T\theta)}{\sum_{a'\in A}\exp(\phi(s,a')^T\theta)}\right] &= \nabla_\theta(\ln \exp(\phi(s,a)^T\theta) - (\ln \sum_{a'\in A}\exp(\phi(s,a')^T\theta))) \\
-&= \nabla_\theta\left(\phi(s,a)^T\theta -\frac{\phi(s,a)\sum_{a'\in A}\exp(\phi(s,a')^T\theta)}{\sum_{a'\in A}\exp(\phi(s,a')^T\theta)}\right) \\
+\nabla_\theta \ln\left[\frac{\exp(\phi(s,a)^\top\theta)}{\sum_{a'\in A}\exp(\phi(s,a')^\top\theta)}\right] &= \nabla_\theta(\ln \exp(\phi(s,a)^\top\theta) - (\ln \sum_{a'\in A}\exp(\phi(s,a')^\top\theta))) \\
+&= \nabla_\theta\left(\phi(s,a)^\top\theta -\frac{\phi(s,a)\sum_{a'\in A}\exp(\phi(s,a')^\top\theta)}{\sum_{a'\in A}\exp(\phi(s,a')^\top\theta)}\right) \\
 &=\phi(s,a) - \sum_{a'\in A} \prod_\theta(s,a') \phi(s,a')
 &= \phi(s,a) - \mathbb{E}_{a'\sim \pi_\theta(s,a')}[\phi(s,a')]
 \end{aligned}
@@ -221,7 +221,7 @@ $$
 
 In continuous action spaces, a Gaussian policy is natural
 
-Mean is a linear combination of state features $\mu(s) = \phi(s)^T\theta$
+Mean is a linear combination of state features $\mu(s) = \phi(s)^\top\theta$
 
 Variance may be fixed $\sigma^2$, or can also parametrized
 

content/CSE510/CSE510_L12.md

@@ -53,7 +53,7 @@ $$
 Action-Value Actor-Critic
 
 - Simple actor-critic algorithm based on action-value critic
-- Using linear value function approximation $Q_w(s,a)=\phi(s,a)^T w$
+- Using linear value function approximation $Q_w(s,a)=\phi(s,a)^\top w$
 
 Critic: updates $w$ by linear $TD(0)$
 Actor: updates $\theta$ by policy gradient

content/CSE510/CSE510_L13.md

@@ -193,7 +193,7 @@ $$
 
 Make linear approximation to $L_{\pi_{\theta_{old}}}$ and quadratic approximation to KL term.
 
-Maximize $g\cdot(\theta-\theta_{old})-\frac{\beta}{2}(\theta-\theta_{old})^T F(\theta-\theta_{old})$
+Maximize $g\cdot(\theta-\theta_{old})-\frac{\beta}{2}(\theta-\theta_{old})^\top F(\theta-\theta_{old})$
 
 where $g=\frac{\partial}{\partial \theta}L_{\pi_{\theta_{old}}}(\pi_{\theta})\vert_{\theta=\theta_{old}}$ and $F=\frac{\partial^2}{\partial \theta^2}\overline{KL}_{\pi_{\theta_{old}}}(\pi_{\theta})\vert_{\theta=\theta_{old}}$
 
@@ -201,7 +201,7 @@ where $g=\frac{\partial}{\partial \theta}L_{\pi_{\theta_{old}}}(\pi_{\theta})\ve
 <summary>Taylor Expansion of KL Term</summary>
 
 $$
-D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\approx D_{KL}(\pi_{\theta_{old}}|\pi_{\theta_{old}})+d^T \nabla_\theta D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\vert_{\theta=\theta_{old}}+\frac{1}{2}d^T \nabla_\theta^2 D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\vert_{\theta=\theta_{old}}d
+D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\approx D_{KL}(\pi_{\theta_{old}}|\pi_{\theta_{old}})+d^\top \nabla_\theta D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\vert_{\theta=\theta_{old}}+\frac{1}{2}d^\top \nabla_\theta^2 D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\vert_{\theta=\theta_{old}}d
 $$
 
 $$
@@ -220,9 +220,9 @@ $$
 \begin{aligned}
 \nabla_\theta^2 D_{KL}(\pi_{\theta_{old}}|\pi_{\theta})\vert_{\theta=\theta_{old}}&=-\mathbb{E}_{x\sim \pi_{\theta_{old}}}\nabla_\theta^2 \log P_\theta(x)\vert_{\theta=\theta_{old}}\\
 &=-\mathbb{E}_{x\sim \pi_{\theta_{old}}}\nabla_\theta \left(\frac{\nabla_\theta P_\theta(x)}{P_\theta(x)}\right)\vert_{\theta=\theta_{old}}\\
-&=-\mathbb{E}_{x\sim \pi_{\theta_{old}}}\left(\frac{\nabla_\theta^2 P_\theta(x)-\nabla_\theta P_\theta(x)\nabla_\theta P_\theta(x)^T}{P_\theta(x)^2}\right)\vert_{\theta=\theta_{old}}\\
-&=-\mathbb{E}_{x\sim \pi_{\theta_{old}}}\left(\frac{\nabla_\theta^2 P_\theta(x)\vert_{\theta=\theta_{old}}}P_{\theta_{old}}(x)\right)+\mathbb{E}_{x\sim \pi_{\theta_{old}}}\left(\nabla_\theta \log P_\theta(x)\nabla_\theta \log P_\theta(x)^T\right)\vert_{\theta=\theta_{old}}\\
-&=\mathbb{E}_{x\sim \pi_{\theta_{old}}}\nabla_\theta\log P_\theta(x)\nabla_\theta\log P_\theta(x)^T\vert_{\theta=\theta_{old}}\\
+&=-\mathbb{E}_{x\sim \pi_{\theta_{old}}}\left(\frac{\nabla_\theta^2 P_\theta(x)-\nabla_\theta P_\theta(x)\nabla_\theta P_\theta(x)^\top}{P_\theta(x)^2}\right)\vert_{\theta=\theta_{old}}\\
+&=-\mathbb{E}_{x\sim \pi_{\theta_{old}}}\left(\frac{\nabla_\theta^2 P_\theta(x)\vert_{\theta=\theta_{old}}}P_{\theta_{old}}(x)\right)+\mathbb{E}_{x\sim \pi_{\theta_{old}}}\left(\nabla_\theta \log P_\theta(x)\nabla_\theta \log P_\theta(x)^\top\right)\vert_{\theta=\theta_{old}}\\
+&=\mathbb{E}_{x\sim \pi_{\theta_{old}}}\nabla_\theta\log P_\theta(x)\nabla_\theta\log P_\theta(x)^\top\vert_{\theta=\theta_{old}}\\
 \end{aligned}
 $$
 

content/CSE510/CSE510_L14.md

@@ -27,7 +27,7 @@ $\theta_{new}=\theta_{old}+d$
 First order Taylor expansion for the loss and second order for the KL:
 
 $$
-\approx \arg\max_{d} J(\theta_{old})+\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d-\frac{1}{2}\lambda(d^T\nabla_\theta^2 D_{KL}\left[\pi_{\theta_{old}}||\pi_{\theta}\right]\mid_{\theta=\theta_{old}}d)+\lambda \delta
+\approx \arg\max_{d} J(\theta_{old})+\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d-\frac{1}{2}\lambda(d^\top\nabla_\theta^2 D_{KL}\left[\pi_{\theta_{old}}||\pi_{\theta}\right]\mid_{\theta=\theta_{old}}d)+\lambda \delta
 $$
 
 If you are really interested, try to fill the solving the KL Constrained Problem section.
@@ -38,7 +38,7 @@ Setting the gradient to zero:
 
 $$
 \begin{aligned}
-0&=\frac{\partial}{\partial d}\left(-\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d+\frac{1}{2}\lambda(d^T F(\theta_{old})d\right)\\
+0&=\frac{\partial}{\partial d}\left(-\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}d+\frac{1}{2}\lambda(d^\top F(\theta_{old})d\right)\\
 &=-\nabla_\theta J(\theta)\mid_{\theta=\theta_{old}}+\frac{1}{2}\lambda F(\theta_{old})d
 \end{aligned}
 $$
@@ -58,15 +58,15 @@ $$
 $$
 
 $$
-D_{KL}(\pi_{\theta_{old}}||\pi_{\theta})\approx \frac{1}{2}(\theta-\theta_{old})^T F(\theta_{old})(\theta-\theta_{old})
+D_{KL}(\pi_{\theta_{old}}||\pi_{\theta})\approx \frac{1}{2}(\theta-\theta_{old})^\top F(\theta_{old})(\theta-\theta_{old})
 $$
 
 $$
-\frac{1}{2}(\alpha g_N)^T F(\alpha g_N)=\delta
+\frac{1}{2}(\alpha g_N)^\top F(\alpha g_N)=\delta
 $$
 
 $$
-\alpha=\sqrt{\frac{2\delta}{g_N^T F g_N}}
+\alpha=\sqrt{\frac{2\delta}{g_N^\top F g_N}}
 $$
 
 However, due to the quadratic approximation, the KL constrains may be violated.

content/CSE510/CSE510_L18.md

@@ -16,7 +16,7 @@ So we can learn $f(s_t,a_t)$ from data, and _then_ plan through it.
 
 Model-based reinforcement learning version **0.5**:
 
-1. Run base polity $\pi_0$ (e.g. random policy) to collect $\mathcal{D} = \{(s_t, a_t, s_{t+1})\}_{t=0}^T$
+1. Run base polity $\pi_0$ (e.g. random policy) to collect $\mathcal{D} = \{(s_t, a_t, s_{t+1})\}_{t=0}^\top$
 2. Learn dynamics model $f(s_t,a_t)$ to minimize $\sum_{i}\|f(s_i,a_i)-s_{i+1}\|^2$
 3. Plan through $f(s_t,a_t)$ to choose action $a_t$
 
@@ -52,10 +52,10 @@ Version 2.0: backpropagate directly into policy
 
 Final version:
 
-1. Run base polity $\pi_0$ (e.g. random policy) to collect $\mathcal{D} = \{(s_t, a_t, s_{t+1})\}_{t=0}^T$
+1. Run base polity $\pi_0$ (e.g. random policy) to collect $\mathcal{D} = \{(s_t, a_t, s_{t+1})\}_{t=0}^\top$
 2. Learn dynamics model $f(s_t,a_t)$ to minimize $\sum_{i}\|f(s_i,a_i)-s_{i+1}\|^2$
 3. Backpropagate through $f(s_t,a_t)$ into the policy to optimized $\pi_\theta(s_t,a_t)$
-4. Run the policy $\pi_\theta(s_t,a_t)$ to collect $\mathcal{D} = \{(s_t, a_t, s_{t+1})\}_{t=0}^T$
+4. Run the policy $\pi_\theta(s_t,a_t)$ to collect $\mathcal{D} = \{(s_t, a_t, s_{t+1})\}_{t=0}^\top$
 5. Goto 2
 
 ## Model Learning with High-Dimensional Observations

content/CSE5313/CSE5313_L10.md

@@ -40,20 +40,20 @@ Let $G$ and $H$ be the generator and parity-check matrices of (any) linear code
 #### Lemma 1
 
 $$
-H G^T = 0
+H G^\top = 0
 $$
 
 <details>
 <summary>Proof</summary>
 
-By definition of generator matrix and parity-check matrix, $forall e_i\in H$, $e_iG^T=0$.
+By definition of generator matrix and parity-check matrix, $forall e_i\in H$, $e_iG^\top=0$.
 
-So $H G^T = 0$.
+So $H G^\top = 0$.
 </details>
 
 #### Lemma 2
 
-Any matrix $M\in \mathbb{F}_q^{(n-k)\times n}$ such that $\operatorname{rank}(M) = n - k$ and $M G^T = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).
+Any matrix $M\in \mathbb{F}_q^{(n-k)\times n}$ such that $\operatorname{rank}(M) = n - k$ and $M G^\top = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).
 
 <details>
 <summary>Proof</summary>
@@ -62,7 +62,7 @@ It is sufficient to show that the two statements
 
 1. $\forall c\in C, c=uG, u\in \mathbb{F}^k$
 
-$M c^T = M(uG)^T = M(G^T u^T) = 0$ since $M G^T = 0$.
+$M c^\top = M(uG)^\top = M(G^\top u^\top) = 0$ since $M G^\top = 0$.
 
 Thus $C \subseteq \ker M$.
 
@@ -84,15 +84,15 @@ We proceed by applying the lemma 2.
 
 1. $\operatorname{rank}(H) = n - k$ since $H$ is a Vandermonde matrix times a diagonal matrix with no zero entries, so $H$ is invertible.
 
-2. $H G^T = 0$.
+2. $H G^\top = 0$.
 
-note that $\forall$ row $i$ of $H$, $0\leq i\leq n-k-1$, $\forall$ column $j$ of $G^T$, $0\leq j\leq k-1$
+note that $\forall$ row $i$ of $H$, $0\leq i\leq n-k-1$, $\forall$ column $j$ of $G^\top$, $0\leq j\leq k-1$
 
 So
 
 $$
 \begin{aligned}
-H G^T &= \begin{bmatrix}
+H G^\top &= \begin{bmatrix}
 1 & 1 & \cdots & 1\\
 \alpha_1 & \alpha_2 & \cdots & \alpha_n\\
 \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\

content/CSE5313/CSE5313_L11.md

@@ -101,7 +101,7 @@ $$
 
 Let $\mathcal{C}=[n,k,d]_q$.
 
-The dual code of $\mathcal{C}$ is $\mathcal{C}^\perp=\{x\in \mathbb{F}^n_q|xc^T=0\text{ for all }c\in \mathcal{C}\}$.
+The dual code of $\mathcal{C}$ is $\mathcal{C}^\perp=\{x\in \mathbb{F}^n_q|xc^\top=0\text{ for all }c\in \mathcal{C}\}$.
 
 <details>
 <summary>Example</summary>
@@ -151,7 +151,7 @@ So $\langle f,h\rangle=0$.
 <details>
 <summary>Proof for the theorem</summary>
 
-Recall that the dual code of $\operatorname{RM}(r,m)^\perp=\{x\in \mathbb{F}_2^m|xc^T=0\text{ for all }c\in \operatorname{RM}(r,m)\}$.
+Recall that the dual code of $\operatorname{RM}(r,m)^\perp=\{x\in \mathbb{F}_2^m|xc^\top=0\text{ for all }c\in \operatorname{RM}(r,m)\}$.
 
 So $\operatorname{RM}(m-r-1,m)\subseteq \operatorname{RM}(r,m)^\perp$.
 

content/CSE5313/CSE5313_L14.md

@@ -230,7 +230,7 @@ Step 1: Arrange the $B=\binom{k+1}{2}+k(d-k)$ symbols in a matrix $M$ follows:
 $$
 M=\begin{pmatrix}
 S & T\\
-T^T & 0
+T^\top & 0
 \end{pmatrix}\in \mathbb{F}_q^{d\times d}
 $$
 
@@ -267,15 +267,15 @@ Repair from (any) nodes $H = \{h_1, \ldots, h_d\}$.
 
 Newcomer contacts each $h_j$: “My name is $i$, and I’m lost.”
 
-Node $h_j$ sends $c_{h_j}M c_i^T$ (inner product).
+Node $h_j$ sends $c_{h_j}M c_i^\top$ (inner product).
 
-Newcomer assembles $C_H Mc_i^T$.
+Newcomer assembles $C_H Mc_i^\top$.
 
 $CH$ invertible by construction!
 
-- Recover $Mc_i^T$.
+- Recover $Mc_i^\top$.
 
-- Recover $c_i^TM$ ($M$ is symmetric)
+- Recover $c_i^\topM$ ($M$ is symmetric)
 
 #### Reconstruction on Product-Matrix MBR codes
 
@@ -292,9 +292,9 @@ DC assembles $C_D M$.
 
 $\Psi_D$ invertible by construction.
 
-- DC computes $\Psi_D^{-1}C_DM = (S+\Psi_D^{-1}\Delta_D^T, T)$
+- DC computes $\Psi_D^{-1}C_DM = (S+\Psi_D^{-1}\Delta_D^\top, T)$
 - DC obtains $T$.
-- Subtracts $\Psi_D^{-1}\Delta_D T^T$ from $S+\Psi_D^{-1}\Delta_D T^T$ to obtain $S$.
+- Subtracts $\Psi_D^{-1}\Delta_D T^\top$ from $S+\Psi_D^{-1}\Delta_D T^\top$ to obtain $S$.
 
 <details>
 <summary>Fill an example here please.</summary>

content/CSE5313/CSE5313_L19.md

@@ -0,0 +1,232 @@
+# CSE5313 Coding and information theory for data science (Lecture 19)
+
+## Private information retrieval
+
+### Problem setup
+
+Premise:
+
+- Database $X = \{x_1, \ldots, x_m\}$, each $x_i \in \mathbb{F}_q^k$ is a "file" (e.g., medical record).
+- $X$ is coded $X \mapsto \{y_1, \ldots, y_n\}$, $y_j$ stored at server $j$.
+- The user (physician) wants $x_i$.
+- The user sends a query $q_j \sim Q_j$ to server $j$.
+- Server $j$ responds with $a_j \sim A_j$.
+
+Decodability:
+
+- The user can retrieve the file: $H(X_i | A_1, \ldots, A_n) = 0$.
+
+Privacy:
+
+- $i$ is seen as $i \sim U = U_{m}$, reflecting server's lack of knowledge.
+- $i$ must be kept private: $I(Q_j; U) = 0$ for all $j \in n$.
+
+> In short, we want to retrieve $x_i$ from the servers without revealing $i$ to the servers.
+
+### Private information retrieval from Replicated Databases
+
+#### Simple case, one server
+
+Say $n = 1, y_1 = X$.
+
+- All data is stored in one server.
+- Simple solution:
+- $q_1 =$ "send everything".
+- $a_1 = y_1 = X$.
+
+Theorem: Information Theoretic PIR with $n = 1$ can only be achieved by downloading the entire database.
+
+- Can we do better if $n > 1$?
+
+#### Collusion parameter
+
+Key question for $n > 1$: Can servers collude?
+
+- I.e., does server $j$ see any $Q_\ell$, $\ell \neq j$?
+- Key assumption:
+  - Privacy parameter $z$.
+  - At most $z$ servers can collude.
+  - $z = 1\implies$ No collusion.
+- Requirement for $z = 1$: $I(Q_j; U) = 0$ for all $j \in n$.
+- Requirement for a general $z$:
+  - $I(Q_\mathcal{T}; U) = 0$ for all $\mathcal{T} \in n$, $|\mathcal{T}| \leq z$, where $Q_\mathcal{T} = Q_\ell$ for all $\ell \in \mathcal{T}$.
+- Motivation:
+  - Interception of communication links.
+  - Data breaches.
+
+Other assumptions:
+
+- Computational Private information retrieval (even all the servers are hacked, still cannot get the information -> solve np-hard problem):
+- Non-zero MI
+
+#### Private information retrieval from 2-replicated databases
+
+First PIR protocol: Chor et al. FOCS ‘95.
+
+- The data $X = \{x_1, \ldots, x_m\}$ is replicated on two servers.
+  - $z = 1$, i.e., no collusion.
+- Protocol: User has $i \sim U_{m}$.
+  - User generates $r \sim U_{\mathbb{F}_q^m}$.
+  - $q_1 = r, q_2 = r + e_i$ ($e_i \in \mathbb{F}_q^m$ is the $i$-th unit vector, $q_2$ is equivalent to one-time pad encryption of $x_i$ with key $r$).
+  - $a_j = q_j X^\top = \sum_{\ell \in m} q_j, \ell x_\ell$
+  - Linear combination of the files according to the query vector $q_j$.
+- Decoding?
+  - $a_2 - a_1 = q_2 - q_1 X^\top = e_i X^\top = x_i$.
+- Download?
+  - $a_j =$ size of file $\implies$ downloading **twice** the size of the file.
+- Privacy?
+  - Since $z = 1$, need to show $I(U; Q_i) = 0$.
+    - $I(U; Q_1) = I(e_U; F) = 0$ since $U$ and $F$ are independent.
+    - $I(U; Q_2) = I(e_U; F + e_U) = 0$ since this is one-time pad!
+
+##### Parameters and notations in PIR
+
+Parameters of the system:
+
+- $n =$ # servers (as in storage).
+- $m =$ # files.
+- $k =$ size of each file (as in storage).
+- $z =$ max. collusion (as in secret sharing).
+- $t =$ # of answers required to obtain $x_i$ (as in secret sharing).
+  - $n - t$ servers are “stragglers”, i.e., might not respond.
+
+Figures of merit:
+
+- PIR-rate = $\#$ desired symbols / $\#$ downloaded symbols
+- PIR-capacity = largest possible rate.
+
+Notaional conventions:
+
+-The dataset $X = \{x_j\}_{j \in m} = \{x_{j, \ell}\}_{(j, \ell) \in [m] \times [k]}$ is seen as a vector in $\mathbb{F}_q^{mk}$.
+
+- Index $\mathbb{F}_q^{mk}$ using $[m] \times [k]$, i.e., $x_{j, \ell}$ is the $\ell$-th symbol of the $j$-th file.
+
+#### Private information retrieval from 4-replicated databases
+
+Consider $n = 4$ replicated servers, file size $k = 2$, collusion $z = 1$.
+
+Protocol: User has $i \sim U_{m}$.
+
+- Fix distinct nonzero $\alpha_1, \ldots, \alpha_4 \in \mathbb{F}_q$.
+- Choose $r \sim U_{\mathbb{F}_q^{2m}}$.
+- User sends $q_j = e_{i, 1} + \alpha_j e_{i, 2} + \alpha_j^2 r$ to each server $j$.
+- Server $j$ responds with
+  $$
+  a_j = q_j X^\top = e_{i, 1} X^\top + \alpha_j e_{i, 2} X^\top + \alpha_j^2 r X^\top
+  $$
+  - This is an evaluation at $\alpha_j$ of the polynomial $f_i(w) = x_{i, 1} + x_{i, 2} \cdot w + r \cdot w^2$.
+  - Where $r$ is some random combination of the entries of $X$.
+- Decoding?
+  - Any 3 responses suffice to interpolate $f_i$ and obtain $x_i = x_{i, 1}, x_{i, 2}$.
+  - $\implies t = 3$, (one straggler is allowed)
+- Privacy?
+  - Does $q_j = e_{i, 1} + \alpha_j e_{i, 2} + \alpha_j^2 r$ look familiar?
+  - This is a share in [ramp scheme](CSE5313_L18.md#scheme-2-ramp-secret-sharing-scheme-mceliece-sarwate-scheme) with vector messages $m_1 = e_{i, 1}, m_2 = e_{i, 2}, m_i \in \mathbb{F}_q^{2m}$.
+  - This is equivalent to $2m$ "parallel" ramp scheme over $\mathbb{F}_q$.
+  - Each one reveals nothing to any $z = 1$ shareholders $\implies$ Private!
+
+### Private information retrieval from general replicated databases
+
+$n$ servers, $m$ files, file size $k$, $X \in \mathbb{F}_q^{mk}$.
+
+Server decodes $x_i$ from any $t$ responses.
+
+Any $\leq z$ servers might collude to infer $i$ ($z < t$).
+
+Protocol: User has $i \sim U_{m}$.
+
+- User chooses $r_1, \ldots, r_z \sim U_{\mathbb{F}_q^{mk}}$.
+- User sends $q_j = \sum_{\ell=1}^k e_{i, \ell} \alpha_j^{\ell-1} + \sum_{\ell=1}^z r_\ell \alpha_j^{k+\ell-1}$ to each server $j$.
+- Server $j$ responds with $a_j = q_j X^\top = f_i(\alpha_j)$.
+  - $f_i(w) = \sum_{\ell=1}^k e_{i, \ell} X^\top w^{\ell-1} + \sum_{\ell=1}^z r_\ell X^\top w^{k+\ell-1}$ (random combinations of $X$).
+  - Caveat: must have $t = k + z$.
+  - $\implies \deg f_i = k + z - 1 = t - 1$.
+- Decoding?
+  - Interpolation from any $t$ evaluations of $f_i$.
+- Privacy?
+  - Against any $z = t - k$ colluding servers, immediate from the proof of the ramp scheme.
+
+PIR-rate?
+
+- Each $a_j$ is a single field element.
+- Download $t = k + z$ elements in $\mathbb{F}_q$ in order to obtain $x_i \in \mathbb{F}_q^k$.
+- $\implies$ PIR-rate = $\frac{k}{k+z} = \frac{k}{t}$.
+
+#### Theorem: PIR-capacity for general replicated databases
+
+The PIR-capacity for $n$ replicated databases with $z$ colluding servers, $n - t$ unresponsive servers, and $m$ files is $C = \frac{1-\frac{z}{t}}{1-(\frac{z}{t})^m}$.
+
+- When $m \to \infty$, $C \to 1 - \frac{z}{t} = \frac{t-z}{t} = \frac{k}{t}$.
+- The above scheme achieves PIR-capacity as $m \to \infty$
+
+### Private information retrieval from coded databases
+
+#### Problem setup:
+
+Example:
+
+- $n = 3$ servers, $m$ files $x_j$, $x_j = x_{j, 1}, x_{j, 2}$, $k = 2$, and $q = 2$.
+- Code each file with a parity code: $x_{j, 1}, x_{j, 2} \mapsto x_{j, 1}, x_{j, 2}, x_{j, 1} + x_{j, 2}$.
+- Server $j \in 3$ stores all $j$-th symbols of all coded files.
+
+Queries, answers, decoding, and privacy must be tailored for the code at hand.
+
+With respect to a code $C$ and parameters $n, k, t, z$, such scheme is called coded-PIR.
+
+- The content for server $j$ is denoted by $c_j = c_{j, 1}, \ldots, c_{j, m}$.
+- $C$ is usually an MDS code.
+
+#### Private information retrieval from parity-check codes
+
+Example:
+
+ Say $z = 1$ (no collusion).
+
+- Protocol: User has $i \sim U_{m}$.
+- User chooses $r_1, r_2 \sim U_{\mathbb{F}_2^m}$.
+- Two queries to each server:
+  - $q_{1, 1} = r_1 + e_i$, $q_{1, 2} = r_2$.
+  - $q_{2, 1} = r_1$, $q_{2, 2} = r_2 + e_i$.
+  - $q_{3, 1} = r_1$, $q_{3, 2} = r_2$.
+- Server $j$ responds with $q_{j, 1} c_j^\top$ and $q_{j, 2} c_j^\top$.
+- Decoding?
+  - $q_{1, 1} c_1^\top + q_{2, 1} c_2^\top + q_{3, 1} c_3^\top = r_1 c_1 + c_2 + c_3 + e_i c_1^\top = r_1 \cdot 0^\top + x_{i, 1} = x_{i, 1}$.
+  - $q_{1, 1} c_1^\top + q_{2, 1} c_2^\top + q_{3, 1} c_3^\top = r_1 \cdot 0^\top + x_{i, 1} = x_{i, 1}$.
+  - $q_{1, 2} c_1^\top + q_{2, 2} c_2^\top + q_{3, 2} c_3^\top = r_2 c_1 + c_2 + c_3^\top + e_i c_2^\top = x_{i, 2}$.
+- Privacy?
+  - Every server sees two uniformly random vectors in $\mathbb{F}_2^m$.
+
+<details>
+<summary>Proof from coding-theoretic interpretation</summary>
+
+Let $G = g_1^\top, g_2^\top, g_3^\top$ be the generator matrix. 
+
+- For every file $x_j = x_{j, 1}, x_{j, 2}$ we encode $x_j G = (x_{j, 1} g_1^\top, x_{j, 2} g_2^\top, x_{j, 1} g_3^\top) = (c_{j, 1}, c_{j, 2}, c_{j, 3})$.
+- Server $j$ stores $X g_j^\top = (x_1^\top, \ldots, x_m^\top)^\top g_j^\top = (c_{j, 1}, \ldots, c_{j, m})^\top$.
+
+- By multiplying by $r_1$, the servers together store a codeword in $C$:
+  - $r_1 X g_1^\top, r_1 X g_2^\top, r_1 X g_3^\top = r_1 X G$.
+- By replacing one of the $r_1$’s by $r_1 + e_i$, we introduce an error in that entry:
+  - $\left((r_1 + e_i) X g_1^\top, r_1 X g_2^\top, r_1 X g_3^\top\right) = r_1 X G + (e_i X g_1^\top, 0,0)$.
+- Downloading this “erroneous” word from the servers and multiply by $H = h_1^\top, h_2^\top, h_3^\top$ be the parity-check matrix.
+
+$$
+\begin{aligned}
+\left((r_1 + e_i) X g_1^\top, r_1 X g_2^\top, r_1 X g_3^\top\right) H^\top &= \left(r_1 X G + (e_i X g_1^\top, 0,0)\right) H^\top \\
+&= r_1 X G H^\top + (e_i X g_1^\top, 0,0) H^\top \\
+&= 0 + x_{i, 1} g_1^\top \\
+&= x_{i, 1}.
+\end{aligned}
+$$
+
+> In homework we will show tha this work with any MDS code ($z=1$).
+
+- Say we obtained $x_{i, 1} g_1^\top, \ldots, x_{i, k} g_k^\top$ (𝑑 − 1 at a time, how?).
+- $x_{i, 1} g_1^\top, \ldots, x_{i, k} g_k^\top = x_{i, B}$, where $B$ is a $k \times k$ submatrix of $G$.
+- $B$ is a $k \times k$ submatrix of $G$ $\implies$ invertible! $\implies$ Obtain $x_{i}$.
+
+</details>
+
+> [!TIP]
+>
+> error + known location $\implies$ erasure. $d = 2 \implies$ 1 erasure is correctable.

content/CSE5313/CSE5313_L6.md

@@ -92,10 +92,10 @@ Two equivalent ways to constructing a linear code:
 
 - A **parity check** matrix $H\in \mathbb{F}^{(n-k)\times n}$ with $(n-k)$ rows and $n$ columns.
   $$
-  \mathcal{C}=\{c\in \mathbb{F}^n:Hc^T=0\}
+  \mathcal{C}=\{c\in \mathbb{F}^n:Hc^\top=0\}
   $$
   - The right kernel of $H$ is $\mathcal{C}$.
-  - Multiplying $c^T$ by $H$ "checks" if $c\in \mathcal{C}$.
+  - Multiplying $c^\top$ by $H$ "checks" if $c\in \mathcal{C}$.
 
 ### Encoding of linear codes
 
@@ -144,7 +144,7 @@ Decoding: $(y+e)\to x$, $y=xG$.
 Use **syndrome** to identify which coset $\mathcal{C}_i$ that the noisy-code to $\mathcal{C}_i+e$ belongs to.
 
 $$
-H(y+e)^T=H(y+e)=Hx+He=He
+H(y+e)^\top=H(y+e)=Hx+He=He
 $$
 
 ### Syndrome decoding
@@ -215,7 +215,7 @@ Fourth row is $\mathcal{C}+(00100)$.
 
 Any two elements in a row are of the form $y_1'=y_1+e$ and $y_2'=y_2+e$ for some $e\in \mathbb{F}^n$.
 
-Same syndrome if $H(y_1'+e)^T=H(y_2'+e)^T$.
+Same syndrome if $H(y_1'+e)^\top=H(y_2'+e)^\top$.
 
 Entries in different rows have different syndrome.
 

content/CSE5313/CSE5313_L7.md

@@ -7,7 +7,7 @@ Let $\mathcal{C}= [n,k,d]_{\mathbb{F}}$ be a linear code.
 There are two equivalent ways to describe a linear code:
 
 1. A generator matrix $G\in \mathbb{F}^{k\times n}_q$ with $k$ rows and $n$ columns, entry taken from $\mathbb{F}_q$. $\mathcal{C}=\{xG|x\in \mathbb{F}^k\}$
-2. A parity check matrix $H\in \mathbb{F}^{(n-k)\times n}_q$ with $(n-k)$ rows and $n$ columns, entry taken from $\mathbb{F}_q$. $\mathcal{C}=\{c\in \mathbb{F}^n:Hc^T=0\}$
+2. A parity check matrix $H\in \mathbb{F}^{(n-k)\times n}_q$ with $(n-k)$ rows and $n$ columns, entry taken from $\mathbb{F}_q$. $\mathcal{C}=\{c\in \mathbb{F}^n:Hc^\top=0\}$
 
 ### Dual code
 
@@ -21,7 +21,7 @@ $$
 
 Also, the alternative definition is:
 
-1. $C^{\perp}=\{x\in \mathbb{F}^n:Gx^T=0\}$ (only need to check basis of $C$)
+1. $C^{\perp}=\{x\in \mathbb{F}^n:Gx^\top=0\}$ (only need to check basis of $C$)
 2. $C^{\perp}=\{xH|x\in \mathbb{F}^{n-k}\}$
 
 By rank-nullity theorem, $dim(C^{\perp})=n-dim(C)=n-k$.
@@ -87,7 +87,7 @@ Assume minimum distance is $d$. Show that every $d-1$ columns of $H$ are indepen
 
 - Fact: In linear codes minimum distance is the minimum weight ($d_H(x,y)=w_H(x-y)$).
 
-Indeed, if there exists a $d-1$ columns of $H$ that are linearly dependent, then we have $Hc^T=0$ for some $c\in \mathcal{C}$ with $w_H(c)<d$.
+Indeed, if there exists a $d-1$ columns of $H$ that are linearly dependent, then we have $Hc^\top=0$ for some $c\in \mathcal{C}$ with $w_H(c)<d$.
 
 Reverse are similar.
 
@@ -130,7 +130,7 @@ $k=2^m-m-1$.
 
 Define the code by encoding function:
 
-$E(x): \mathbb{F}_2^m\to \mathbb{F}_2^{2^m}=(xy_1^T,\cdots,xy_{2^m}^T)$ ($y\in \mathbb{F}_2^m$)
+$E(x): \mathbb{F}_2^m\to \mathbb{F}_2^{2^m}=(xy_1^\top,\cdots,xy_{2^m}^\top)$ ($y\in \mathbb{F}_2^m$)
 
 Space of codewords is image of $E$.
 

content/CSE5313/CSE5313_L8.md

@@ -258,7 +258,7 @@ Algorithm:
 - Begin with $(n-k)\times (n-k)$ identity matrix.
 - Assume we choose columns $h_1,h_2,\ldots,h_\ell$ (each $h_i$ is in $\mathbb{F}^n_q$)
 - Then next column $h_{\ell}$ must not be in the space of any previous $d-2$ columns.
-  - $h_{\ell}$ cannot be written as $[h_1,h_2,\ldots,h_{\ell-1}]x^T$ for $x$ of Hamming weight at most $d-2$.
+  - $h_{\ell}$ cannot be written as $[h_1,h_2,\ldots,h_{\ell-1}]x^\top$ for $x$ of Hamming weight at most $d-2$.
   - So the ineligible candidates for $h_{\ell}$ is:
     - $B_{\ell-1}(0,d-2)=\{x\in \mathbb{F}^{\ell-1}_q: d_H(0,x)\leq d-2\}$.
     - $|B_{\ell-1}(0,d-2)|=\sum_{i=0}^{d-2}\binom{\ell-1}{i}(q-1)^i$, denoted by $V_q(\ell-1, d-2)$.

content/CSE5313/CSE5313_L9.md

@@ -148,15 +148,15 @@ The generator matrix for Reed-Solomon code is a Vandermonde matrix $V(a_1,a_2,\l
 
 Fact: $V(a_1,a_2,\ldots,a_n)$ is invertible if and only if $a_1,a_2,\ldots,a_n$ are distinct. (that's how we choose $a_1,a_2,\ldots,a_n$)
 
-The parity check matrix for Reed-Solomon code is also a Vandermonde matrix $V(a_1,a_2,\ldots,a_n)^T$ with scalar multiples of the columns.
+The parity check matrix for Reed-Solomon code is also a Vandermonde matrix $V(a_1,a_2,\ldots,a_n)^\top$ with scalar multiples of the columns.
 
 Some technical lemmas:
 
 Let $G$ and $H$ be the generator and parity-check matrices of (any) linear code
 $C = [n, k, d]_{\mathbb{F}_q}$. Then:
 
-I. Then $H G^T = 0$.
-II. Any matrix $M \in \mathbb{F}_q^{n-k \times k}$ such that $\rank(M) = n - k$ and $M G^T = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).
+I. Then $H G^\top = 0$.
+II. Any matrix $M \in \mathbb{F}_q^{n-k \times k}$ such that $\rank(M) = n - k$ and $M G^\top = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).
 
 ## Reed-Muller code
 

content/CSE5313/_meta.js

@@ -22,4 +22,5 @@ export default {
     CSE5313_L16: "CSE5313 Coding and information theory for data science (Exam Review)",
     CSE5313_L17: "CSE5313 Coding and information theory for data science (Lecture 17)",
     CSE5313_L18: "CSE5313 Coding and information theory for data science (Lecture 18)",
+    CSE5313_L19: "CSE5313 Coding and information theory for data science (Exam Review)",
 }

content/CSE5519/CSE5519_E1.md

@@ -469,7 +469,7 @@ $$
 Then we use $\mathcal{L}_{ds}$ to enforce the smoothness of the disparity map.
 
 $$
-\mathcal{L}_{ds}=\sum_{p\in I^l}\left|\partial_x d^l_p\right|e^{-\left|\partial_x d^l_p\right|}+\left|\partial_y d^l_p\right|e^{-\left|\partial_y d^l_p\right|}=\sum_{p_t}|\nabla D(p_t)|\cdot \left(e^{-|\nabla I(p_t)|}\right)^T\tag{2}
+\mathcal{L}_{ds}=\sum_{p\in I^l}\left|\partial_x d^l_p\right|e^{-\left|\partial_x d^l_p\right|}+\left|\partial_y d^l_p\right|e^{-\left|\partial_y d^l_p\right|}=\sum_{p_t}|\nabla D(p_t)|\cdot \left(e^{-|\nabla I(p_t)|}\right)^\top\tag{2}
 $$
 
 Replacing $\hat{I}^{rig}_s$ with $\hat{I}^{full}_s$, in (1) and (2), we get the $\mathcal{L}_{fw}$ and $\mathcal{L}_{fs}$ for the non-rigid motion localizer.

content/CSE559A/CSE559A_L21.md

@@ -64,7 +64,7 @@ $d = \begin{bmatrix}
 u \\ v
 \end{bmatrix}$
 
-The solution is $d=(A^T A)^{-1} A^T b$
+The solution is $d=(A^\top A)^{-1} A^\top b$
 
 Lucas-Kanade flow: 
 
@@ -170,7 +170,7 @@ E = \sum_{i=1}^n (a(x_i-\bar{x})+b(y_i-\bar{y}))^2 = \left\|\begin{bmatrix}x_1-\
 $$
 
 We want to find $N$ that minimizes $\|UN\|^2$ subject to $\|N\|^2= 1$
-Solution is given by the eigenvector of $U^T U$ associated with the smallest eigenvalue
+Solution is given by the eigenvector of $U^\top U$ associated with the smallest eigenvalue
 
 Drawbacks:
 

content/CSE559A/CSE559A_L22.md

@@ -178,7 +178,7 @@ $$
 \begin{pmatrix}a\\b\\c\end{pmatrix} \times \begin{pmatrix}a'\\b'\\c'\end{pmatrix} = \begin{pmatrix}bc'-b'c\\ca'-c'a\\ab'-a'b\end{pmatrix}
 $$
 
-Let $h_1^T, h_2^T, h_3^T$ be the rows of $H$. Then
+Let $h_1^\top, h_2^\top, h_3^\top$ be the rows of $H$. Then
 
 $$
 x_i' × Hx_i=\begin{pmatrix}
@@ -186,15 +186,15 @@ x_i' × Hx_i=\begin{pmatrix}
     y_i' \\
     1
 \end{pmatrix} \times \begin{pmatrix}
-    h_1^T x_i \\
-    h_2^T x_i \\
-    h_3^T x_i
+    h_1^\top x_i \\
+    h_2^\top x_i \\
+    h_3^\top x_i
 \end{pmatrix}
 =
 \begin{pmatrix}
-    y_i' h_3^T x_i−h_2^T x_i \\
-    h_1^T x_i−x_i' h_3^T x_i \\
-    x_i' h_2^T x_i−y_i' h_1^T x_i
+    y_i' h_3^\top x_i−h_2^\top x_i \\
+    h_1^\top x_i−x_i' h_3^\top x_i \\
+    x_i' h_2^\top x_i−y_i' h_1^\top x_i
 \end{pmatrix}
 $$
 
@@ -206,15 +206,15 @@ x_i' × Hx_i=\begin{pmatrix}
     y_i' \\
     1
 \end{pmatrix} \times \begin{pmatrix}
-    h_1^T x_i \\
-    h_2^T x_i \\
-    h_3^T x_i
+    h_1^\top x_i \\
+    h_2^\top x_i \\
+    h_3^\top x_i
 \end{pmatrix}
 =
 \begin{pmatrix}
-    y_i' h_3^T x_i−h_2^T x_i \\
-    h_1^T x_i−x_i' h_3^T x_i \\
-    x_i' h_2^T x_i−y_i' h_1^T x_i
+    y_i' h_3^\top x_i−h_2^\top x_i \\
+    h_1^\top x_i−x_i' h_3^\top x_i \\
+    x_i' h_2^\top x_i−y_i' h_1^\top x_i
 \end{pmatrix}
 $$
 
@@ -222,9 +222,9 @@ Rearranging the terms:
 
 $$
 \begin{bmatrix}
-    0^T &-x_i^T &y_i' x_i^T \\
-    x_i^T &0^T &-x_i' x_i^T \\
-    y_i' x_i^T &x_i' x_i^T &0^T
+    0^\top &-x_i^\top &y_i' x_i^\top \\
+    x_i^\top &0^\top &-x_i' x_i^\top \\
+    y_i' x_i^\top &x_i' x_i^\top &0^\top
 \end{bmatrix}
 \begin{bmatrix}
     h_1 \\

content/CSE559A/CSE559A_L26.md

@@ -17,16 +17,16 @@ If we set the config for the first camera as the world origin and $[I|0]\begin{p
 Notice that $x'\cdot [t\times (Ry)]=0$
 
 $$
-x'^T E x_1 = 0
+x'^\top E x_1 = 0
 $$
 
 We denote the constraint defined by the Essential Matrix as $E$.
 
 $E x$ is the epipolar line associated with $x$ ($l'=Ex$)
 
-$E^T x'$ is the epipolar line associated with $x'$ ($l=E^T x'$)
+$E^\top x'$ is the epipolar line associated with $x'$ ($l=E^\top x'$)
 
-$E e=0$ and $E^T e'=0$ ($x$ and $x'$ don't matter)
+$E e=0$ and $E^\top e'=0$ ($x$ and $x'$ don't matter)
 
 $E$ is singular (rank 2) and have five degrees of freedom.
 
@@ -35,13 +35,13 @@ $E$ is singular (rank 2) and have five degrees of freedom.
 If the calibration matrices $K$ and $K'$ are unknown, we can write the epipolar constraint in terms of unknown normalized coordinates:
 
 $$
-x'^T_{norm} E x_{norm} = 0
+x'^\top_{norm} E x_{norm} = 0
 $$
 
 where $x_{norm}=K^{-1} x$, $x'_{norm}=K'^{-1} x'$
 
 $$
-x'^T_{norm} E x_{norm} = 0\implies x'^T_{norm} Fx=0
+x'^\top_{norm} E x_{norm} = 0\implies x'^\top_{norm} Fx=0
 $$
 
 where $F=K'^{-1}EK^{-1}$ is the **Fundamental Matrix**.
@@ -60,17 +60,17 @@ Properties of $F$:
 
 $F x$ is the epipolar line associated with $x$ ($l'=F x$)
 
-$F^T x'$ is the epipolar line associated with $x'$ ($l=F^T x'$)
+$F^\top x'$ is the epipolar line associated with $x'$ ($l=F^\top x'$)
 
-$F e=0$ and $F^T e'=0$
+$F e=0$ and $F^\top e'=0$
 
 $F$ is singular (rank two) and has seven degrees of freedom
 
 #### Estimating the fundamental matrix
 
-Given: correspondences $x=(x,y,1)^T$ and $x'=(x',y',1)^T$
+Given: correspondences $x=(x,y,1)^\top$ and $x'=(x',y',1)^\top$
 
-Constraint: $x'^T F x=0$
+Constraint: $x'^\top F x=0$
 
 $$
 (x',y',1)\begin{bmatrix}
@@ -95,7 +95,7 @@ F=U\begin{bmatrix}
 \sigma_1 & 0 \\
 0 & \sigma_2 \\
 0 & 0
-\end{bmatrix}V^T
+\end{bmatrix}V^\top
 $$
 
 ## Structure from Motion
@@ -126,7 +126,7 @@ a_{21} & a_{22} & a_{23} & t_2 \\
 0 & 0 & 0 & 1
 \end{bmatrix}=\begin{bmatrix}
 A & t \\
-0^T & 1
+0^\top & 1
 \end{bmatrix}
 $$
 
@@ -160,10 +160,10 @@ The reconstruction is defined up to an arbitrary affine transformation $Q$ (12 d
 $$
 \begin{bmatrix}
 A & t \\
-0^T & 1
+0^\top & 1
 \end{bmatrix}\rightarrow\begin{bmatrix}
 A & t \\
-0^T & 1
+0^\top & 1
 \end{bmatrix}Q^{-1}, \quad \begin{pmatrix}X_j\\1\end{pmatrix}\rightarrow Q\begin{pmatrix}X_j\\1\end{pmatrix}
 $$
 

content/CSE559A/CSE559A_L5.md

@@ -74,7 +74,7 @@ x\\y
 \end{pmatrix}
 $$
 
-To undo the rotation, we need to rotate the image by $-\theta$. This is equivalent to apply $R^T$ to the image.
+To undo the rotation, we need to rotate the image by $-\theta$. This is equivalent to apply $R^\top$ to the image.
 
 #### Affine transformation
 

content/CSE559A/CSE559A_L7.md

@@ -96,7 +96,7 @@ Example: Linear classification models
 Find a linear function that separates the data.
 
 $$
-f(x) = w^T x + b
+f(x) = w^\top x + b
 $$
 
 [Linear classification models](http://cs231n.github.io/linear-classify/)
@@ -144,13 +144,13 @@ This is a convex function, so we can find the global minimum.
 The gradient is:
 
 $$
-\nabla_w||Xw-Y||^2 = 2X^T(Xw-Y)
+\nabla_w||Xw-Y||^2 = 2X^\top(Xw-Y)
 $$
 
 Set the gradient to 0, we get:
 
 $$
-w = (X^T X)^{-1} X^T Y
+w = (X^\top X)^{-1} X^\top Y
 $$
 
 From the maximum likelihood perspective, we can also derive the same result.

content/CSE559A/CSE559A_L8.md

@@ -59,7 +59,7 @@ Suppose $k=1$, $e=l(f_1(x,w_1),y)$
 
 Example: $e=(f_1(x,w_1)-y)^2$
 
-So $h_1=f_1(x,w_1)=w^T_1x$, $e=l(h_1,y)=(y-h_1)^2$
+So $h_1=f_1(x,w_1)=w^\top_1x$, $e=l(h_1,y)=(y-h_1)^2$
 
 $$
 \frac{\partial e}{\partial w_1}=\frac{\partial e}{\partial h_1}\frac{\partial h_1}{\partial w_1}

content/CSE559A/CSE559A_L9.md

@@ -20,7 +20,7 @@ Suppose $k=1$, $e=l(f_1(x,w_1),y)$
 
 Example: $e=(f_1(x,w_1)-y)^2$
 
-So $h_1=f_1(x,w_1)=w^T_1x$, $e=l(h_1,y)=(y-h_1)^2$
+So $h_1=f_1(x,w_1)=w^\top_1x$, $e=l(h_1,y)=(y-h_1)^2$
 
 $$
 \frac{\partial e}{\partial w_1}=\frac{\partial e}{\partial h_1}\frac{\partial h_1}{\partial w_1}

content/Math401/Extending_thesis/Math401_R1.md

@@ -262,10 +262,10 @@ Basic definitions
 
 The special orthogonal group $SO(n)$ is the set of all **distance preserving** linear transformations on $\mathbb{R}^n$.
 
-It is the group of all $n\times n$ orthogonal matrices ($A^T A=I_n$) on $\mathbb{R}^n$ with determinant $1$.
+It is the group of all $n\times n$ orthogonal matrices ($A^\top A=I_n$) on $\mathbb{R}^n$ with determinant $1$.
 
 $$
-SO(n)=\{A\in \mathbb{R}^{n\times n}: A^T A=I_n, \det(A)=1\}
+SO(n)=\{A\in \mathbb{R}^{n\times n}: A^\top A=I_n, \det(A)=1\}
 $$
 
 <details>
@@ -276,7 +276,7 @@ In [The random Matrix Theory of the Classical Compact groups](https://case.edu/a
 $O(n)$ (the group of all $n\times n$ **orthogonal matrices** over $\mathbb{R}$),
 
 $$
-O(n)=\{A\in \mathbb{R}^{n\times n}: AA^T=A^T A=I_n\}
+O(n)=\{A\in \mathbb{R}^{n\times n}: AA^\top=A^\top A=I_n\}
 $$
 
 $U(n)$ (the group of all $n\times n$ **unitary matrices** over $\mathbb{C}$), 
@@ -296,7 +296,7 @@ $$
 $Sp(2n)$ (the group of all $2n\times 2n$ symplectic matrices over $\mathbb{C}$),
 
 $$
-Sp(2n)=\{U\in U(2n): U^T J U=UJU^T=J\}
+Sp(2n)=\{U\in U(2n): U^\top J U=UJU^\top=J\}
 $$
 
 where $J=\begin{pmatrix}

content/Math401/Freiwald_summer/Math401_P1_2.md

@@ -8,10 +8,10 @@ The page's lemma is a fundamental result in quantum information theory that prov
 
 The special orthogonal group $SO(n)$ is the set of all **distance preserving** linear transformations on $\mathbb{R}^n$.
 
-It is the group of all $n\times n$ orthogonal matrices ($A^T A=I_n$) on $\mathbb{R}^n$ with determinant $1$.
+It is the group of all $n\times n$ orthogonal matrices ($A^\top A=I_n$) on $\mathbb{R}^n$ with determinant $1$.
 
 $$
-SO(n)=\{A\in \mathbb{R}^{n\times n}: A^T A=I_n, \det(A)=1\}
+SO(n)=\{A\in \mathbb{R}^{n\times n}: A^\top A=I_n, \det(A)=1\}
 $$
 
 <details>
@@ -22,7 +22,7 @@ In [The random Matrix Theory of the Classical Compact groups](https://case.edu/a
 $O(n)$ (the group of all $n\times n$ **orthogonal matrices** over $\mathbb{R}$),
 
 $$
-O(n)=\{A\in \mathbb{R}^{n\times n}: AA^T=A^T A=I_n\}
+O(n)=\{A\in \mathbb{R}^{n\times n}: AA^\top=A^\top A=I_n\}
 $$
 
 $U(n)$ (the group of all $n\times n$ **unitary matrices** over $\mathbb{C}$), 
@@ -42,7 +42,7 @@ $$
 $Sp(2n)$ (the group of all $2n\times 2n$ symplectic matrices over $\mathbb{C}$),
 
 $$
-Sp(2n)=\{U\in U(2n): U^T J U=UJU^T=J\}
+Sp(2n)=\{U\in U(2n): U^\top J U=UJU^\top=J\}
 $$
 
 where $J=\begin{pmatrix}

content/Math401/Freiwald_summer/Math401_T2.md

@@ -74,7 +74,7 @@ $c\in \mathbb{C}$.
 The matrix transpose is defined by
 
 $$
-u^T=(a_1,a_2,\cdots,a_n)^T=\begin{pmatrix}
+u^\top=(a_1,a_2,\cdots,a_n)^\top=\begin{pmatrix}
 a_1 \\
 a_2 \\
 \vdots \\
@@ -694,7 +694,7 @@ $$
 
 The unitary group $U(n)$ is the group of all $n\times n$ unitary matrices.
 
-Such that $A^*=A$, where $A^*$ is the complex conjugate transpose of $A$. $A^*=(\overline{A})^T$.
+Such that $A^*=A$, where $A^*$ is the complex conjugate transpose of $A$. $A^*=(\overline{A})^\top$.
 
 #### Cyclic group $\mathbb{Z}_n$
 

content/Math429/Math429_L12.md

@@ -25,7 +25,7 @@ Let $A$ be an $m \times n$ matrix, then
 * The column rank of $A$ is the dimension of the span of the columns in $\mathbb{F}^{m,1}$.
 * The row range of $A$ is the dimension of the span of the row in $\mathbb{F}^{1,n}$.
 
-> Transpose: $A^t=A^T$ refers to swapping rows and columns
+> Transpose: $A^t=A^\top$ refers to swapping rows and columns
 
 #### Theorem 3.56 (Column-Row Factorization)
 
@@ -64,7 +64,7 @@ Proof:
 
 Note that by **Theorem 3.56**, if $A$ is $m\times n$ and has column rank $c$. $A=CR$ for some $C$ is a $m\times c$ matrix, $R$ is a $c\times n$ matrices, ut the rows of $CR$ are a linear combination of the rows of $R$, and row rank of $R\leq C$. So row rank $A\leq$ column rank of $A$.
 
-Taking a transpose of matrix, then row rank of $A^T$ (column rank of $A$) $\leq$ column rank of $A^T$ (row rank $A$).
+Taking a transpose of matrix, then row rank of $A^\top$ (column rank of $A$) $\leq$ column rank of $A^\top$ (row rank $A$).
 
 So column rank is equal to row rank.
 

content/Math429/Math429_L18.md

@@ -39,13 +39,13 @@ $T$ is surjective $\iff range\ T=W\iff null\ T'=0\iff T'$ injective
 
 Let $V,W$ be a finite dimensional vector space, $T\in \mathscr{L}(V,W)$
 
-Then $M(T')=(M(T))^T$. Where the basis for $M(T)'$ are the dual basis to the ones for $M(T)$
+Then $M(T')=(M(T))^\top$. Where the basis for $M(T)'$ are the dual basis to the ones for $M(T)$
 
 #### Theorem 3.133
 
 $col\ rank\ A=row\ rank\ A$
 
-Proof: $col\ rank\ A=col\ rank\ (M(T))=dim\ range\ T=dim\ range\ T'=dim\ range\ T'=col\ rank\ (M(T'))=col\ rank\ (M(T)^T)=row\ rank\ (M(T))$
+Proof: $col\ rank\ A=col\ rank\ (M(T))=dim\ range\ T=dim\ range\ T'=dim\ range\ T'=col\ rank\ (M(T'))=col\ rank\ (M(T)^\top)=row\ rank\ (M(T))$
 
 ## Chapter V Eigenvalue and Eigenvectors