# ML Wiki

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## Gram-Schmidt Process

In Linear Algebra, Gram-Schmidt process is a method for orthogonalization:

### 2D Case

Suppose we have two linearly independent vectors $\mathbf a$ and $\mathbf b$

• we want to make them orthogonal $\mathbf a \Rightarrow \mathbf v_1$, $\mathbf b \Rightarrow \mathbf v_2$
• and then we normalize them: $\mathbf v_1 \Rightarrow \mathbf q_1$ and $\mathbf v_2 \Rightarrow \mathbf q_2$
• • $\mathbf a$ is already fine, let's keep its direction for $\mathbf q_1$ as well
• $\mathbf b$ is not fine: we want it to be orthogonal to $\mathbf q_1$

How do we produce a vector $\mathbf v_2$ from $\mathbf b$ such that $\mathbf v_2 \; \bot \; \mathbf v_1 = \mathbf a$

• • let's project $\mathbf b$ on $\mathbf v_1$, and we get $P \cdot \mathbf v_1 = \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1} \cdot \mathbf v_1$
• $\mathbf e$ is our projection error, so $\mathbf e = \mathbf b - P \mathbf v_1 = \mathbf b - P \mathbf v_1 = \mathbf b - \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1} \cdot \mathbf v_1$
• note that $\mathbf v_2 = \mathbf e$! it has the same length and the same direction
• so $\mathbf v_2 = \mathbf b - \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1} \cdot \mathbf v_1$
• interpretation: we take the original vector $\mathbf b$ and remove the projection of this vector onto $\mathbf v_1$, and it leaves only the orthogonal part
• now $\mathbf v_1 \; \bot \mathbf v_2$
• let's check: $\mathbf v_1^T \mathbf v_2 = \mathbf v_1^T \left( \mathbf b - \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1} \cdot \mathbf v_1 \right) = \mathbf v_1^T \mathbf b - \mathbf v_1^T \cdot \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1} \cdot \mathbf v_1 = \mathbf v_1^T \mathbf b - \mathbf v_1^T \mathbf b = 0$

To find the final solution, we just normalize $\mathbf v_1$ and $\mathbf v_2$:

• $\mathbf q_1 = \mathbf v_1 / \| \mathbf v_1 \|$
• $\mathbf q_2 = \mathbf v_2 / \| \mathbf v_2 \|$

### 3D Case

What if we have 3 vectors $\mathbf a, \mathbf b, \mathbf c$?

• need to find (pair-wise) orthogonal vectors $\mathbf v_1, \mathbf v_2, \mathbf v_3$ and then normalize them
• have $\mathbf v_1 = a$, $\mathbf v_2 = \mathbf b - \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1} \mathbf v_1$
• what about $\mathbf v_3$? Know that we want to have $\mathbf v_3 \; \bot \; \mathbf v_1$ and $\mathbf v_3 \; \bot \; \mathbf v_2$
• for $\mathbf v_3$, we want to subtract its components in direction $\mathbf v_1$ as well as in direction $\mathbf v_2$
• so $\mathbf v_3 = \mathbf c - \cfrac{\mathbf v_1^T \mathbf c}{\mathbf v_1^T \mathbf v_1} \cdot \mathbf v_1 - \cfrac{\mathbf v_2^T \mathbf c}{\mathbf v_2^T \mathbf v_2} \cdot \mathbf v_2$

To get $\mathbf q_1, \mathbf q_2, \mathbf q_3$, we just normalize:

• $\mathbf q_1 = \mathbf v_1 / \| \mathbf v_1 \|$
• $\mathbf q_2 = \mathbf v_2 / \| \mathbf v_2 \|$
• $\mathbf q_3 = \mathbf v_3 / \| \mathbf v_3 \|$

### Column Spaces

Claim: The column space of $A$ does not change when we orthogonalize it

Suppose that we take a matrix $A = \Bigg[ \mathop{\mathbf a_1}\limits_|^| \ \mathop{\mathbf a_2}\limits_|^| \ \cdots \ \mathop{\mathbf a_n}\limits_|^| \Bigg]$ and orthogonalize its columns into $Q = \Bigg[ \mathop{\mathbf q_1}\limits_|^| \ \mathop{\mathbf q_2}\limits_|^| \ \cdots \ \mathop{\mathbf q_n}\limits_|^| \Bigg]$

• Why $C(A) = C(Q)$?
• at each step of the Gram-Schmidt process we take linear combinations from $C(A)$!
• e.g. $\mathbf v_3 = \mathbf c - \alpha_1 \mathbf v_1 - \alpha_2 \mathbf v_2 = \mathbf c - \alpha_1\mathbf a - \alpha_2 \cdot \left(\mathbf b - \alpha_3 \mathbf a \right) = \mathbf c - \alpha_1 \mathbf a - \alpha_2 \mathbf b - \alpha_2 \alpha_3 \mathbf a$
• $\alpha_1 = \cfrac{\mathbf v_1^T \mathbf c}{\mathbf v_1^T \mathbf v_1}, \alpha_2 = \cfrac{\mathbf v_2^T \mathbf c}{\mathbf v_2^T \mathbf v_2}, \alpha_3 = \cfrac{\mathbf v_1^T \mathbf b}{\mathbf v_1^T \mathbf v_1}$ are just scalars

## Implementation

### Java

The implementation is taken from Smile: https://github.com/haifengl/smile/blob/master/core/src/main/java/smile/projection/RandomProjection.java

double[][] X;
Math.unitize(X);
for (int i = 1; i < p; i++) {
for (int j = 0; j < i; j++) {
double t = -Math.dot(X[i], X[j]);
Math.axpy(t, X[j], X[i]);
}
Math.unitize(X[i]);
}


where

• Math.unitize unit-normalizes the vector
• Math.axpy updates an array by adding a multiple of another array y = a * x + y.
• here X[i] is $i$th column of $X$

### Python

def normalize(v):
return v / np.sqrt(v.dot(v))

n = len(A)

A[:, 0] = normalize(A[:, 0])

for i in range(1, n):
Ai = A[:, i]
for j in range(0, i):
Aj = A[:, j]
t = Ai.dot(Aj)
Ai = Ai - t * Aj
A[:, i] = normalize(Ai)


## QR Factorization

We can think of the Gram-Schmidt Process in the matrix language (like for Gaussian Elimination that brings us to LU Factorization)

• recall that $C(Q) = C(A)$
• because of this, there exists a third matrix $R$ that brings $A$ to $Q$
• or, $A = Q R$
• (so we want Gram-Schmidt applied to the columns of $A$)

How to find this $R$?

• $A = QR$, let's multiply both sides by $Q^T$:
• $Q^T A = Q^T Q R$
• since $Q^T Q = I$, we have $R = Q^T A$
• let $A = \Bigg[ \mathop{\mathbf a_1}\limits_|^| \ \mathop{\mathbf a_2}\limits_|^| \ \cdots \ \mathop{\mathbf a_n}\limits_|^| \Bigg]$ and $Q = \Bigg[ \mathop{\mathbf q_1}\limits_|^| \ \mathop{\mathbf q_2}\limits_|^| \ \cdots \ \mathop{\mathbf q_n}\limits_|^| \Bigg]$
• thus we have $R = Q^T A = \begin{bmatrix} \mathbf q_1^T \mathbf a_1 & \mathbf q_1^T \mathbf a_2 & \cdots & \mathbf q_1^T \mathbf a_n \\ \mathbf q_2^T \mathbf a_1 & \mathbf q_2^T \mathbf a_2 & \cdots & \mathbf q_2^T \mathbf a_n \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf q_n^T \mathbf a_1 & \mathbf q_n^T \mathbf a_2 & \cdots & \mathbf q_n^T \mathbf a_n \\ \end{bmatrix}$

Now recall the way we constructed $Q$

• we took $\mathbf q_1 = \mathbf a_1 / \| \mathbf a_1 \|$ and make all other $\mathbf a_i$ orthogonal to it, so $\mathbf q_2^T \mathbf a_1 = \ ... \ = \mathbf q_n^T \mathbf a_1 = 0$
• then we took $\mathbf q_1$ and $\mathbf q_2$ and made all $\mathbf a_3, ..., \mathbf a_n$ orthogonal to them, so $\mathbf q_3^T \mathbf a_2 = \ ... \ = \mathbf q_n^T \mathbf a_2 = 0$
• proceeding this way till the end we see that $R$ is upper-diagonal:
• thus we have $R = Q^T A = \begin{bmatrix} \mathbf q_1^T \mathbf a_1 & \mathbf q_1^T \mathbf a_2 & \cdots & \mathbf q_1^T \mathbf a_n \\ 0 & \mathbf q_2^T \mathbf a_2 & \cdots & \mathbf q_2^T \mathbf a_n \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf q_n^T \mathbf a_n \\ \end{bmatrix}$

Note that for the diagonal elements of $R$, $\mathbf q_i^T \mathbf a_i = \| \mathbf v_i \|$

• why?
• • $\mathbf q_i^T \mathbf a_i = \| \mathbf q_i \| \| \mathbf a_i \| \cos \theta$ by the Dot Product definition
• $\| \mathbf q_i \| = 1$ and $\cos \theta = \cfrac{\| \mathbf v_i \|}{\| \mathbf a_i \|}$, so
• $\mathbf q_i^T \mathbf a_i = \| \mathbf q_i \| \| \mathbf a_i \| \cos \theta = 1 \cdot \cancel{\| \mathbf a_i \|} \cfrac{\| \mathbf v_i \|}{ \cancel{\| \mathbf a_i \|} } = \| \mathbf v_i \|$
• this can be uses to speed up the computation a bit

### Usage

This is often used for Linear Least Squares - to make the Normal Equation faster

### Implementation code

#### Python

Implementation of the pseudo-code from the Strang's book:

import numpy as np

def qr_factorization(A):
m, n = A.shape
Q = np.zeros((m, n))
R = np.zeros((n, n))

for j in range(n):
v = A[:, j]

for i in range(j - 1):
q = Q[:, i]
R[i, j] = q.dot(v)
v = v - R[i, j] * q

norm = np.linalg.norm(v)
Q[:, j] = v / norm
R[j, j] = norm
return Q, R

A = np.random.rand(13, 10) * 1000
Q, R = qr_factorization(A)

Q.shape, R.shape
np.abs((A - Q.dot(R)).sum()) < 1e-6