## What

Given a matrix \(A\), its QR decomposition is to find matrices \(Q\) and \(R\) such that

\[\begin{aligned} A &= [x_1\ x_2\ \cdots\ x_m] \\ Q &= [\epsilon_1\ \epsilon_2\ \cdots\ \epsilon_m] \\ R &= \left[\begin{matrix} r_{11} & r_{12} & \cdots & r_{1m} \\ 0 & r_{22} & \cdots & r_{2m} \\ \vdots & & \ddots & \\ 0 & 0 & \cdots & r_{mm} \end{matrix}\right] \\ A &= QR \\ Q^TQ &= I \\ \end{aligned}\]i.e., \(Q\) is an orthogonal matrix and \(R\) is a upper-triangular matrix.

## How

Gramm-Schmidt orthogonalization is a method to perform QR decomposition: Consider matrix \(A\) of \(m\) columns, as denoted above. We first normalize the first column \(x_1\) into unit length, \(\epsilon_1 = \frac{1}{r_{11}}x_1 \textrm{, where } r_{11}=||x_1||.\) Then decompose the second column \(x_2\) into components parallel and orthogonal to column vector \(\epsilon_1\),

\[\begin{aligned} r_{12} &= \epsilon_1^T x_2 \\ r_{22} &= ||x_2 - r_{12}\epsilon_1|| \\ \epsilon_2 &= \frac{1}{r_{22}}(x_2 - r_{12}\epsilon_1) \\ x_2 &= r_{12}\epsilon_1 + (x_2 - r_{12}\epsilon_1) = r_{12}\epsilon_1 + r_{22}\epsilon_2 \end{aligned}\]\(r_{12}\) is measuring the length of \(x_2\) projected to the unit vector \(\epsilon_1\). Thus \(x_2 - r_{12}\epsilon_1\) is a vector orthogonal to \(x_1\), which the unit vector of such is represented by \(\epsilon_2\).

Subsequently, we decompose column vector \(x_k\) into components parallel to \(x_1,x_2,\cdots,x_{k-1}\) and a component that is orthogonal to all of them. For example, on \(x_3\), we have

\[\begin{aligned} r_{13} &= \epsilon_1^T x_3 \\ r_{23} &= \epsilon_2^T x_3 \\ r_{33} &= ||x_3 - r_{13}\epsilon_1 - r_{23}\epsilon_2|| \\ \epsilon_3 &= \frac{1}{r_{33}}(x_3 - r_{13}\epsilon_1 - r_{23}\epsilon_2) \\ x_3 &= r_{13}\epsilon_1 + r_{23}\epsilon_2 + (x_3 - r_{13}\epsilon_1 - r_{23}\epsilon_2) = r_{13}\epsilon_1 + r_{23}\epsilon_2 + r_{33}\epsilon_3 \end{aligned}\]Note, matrices \(A\) and \(Q\) are of the same dimension, and may not be a square shape, but matrix \(R\) is always a square matrix.

## Why

Assume we have a system of linear equations \(Ax=b\), for which \(x\) is the vector of unknown and \(b\) is a vector of constants. If we decompose \(A=QR\), we have

\[\begin{aligned} QRx &= b \\ Q^TQRx &= Q^Tb \\ Rx &= Q^Tb \end{aligned}\]Now since \(R\) is a upper-triangular matrix, we can use Gaussian elimination to find \(x\).