# QR Factorization
* numerical method (standard) for LS probs
* properties of QR
* numerical benefits
* computation
In general, $$\lVert(a,b)\rVert^2=(a,b)^T(a,b)=a^Ta+b^Tb=\lVert a\rVert^2+\lVert b\rVert^2$$
### Linear Least Squares
$$\min \frac{1}{2}\lVert Ax-b\rVert^2\_2$$
if A has full col rank ($$m>n$$) then $$X\_{LS}$$ uniquely solves Normal Equation’s
$$A^TAx\_{LS} = A^Tb$$
$$x\_{LS} = (A^TA)^{-1}A^Tb$$
$$\bar{y}=Ax\_{LS}=A(A^TA)^{-1}A^Tb$$ (projector onto $$range(A)$$)
“Conditioning” of a matrix determines the accuracy of a linear system solve: $$cond(M)=\frac{\lambda\_{max}(M)}{\lambda\_{min}(M)}$$
If $$M$$ is not square:
$$cond(M)=\frac{\nabla{max}(M)}{\nabla{min}(M)}$$, where $$\nabla$$ is the gradient
For $$M = A^TA$$, $$\nabla(M) = \nabla(A^TA)=\nabla(A)^2$$
$$\lambda\_i(A^TA)=\nabla\_i(A)^2$$
With $$Q$$ be orthogonal basis for $$range(A)$$
$$\underbrace{Q^T}*{n \times m}\underbrace{Q}*{m \times n}=\underbrace{I}\_{n\times n}$$
$$P$$ is an orthogonal projector onto $$range(P)$$ if
1. $$P^2=P$$ idempotent
2. $$P=P^T$$ symmetric
Take $$P=QQ^T$$: $$range(P)=range(Q)=range(A)$$
$$P^2=QQ^TQQ^T=QQ^T=P$$
$$range(Q) = range(A)$$
$$Q=\[q\_1,q\_2,…,q\_n]$$
$$span{q\_1,…,q\_n}=span{a\_1,…,a\_n}$$
Every matrix $$A\_{m\times n}$$ has a QR Factorization:
$$A=Q\begin{bmatrix} R \ 0 \end{bmatrix}$$, where $$Q$$ is orthogonal ($$QQ^T=Q^TQ=I$$) and R=upper triangular
$$Q$$ is orthogonal (and square)
$$Q^TQ=Q^TQ=I\_m$$
$$Q\_1$$ is orthonormal (not orthogonal)
$$Q\_1^TQ\_1=I\_n$$
$$Q\_1Q\_1^T \ne I\_m$$
orthonormal matrices are rotations:
1. for any vector $$x \in \R^n$$, $$\lVert Q\_1x\rVert\_2=\lVert x\rVert\_2$$ ⇒ $$\lVert Q\_1x\rVert^2=x^TQ\_1^TQ\_1x=x^Tx=\lVert x\rVert^2$$
2. any $$y \in \R^m$$, $$(Qx)^T(Qy)=x^TQ^TQy=x^Ty$$
### Solving Least Squares
$$\underset{x}{\min}\lVert Ax-b\rVert^2$$
$$\begin{aligned} \lVert Ax-b \rVert ^2 &=\lVert Q^T(Ax-b)\rVert ^2 \ &=\lVert Q^TAx-Q^Tb\rVert ^2 \ &=\lVert \begin{bmatrix} Rx\ \hdashline \ 0\end{bmatrix}-\begin{bmatrix} Q^T\_1b\ \hdashline \ Q^T\_2 b\end{bmatrix}\rVert ^2 \ &= \lVert Rx-Q^Tb\rVert^2+\lVert Q^T\_2b\rVert^2\end{aligned}$$
$$A = Q\begin{bmatrix}R \ 0\end{bmatrix}$$
Multiply both sides on the left with $$Q^T$$
$$Q^TA=\underbrace{Q^TQ}\_I\begin{bmatrix}R\0\end{bmatrix}$$
$$\begin{bmatrix}Q^T\_1A\Q\_2^TA\end{bmatrix}=\begin{bmatrix}R\0\end{bmatrix}$$
$$Q\_1^TA=R$$
$$Q\_2^TA=0$$
$$R\_1x=Q\_1^Tb$$ ⇒ solving least squares using QR
$$\begin{aligned}Cond(A) &= Cond(Q,R)\ &= Cond(R)\end{aligned}$$
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