Linear Algebra: VIII Gram-Schmidt and QR Decomposition

Gram–Schmidt and QR Decomposition


Suppose that $\{ \vec{v}_1, \ldots, \vec{v}_n \}$ is a set of linearly independent column vectors in $\mathbb{R}^n$, and let $A$ be an $n$ by $n$ square matrix:


$$ A = [\vec{v}_1 \ \cdots \ \vec{v}_n] $$

Then the Gram--Schmidt process produces an orthonormal basis


$$ \{ \hat{q}_1, \ldots, \hat{q}_n \} $$

which can be written as the $n$ by $n$ orthonormal matrix:


$$ Q = [\hat{q}_1 \ \cdots \ \hat{q}_n] $$

Let $\hat{v}_1$ be a unit vector. Then the vector


$$ (I - \hat{v}_1 \hat{v}_1^\top)\hat{v}_2 $$

Is orthogonal to $\hat{v}_1$ since


$$ \hat{v}_1^\top (I - \hat{v}_1 \hat{v}_1^\top)\hat{v}_2 = \hat{v}_1^\top\hat{v}_2 - (\hat{v}_1^\top\hat{v}_1)\hat{v}_1^\top\hat{v}_2 = \hat{v}_1^\top\hat{v}_2 - \hat{v}_1^\top\hat{v}_2 = 0 $$

Using this idea, we construct an orthonormal basis:


$$ \hat{q}_1 = \frac{\vec{v}_1}{\|\vec{v}_1\|} $$
$$ \vec{q}_2 = (I - \hat{q}_1 \hat{q}_1^\top)\vec{v}_2, \qquad \hat{q}_2 = \frac{\vec{q}_2}{\|\vec{q}_2\|} $$
$$ \vec{q}_3 = (I - \hat{q}_1 \hat{q}_1^\top - \hat{q}_2 \hat{q}_2^\top)\vec{v}_3, \qquad \hat{q}_3 = \frac{\vec{q}_3}{\|\vec{q}_3\|} $$

In general:


$$ \vec{q}_k = \left(I - \sum_{i=1}^{k-1} \hat{q}_i \hat{q}_i^\top \right)\vec{v}_k, \qquad \hat{q}_k = \frac{\vec{q}_k}{\|\vec{q}_k\|} $$

Now define


$$ R = Q^\top A $$

With entries


$$ r_{ij} = \hat{q}_i^\top \vec{v}_j $$

Because $\vec{v}_j$ is orthogonal to $\hat{q}_{j+1}, \ldots \hat{q}_j$, we have:


$$ r_{ij} = 0 \quad \text{if } i > j $$

Thus, $R$ is upper triangular.


Since $Q^{-1} = Q^\top$, this gives the QR decomposition:


$$ R = Q^\top A \quad \Rightarrow \quad A = QR $$

Consider solving $A\vec{x} = \vec{b}$. Then,


$$ QR\vec{x} = \vec{b} \quad \Rightarrow \quad R\vec{x} = Q^\top \vec{b} $$

Since $R$ is upper triangular, we can solve this system using backward substitution.

Comments

Post a Comment

Popular posts from this blog

Linear Algebra: III Square and Inverse Matrices

Square and Inverse Matrices An $n$ by $n$ square matrix has the same number of rows and columns: $$ A = \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} $$ If $A$ is an $n$ by $n$ square matrix and $\vec{x} \in \mathbb{R}^n$, then the product $A\vec{x}$ is also a vector in $\mathbb{R}^n$, defining a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^n$. The identity matrix $I$ is an $n$ by $n$ square matrix with 1s on the main diagonal and 0s elsewhere: $$ I = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} $$ It satisfies: $$ AI = IA = A $$ The inverse of a square matrix $A$, denoted $A^{-1}$, is a matrix such that: $$ A A^{-1} = A^{-1} A = I $$ Which may not exist. If $A^{-1}$ exists, then $A$ is called invertible . If $A$ and $B$ are inve...
Read more

Linear Algebra: VII Orthonormal Matrices

Orthonormal Matrices A square $n$ by $n$ matrix $Q$ is called orthonormal if: $$ QQ^\top = Q^\top Q = I $$ $$ Q^{-1} = Q^\top $$ The columns $\vec{q}_1, \ldots, \vec{q}_n$ of $Q$ (and the rows) form an orthonormal basis of $\mathbb{R}^n$: $$ \vec{q}_i^\top \vec{q}_j = 0 \text{ if } i \neq j, \quad \|\vec{q}_i\| = 1 $$ An orthonormal matrix preserves inner products / dot products: $$ (Q\vec{x})^\top(Q\vec{y}) = \vec{x}^\top Q^\top Q \vec{y} = \vec{x}^\top I \vec{y} = \vec{x}^\top \vec{y} $$ An orthonormal matrix also preserves magnitude: $$ \|Q\vec{x}\|^2 = (Q\vec{x})^\top Q\vec{x} = \vec{x}^\top Q^\top Q \vec{x} = \vec{x}^\top I \vec{x} = \vec{x}^\top \vec{x} = \|\vec{x}\|^2 $$ So an orthonormal matrix preserves angles between vectors: $$ \cos{\theta} = \frac{(Q\vec{x})^\top(Q\vec{y})}{\|Q\vec{x}\|\|Q\vec{y}\|} = \frac{\vec{x}^\top \vec{y}}{\|\vec{x}\|\|\vec{y}\|} $$ There are two special types of orthogonal matrices. First, a...
Read more

Linear Algebra: VI Triangular Matrices and LU Decomposition

Triangular Matrices and LU Decomposition A square matrix $A$ is called upper triangular if: $$ a_{ij} = 0 \quad \text{for all } i > j $$ A square matrix $A$ is called lower triangular if: $$ a_{ij} = 0 \quad \text{for all } i Upper and lower triangular $n \times n$ square matrices both contain at least $$ \frac{n(n-1)}{2} $$ Zeros and $$ \frac{n(n+1)}{2} $$ Other entries. The identity matrix $I$ is both upper triangular and lower triangular. Scaling elementary matrices and their products are also both upper and lower triangular. Shearing elementary matrices are either upper triangular or lower triangular but never both. Swapping elementary matrices are neither upper nor lower triangular. If $A$ and $B$ are upper triangular $n \times n$ square matrices, then $AB$ is also upper triangular. Indeed, for $i > j$: $$ (ab)_{ij} = \sum_{k=1}^n a_{ik} b_{kj} $$ If $i > k$, then $a_{ik} = 0$. If $k \ge i$, then $k > j...
Read more