Question

Given $A=QR$ as in Theorem 12, describe how to find an orthogonal$m\times m$(square) matrix $Q_1$ and an invertible $n\times n$ upper triangular matrix $R$ such that

$A=Q_1\left[\begin{array}{r}R\\ 0\end{array}\right]$

The MATLAB qr command supplies this “full” QR factorization

when rank $A=n$.

Short answer

We find the square matrix $Q_1$ by extending the column vectors of $Q$ to the orthonormal basis of ${\mathbb{R}}^m$.

Step-by-step solution

$QR$ factorization of a Matrix

A matrix with order $m\times n$ can be written as the multiplication of an upper triangular matrix $R$ and a matrix $Q$ which is formed by applying the Gram–Schmidt orthogonalization process to the $\mathrm{c}\mathrm{o}\mathrm{l}\left(A\right)$.

The matrix $R$ can be found by the formula $Q^{T}A=R$.

Finding the matrix $R$ and $Q_1$

Let $q_1,q_2,\dots ,q_n$ be the columns of the matrix $Q$. Here $n\le m$, since $A$ is a $m\times n$ matrix and it has linearly independent columns.

The columns of $Q$ can be extended to an orthogonal basis ${\mathbb{R}}^m$.

Let $v_1$ be the 1^st vector of the standard basis of ${\mathbb{R}}^m$ which is not in the set$W_n=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{q_1,......,q_n\right\}$.

Let $u_1=v_1-\mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{W_n}{v}_1$ and let $q_{n+1}=\frac{u_1}{\Vert u_1\Vert }$. Then $\left\{q_1,......,q_n,q_{n+1}\right\}$ is an orthogonal basis for $$$W_{n+1}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{q_1,......,q_n,q_{n+1}\right\}$

Now, again $v_2$be the 1st vector of the standard basis of ${\mathbb{R}}^m$ which is not in the set $W_{n+1}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{q_1,......,q_n,q_{n+1}\right\}$.

Let $u_2=v_2-\mathrm{p}\mathrm{r}\mathrm{o}{\mathrm{j}}_{W_{n+1}}{v}_2$ and let $q_{n+2}=\frac{u_2}{\Vert u_2\Vert }$. Then $\left\{q_1,......,q_n,q_{n+1},q_{n+2}\right\}$is an orthogonal basis for$W_{n+2}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\left\{q_1,......,q_n,q_{n+1},q_{n+2}\right\}$.

Applying this process to get $m-n$ such vectors, we get an orthogonal basis

$\left\{q_1,......,q_n,q_{n+1},...,q_m\right\}$for ${\mathbb{R}}^m$.

Let $Q_0=\left[\begin{array}{rlrlrl}r{q}_{n+1}& .& .& .& .& q_m\end{array}\right]$ and then let $Q_1=\left[\begin{array}{rl}Q& Q_0\end{array}\right]$.

Then

$\begin{array}{rl}{Q}_1\left[\begin{array}{c}R\\ 0\end{array}\right]& =\left[\begin{array}{rl}Q& Q_0\end{array}\right]\left[\begin{array}{r}R\\ 0\end{array}\right]\\ & =QR\\ & =A\end{array}$

In this way, we can find $Q_1$ and the upper triangular matrix.