# The QR factorization¶

An important property of some groups of vectors is called orthogonality. We say that two vectors $$\mathbf{u}$$ and $$\mathbf{v}$$ in $$\mathbb{R}^n$$ are orthogonal if $$\mathbf{u}^T\mathbf{v}=0$$. For $$n=2$$ or $$n=3$$ this means the vectors are perpendicular. We say that a collection of vectors $$\mathbf{q}_1,\ldots,\mathbf{q}_k$$ is orthogonal if

(67)$\mathbf{q}_i^T\mathbf{q}_j = 0 \quad \text{whenever i\neq j.}$

If also $$\mathbf{q}_i^T\mathbf{q}_i=1$$ for all $$i=1,\ldots,n$$, we say the vectors are orthonormal.

Orthogonal vectors are convenient theoretically and computationally. In theoretical calculations they make many terms of inner products vanish. For example, if $$\mathbf{q}_1$$ and $$\mathbf{q}_2$$ are orthogonal, then (in the 2-norm)

(68)$\| \mathbf{q}_1 - \mathbf{q}_2 \|^2 = (\mathbf{q}_1-\mathbf{q}_2)^T(\mathbf{q}_1-\mathbf{q}_2) = \mathbf{q}_1^T\mathbf{q}_1 - 2 \mathbf{q}_1^T\mathbf{q}_2 + \mathbf{q}_2^T\mathbf{q}_2 = \|\mathbf{q}_1\|^2 + \|\mathbf{q}_2\|^2.$

This calculation is also the key to the computational attractiveness of orthogonality. Fig. 3 shows how nonorthogonal vectors can allow a multidimensional version of subtractive cancellation, in which $$\|\mathbf{x}-\mathbf{y}\|$$ is much smaller than $$\|\mathbf{x}\|$$ and $$\|\mathbf{y}\|$$. Fig. 3 Cancellation and orthogonality.

As the figure and (68) show, orthogonal vectors do not allow this phenomenon. In other words, the addition and subtraction of vectors are guaranteed to be well conditioned when the vectors are orthogonal.

## Orthogonal and ONC matrices¶

Statements about orthogonal vectors are often made more easily in matrix form. Let $$\mathbf{Q}$$ be an $$n\times k$$ matrix whose columns $$\mathbf{q}_1$$, \ldots, $$\mathbf{q}_k$$ are orthogonal vectors. The orthogonality conditions (67) become simply that $$\mathbf{Q}^T\mathbf{Q}$$ is a diagonal matrix, since

$\begin{split}\mathbf{Q}^T \mathbf{Q} = \begin{bmatrix} \mathbf{q}_1^T \\[1mm] \mathbf{q}_2^T \\ \vdots \\ \mathbf{q}_k^T \end{bmatrix} \begin{bmatrix} \mathbf{q}_1 & \mathbf{q}_2 & \cdots & \mathbf{q}_k \end{bmatrix} = \begin{bmatrix} \mathbf{q}_1^T\mathbf{q}_1 & \mathbf{q}_1^T\mathbf{q}_2 & \cdots & \mathbf{q}_1^T\mathbf{q}_k \\[1mm] \mathbf{q}_2^T\mathbf{q}_1 & \mathbf{q}_2^T\mathbf{q}_2 & \cdots & \mathbf{q}_2^T\mathbf{q}_k \\ \vdots & \vdots & & \vdots \\ \mathbf{q}_k^T\mathbf{q}_1 & \mathbf{q}_k^T\mathbf{q}_2 & \cdots & \mathbf{q}_k^T\mathbf{q}_k \end{bmatrix}.\end{split}$

If the columns of $$\mathbf{Q}$$ are orthonormal, then $$\mathbf{Q}^T\mathbf{Q}$$ is the $$k\times k$$ identity matrix. This is such an important property that we will break with common practice here and give this type of matrix a name: an ONC matrix is one whose columns are an orthonormal set of vectors. We summarize their important properties here.

Theorem 25 ((ONC matrix))

Suppose $$\mathbf{Q}$$ is a real $$n\times k$$ ONC matrix (matrix with orthonormal columns). Then:

1. $$\mathbf{Q}^T\mathbf{Q} = \mathbf{I}$$ ($$k\times k$$ identity).

2. $$\| \mathbf{Q}\mathbf{x} \|_2 = \| \mathbf{x} \|_2$$ for all $$k$$-vectors $$\mathbf{x}$$.

3. $$\| \mathbf{Q} \|_2=1$$.

Proof

The first part is derived above. The second part follows a pattern that has become well established by now:

$\| \mathbf{Q}\mathbf{x} \|_2^2 = (\mathbf{Q}\mathbf{x})^T(\mathbf{Q}\mathbf{x}) = \mathbf{x}^T \mathbf{Q}^T \mathbf{Q} \mathbf{x} = \mathbf{x}^T \mathbf{I} \mathbf{x} = \| \mathbf{x} \|_2^2.$

The last part of the theorem is left to the exercises.

Of particular interest is a square ONC matrix, for which $$\mathbf{Q}^T\mathbf{Q}=\mathbf{I}$$, where all three matrices are $$n\times n$$. Hence $$\mathbf{Q}^{-1}=\mathbf{Q}^T$$. Such a matrix is called an orthogonal matrix.1 These matrices have properties beyond the general ONC type. The proofs of these are left to the exercises.

Theorem 26 ((Orthogonal matrix))

Suppose $$\mathbf{Q}$$ is an $$n\times n$$ real orthogonal matrix. Then:

1. $$\mathbf{Q}^T$$ is also an orthogonal matrix.

2. $$\kappa(\mathbf{Q})=1$$ in the 2-norm.

3. For any other $$n\times n$$ matrix $$\mathbf{A}$$, $$\| \mathbf{A}\mathbf{Q} \|_2=\| \mathbf{A} \|_2$$.

4. If $$\mathbf{U}$$ is another $$n\times n$$ orthogonal matrix, then $$\mathbf{Q}\mathbf{U}$$ is also orthogonal.

## Orthogonal factorization¶

Now we come to another important way to factor a matrix, the QR factorization. As we will show below, the QR factorization plays a role in linear least squares analogous to the role of LU factorization in linear systems.

Theorem 27

Every real $$m\times n$$ matrix $$\mathbf{A}$$ ($$m\ge n$$) can be written as $$\mathbf{A}=\mathbf{Q}\mathbf{R}$$, where $$\mathbf{Q}$$ is an $$m\times m$$ orthogonal matrix and $$\mathbf{R}$$ is an $$m\times n$$ upper triangular matrix.

In most introductory books on linear algebra, the QR factorization is derived through a process known as Gram–Schmidt orthogonalization. However, while it is an important tool for theoretical work, the Gram–Schmidt process is numerically unstable. We will consider an alternative construction in the next section.

When $$m$$ is much larger than $$n$$, which is often the case, there is a compressed form of the factorization that is more efficient. In the product

$\begin{split}\mathbf{A} = \begin{bmatrix} \mathbf{q}_1 & \mathbf{q}_2 & \cdots & \mathbf{q}_m \end{bmatrix} \begin{bmatrix} r_{11} & r_{12} & \cdots & r_{1n} \\ 0 & r_{22} & \cdots & r_{2n} \\ \vdots & & \ddots & \vdots\\ 0 & 0 & \cdots & r_{nn} \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & 0 \end{bmatrix},\end{split}$

the vectors $$\mathbf{q}_{n+1},\ldots,\mathbf{q}_m$$ always get multiplied by zero. Nothing about $$\mathbf{A}$$ is lost if we delete them and reduce the factorization to the equivalent form

(69)$\begin{split}\mathbf{A} = \begin{bmatrix} \mathbf{q}_1 & \mathbf{q}_2 & \cdots & \mathbf{q}_n \end{bmatrix} \begin{bmatrix} r_{11} & r_{12} & \cdots & r_{1n} \\ 0 & r_{22} & \cdots & r_{2n} \\ \vdots & & \ddots & \vdots\\ 0 & 0 & \cdots & r_{nn} \end{bmatrix} = \hat{\mathbf{Q}} \hat{\mathbf{R}},\end{split}$

in which $$\hat{\mathbf{Q}}$$ is an $$m\times n$$ ONC matrix and $$\hat{\mathbf{R}}$$ is $$n\times n$$ and upper triangular. We refer to this as a thin QR factorization, as the number of columns in $$\hat{\mathbf{Q}}$$ is $$n$$ rather than $$m$$. Julia returns either type of QR factorization from the qr command.

## Least squares and QR¶

If we substitute the thin factorization (69) into the normal equations (63), we can simplify expressions a great deal.

$\begin{split}\begin{split} \mathbf{A}^T\mathbf{A} \mathbf{x} &= \mathbf{A}^T \mathbf{b} \\ \hat{\mathbf{R}}^T \hat{\mathbf{Q}}^T \hat{\mathbf{Q}} \hat{\mathbf{R}} \mathbf{x} &= \hat{\mathbf{R}}^T \hat{\mathbf{Q}}^T \mathbf{b} \\ \hat{\mathbf{R}}^T \hat{\mathbf{R}} \mathbf{x}& = \hat{\mathbf{R}}^T \hat{\mathbf{Q}}^T \mathbf{b}. \end{split}\end{split}$

In order to have the normal equations be well posed, we require that $$\mathbf{A}$$ is not rank-deficient (as proved in a theorem). This is enough to guarantee that $$\hat{\mathbf{R}}$$ is nonsingular (see this exercise. Therefore, its transpose is nonsingular as well, and we arrive at

(70)$\hat{\mathbf{R}} \mathbf{x}=\hat{\mathbf{Q}}^T \mathbf{b}.$

This is a triangular $$n\times n$$ linear system that is easily solved by backward substitution, as demonstrated in lsqrfact. The function itself is superfluous, however, as this is essentially the algorithm used internally by Julia when \verb+A\b+ is called. Most importantly, even though we derived (70) from the normal equations, the solution of least squares problems via QR factorization does not suffer from the instability seen when the normal equations are solved directly using Cholesky factorization.

Function 28 (lsqrfact)
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 """ lsqrfact(A,b) Solve a linear least squares problem by QR factorization. Returns the minimizer of ||b-Ax||. """ function lsqrfact(A,b) Q,R = qr(A) c = Q'*b x = backsub(R,c) return x end 

## Exercises¶

1. ✍ Prove part 3 of the ONC matrix theorem.

2. ✍ Prove the orthogonal matrix theorem. For the third part, use the definition of the 2-norm as an induced matrix norm, then apply some of our other results as needed.

3. ⌨ Let $$t_0,\ldots,t_m$$ be $$m+1$$ equally spaced points in $$[-1,1]$$. Let $$\mathbf{A}$$ be the matrix in (58) for $$m=400$$ and fitting by polynomials of degree less than 5. Find the thin QR factorization of $$\mathbf{A}$$, and, on a single graph, plot every column of $$\hat{\mathbf{Q}}$$ as a function of the vector $$t$$.

4. ✍ Prove that if the $$m\times n$$ ($$m\ge n$$) matrix $$\mathbf{A}$$ is not rank-deficient, then the factor $$\hat{\mathbf{R}}$$ of the thin QR factorization is nonsingular. (Hint: Suppose on the contrary that $$\hat{\mathbf{R}}$$ is singular. Show using the factored form of $$\mathbf{A}$$ that this would imply that $$\mathbf{A}$$ is rank-deficient.)

5. ✍ Let $$\mathbf{A}$$ be $$m\times n$$ with $$m>n$$. Show that if $$\mathbf{A}=\mathbf{Q}\mathbf{R}$$ is a QR factorization and $$\mathbf{R}$$ has rank $$n$$, then $$\mathbf{A}^+=\mathbf{R}^+\mathbf{Q}^T$$.

6. ✍ Let $$\mathbf{A}$$ be $$m\times n$$ with $$m>n$$. Show that if $$\mathbf{A}=\hat{\mathbf{Q}}\hat{\mathbf{R}}$$ is a QR factorization and $$\hat{\mathbf{R}}$$ is nonsingular, then $$\mathbf{A}^+=\hat{\mathbf{R}}^{-1}\hat{\mathbf{Q}}^T$$.

7. ⌨ Repeat the census fitting problem, but use thin QR factorization rather than the backslash to solve the least-squares problem.

8. ✍ The matrix $$\mathbf{P}=\hat{\mathbf{Q}} \hat{\mathbf{Q}}^T$$ derived from the thin QR factorization has some interesting and important properties.

(a) Show that $$\mathbf{P}=\mathbf{A}\mathbf{A}^+$$.

(b) Prove that $$\mathbf{P}^2=\mathbf{P}$$. (This property defines a projection matrix.)

(c) Any vector $$\mathbf{x}$$ may be written as $$\mathbf{x}=\mathbf{u}+\mathbf{v}$$, where $$\mathbf{u}=\mathbf{P}\mathbf{x}$$ and $$\mathbf{v}=(\mathbf{I}-\mathbf{P})\mathbf{x}$$. Prove that for $$\mathbf{P} =\hat{\mathbf{Q}} \hat{\mathbf{Q}}^T$$, $$\mathbf{u}$$ and $$\mathbf{v}$$ are orthogonal. (Together with part (b), this means that $$\mathbf{P}$$ is an orthogonal projector.)

1

Confusingly, a square matrix whose columns are orthogonal is not necessarily an orthogonal matrix; the columns must be orthonormal, which is a stricter condition.