Least squares: Cholesky, QR and Householder

2 A key concept: orthogonality

Course: Math 535 - Mathematical Methods in Data Science (MMiDS)
Author: Sebastien Roch, Department of Mathematics, University of Wisconsin-Madison
Updated: Jun 14, 2021
Copyright: © 2021 Sebastien Roch

Orthogonality plays a key role in linear algebra for data science thanks to its computational properties and its connection to the least-squares problem.

Definition (Orthogonality): Vectors $\mathbf{u}$ and $\mathbf{v}$ in $V$ are orthogonal if their inner product satisfies $\langle \mathbf{u}, \mathbf{v} \rangle = 0$.

Orthogonality has important implications. The following classical result will be useful below.

Lemma (Pythagoras): Let $\mathbf{u}, \mathbf{v} \in V$ be orthogonal. Then $\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2$.

Proof: Using $\|\mathbf{w}\|^2 = \langle \mathbf{w}, \mathbf{w}\rangle$, we get

$$ \begin{align} \|\mathbf{u} + \mathbf{v}\|^2 &= \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v}\rangle\\ &= \langle \mathbf{u}, \mathbf{u}\rangle + 2 \,\langle \mathbf{u}, \mathbf{v}\rangle + \langle \mathbf{v}, \mathbf{v}\rangle\\ &= \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2. \end{align} $$


Here is an application of Pythagoras.

Lemma (Cauchy-Schwarz): For any $\mathbf{u}, \mathbf{v} \in V$, $|\langle \mathbf{u}, \mathbf{v}\rangle| \leq \|\mathbf{u}\| \|\mathbf{v}\|$.

Proof: Let $\mathbf{q} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$ be the unit vector in the direction of $\mathbf{v}$. We want to show $|\langle \mathbf{u}, \mathbf{q}\rangle| \leq \|\mathbf{u}\|$. Decompose $\mathbf{u}$ into its projection onto $\mathbf{q}$ and what's left:

$$ \mathbf{u} = \left\langle \mathbf{u}, \mathbf{q}\right\rangle \mathbf{q} + \left\{\mathbf{u} - \left\langle \mathbf{u}, \mathbf{q}\right\rangle \mathbf{q}\right\}. $$

The two terms on the right-hand side are orthogonal, so Pythagoras gives

$$ \|\mathbf{u}\|^2 = \left\|\left\langle \mathbf{u}, \mathbf{q}\right\rangle \mathbf{q}\right\|^2 + \left\|\mathbf{u} - \left\langle \mathbf{u}, \mathbf{q}\right\rangle \mathbf{q}\right\|^2 \geq \left\|\left\langle \mathbf{u}, \mathbf{q}\right\rangle \mathbf{q}\right\|^2 = \left\langle \mathbf{u}, \mathbf{q}\right\rangle^2. $$

Taking a square root gives the claim.$\square$

2.1 Basis expansion

To begin to see the power of orthogonality, consider the following. A list of vectors $\{\mathbf{u}_1,\ldots,\mathbf{u}_m\}$ is orthonormal if the $\mathbf{u}_i$'s are pairwise orthogonal and each has norm 1, that is, for all $i$ and all $j \neq i$, $\langle \mathbf{u}_i, \mathbf{u}_j \rangle = 0$ and $\|\mathbf{u}_i\| = 1$.

Lemma (Properties of Orthonormal Lists): Let $\{\mathbf{u}_1,\ldots,\mathbf{u}_m\}$ be an orthonormal list of vectors. Then:

  1. $\|\sum_{j=1}^m \alpha_j \mathbf{u}_j\|^2 = \sum_{j=1}^m \alpha_j^2$ for any $\alpha_j \in \mathbb{R}$, $j \in [m]$, and
  2. $\{\mathbf{u}_1,\ldots,\mathbf{u}_m\}$ are linearly independent.

Proof: For 1, using that $\|\mathbf{x}\|^2 = \langle \mathbf{x}, \mathbf{x} \rangle$ and $\langle \beta \,\mathbf{x}_1 + \mathbf{x}_2, \mathbf{x}_3 \rangle = \beta \,\langle \mathbf{x}_1,\mathbf{x}_3\rangle + \langle \mathbf{x}_2,\mathbf{x}_3\rangle$ (which follow directly from the definition of the inner product), we have

$$ \begin{align} \left\|\sum_{j=1}^m \alpha_j \mathbf{u}_j\right\|^2 &= \left\langle \sum_{i=1}^m \alpha_i \mathbf{u}_i, \sum_{j=1}^m \alpha_j \mathbf{u}_j \right\rangle\\ &= \sum_{i=1}^m \alpha_i \left\langle \mathbf{u}_i, \sum_{j=1}^m \alpha_j \mathbf{u}_j \right\rangle\\ &= \sum_{i=1}^m \sum_{j=1}^m \alpha_i \alpha_j \left\langle \mathbf{u}_i, \mathbf{u}_j \right\rangle\\ &= \sum_{i=1}^m \alpha_i^2 \end{align} $$

where we used orthonormality in the rightmost equation, that is, $\langle \mathbf{u}_i, \mathbf{u}_j \rangle$ is $1$ if $i=j$ and $0$ otherwise.

For 2, suppose $\sum_{i=1}^m \beta_i \mathbf{u}_i = \mathbf{0}$, then we must have by 1 that $\sum_{i=1}^m \beta_i^2 = 0$. That implies $\beta_i = 0$ for all $i$. Hence the $\mathbf{u}_i$'s are linearly independent.$\square$

Given a basis $\{\mathbf{u}_1,\ldots,\mathbf{u}_m\}$ of $\mathcal{U}$, we know that: for any $\mathbf{w} \in \mathcal{U}$, $\mathbf{w} = \sum_{i=1}^m \alpha_i \mathbf{u}_i$ for some $\alpha_i$'s. It is not immediately obvious in general how to find the $\alpha_i$'s. In the orthonormal case, however, it is straighforward.

Theorem (Orthonormal Expansion): Let $\mathbf{q}_1,\ldots,\mathbf{q}_m$ be an orthonormal basis of $\mathcal{U}$ and let $\mathbf{u} \in \mathcal{U}$. Then

$$ \mathbf{u} = \sum_{j=1}^m \langle \mathbf{u}, \mathbf{q}_j\rangle \,\mathbf{q}_j. $$

Proof: Because $\mathbf{u} \in \mathcal{U}$, $\mathbf{u} = \sum_{i=1}^m \alpha_i \mathbf{q}_i$ for some $\alpha_i$. Take the inner product with $\mathbf{q}_j$ and use orthonormality:

$$ \langle \mathbf{u}, \mathbf{q}_j\rangle = \left\langle \sum_{i=1}^m \alpha_i \mathbf{q}_i, \mathbf{q}_j\right\rangle = \sum_{i=1}^m \alpha_i \langle \mathbf{q}_i, \mathbf{q}_j\rangle = \alpha_j. $$


So we've shown that working with orthonormal bases is desirable. What if we don't have one? We review the Gram-Schmidt algorithm below, which will imply that every linear subspace has an orthonormal basis.

But, first, we define the orthogonal projection.

2.2 Orthogonal projection

Let's consider the following problem. We have a linear subspace $\mathcal{U} \subseteq V$ and a vector $\mathbf{v} \notin \mathcal{U}$. We want to find the vector $\mathbf{v}^*$ in $\mathcal{U}$ that is closest to $\mathbf{v}$ in $2$-norm, that is, we want to solve

$$ \min_{\mathbf{v}^* \in \mathcal{U}} \|\mathbf{v}^* - \mathbf{v}\|. $$

Example: Consider the two-dimensional case with a one-dimensional subspace, say $\mathcal{U} = \mathrm{span}(\mathbf{u_1})$ with $\|\mathbf{u}_1\|=1$. The geometrical intuition is in the following figure. The solution $\mathbf{v}^*$ has the property that the difference $\mathbf{v} - \mathbf{v}^*$ makes a right angle with $\mathbf{u}_1$, that is, it is orthogonal to it.

Orthogonal projection (Source)

Letting $\mathbf{v}^* = \alpha^* \,\mathbf{u}_1$, the geometrical condition above translates into

$$ 0 = \langle \mathbf{u}_1, \mathbf{v} - \mathbf{v}^* \rangle = \langle \mathbf{u}_1, \mathbf{v} - \alpha^* \,\mathbf{u}_1 \rangle = \langle \mathbf{u}_1, \mathbf{v} \rangle - \alpha^* \,\langle \mathbf{u}_1, \mathbf{u}_1 \rangle = \langle \mathbf{u}_1, \mathbf{v} \rangle - \alpha^* $$


$$ \mathbf{v}^* = \langle \mathbf{u}_1, \mathbf{v} \rangle \,\mathbf{u}_1. $$

By Pythagoras, we then have for any $\alpha \in \mathbb{R}$

$$ \begin{align} \|\mathbf{v} - \alpha \,\mathbf{u}_1\|^2 &= \|\mathbf{v}- \mathbf{v}^* + \mathbf{v}^* - \alpha \,\mathbf{u}_1\|^2\\ &= \|\mathbf{v}- \mathbf{v}^* + (\alpha^* - \alpha) \,\mathbf{u}_1\|^2\\ &= \|\mathbf{v}- \mathbf{v}^*\|^2 + \| (\alpha^* - \alpha) \,\mathbf{u}_1\|^2\\ &\geq \|\mathbf{v}- \mathbf{v}^*\|^2. \end{align} $$

That confirms the optimality of $\mathbf{v}^*$. The argument in this example carries through in higher dimension, as we show next. $\lhd$

Definition (Orthogonal Projection on an Orthonormal List): Let $\mathbf{q}_1,\ldots,\mathbf{q}_m$ be an orthonormal list. The orthogonal projection of $\mathbf{v} \in V$ on $\{\mathbf{q}_i\}_{i=1}^m$ is defined as

$$ (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} = \sum_{j=1}^m \langle \mathbf{v}, \mathbf{q}_j \rangle \,\mathbf{q}_j. $$


Definition and Theorem (Orthogonal Projection): Let $\mathcal{U} \subseteq V$ be a linear subspace with orthonormal basis $\mathbf{q}_1,\ldots,\mathbf{q}_m$ and let $\mathbf{v} \in V$. Then $(\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} \in \mathcal{U}$ and, for any $\mathbf{u} \in \mathcal{U}$,

$$ (*) \quad \left\langle \mathbf{v} - (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v}, \mathbf{u}\right\rangle =0 $$


$$ (**) \quad \|\mathbf{v} - (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v}\| \leq \|\mathbf{v} - \mathbf{u}\|. $$

Furthermore, if $\mathbf{u} \in \mathcal{U}$ and the inequality above is an equality, then $\mathbf{u} = (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m}) \mathbf{v}$. Hence, for any orthonormal basis $\mathbf{q}'_1,\ldots,\mathbf{q}'_m$ of $\mathcal{U}$, it holds that

$$ (*\!*\!*) \quad \mathcal{P}_\mathcal{U} \mathbf{v} = (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} = (\mathcal{P}_{\{\mathbf{q}'_i\}_{i=1}^m})\, \mathbf{v}, $$

where the first equality is a definition. We refer to $\mathcal{P}_\mathcal{U} \mathbf{v}$ as the orthogonal projection of $\mathbf{v}$ on $\mathcal{U}$.

Proof: By definition, it is immediate that $(\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} \in \mathrm{span}(\{\mathbf{q}_i\}_{i=1}^m) = \mathcal{U}$. We first prove $(*)$. We can write any $\mathbf{u} \in \mathcal{U}$ as $\sum_{j=1}^m \alpha_j \mathbf{q}_j$ for some $\alpha_j$'s. Then

$$ \left\langle \mathbf{v} - \sum_{j=1}^m \langle \mathbf{v}, \mathbf{q}_j \rangle \,\mathbf{q}_j, \sum_{j=1}^m \alpha_j' \mathbf{q}_j \right\rangle = \sum_{j=1}^m \langle \mathbf{v}, \mathbf{q}_j \rangle \,\alpha_j' - \sum_{j=1}^m \alpha_j' \langle \mathbf{v}, \mathbf{q}_j \rangle = 0 $$

where we used the orthonormality of the $\mathbf{q}_j$'s in the leftmost equality.

To prove $(**)$, note that for any $\mathbf{u} \in \mathcal{U}$ the vector $\mathbf{u}' = (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} - \mathbf{u}$ is also in $\mathcal{U}$. Hence by $(*)$ and Pythagoras,

$$ \begin{align} \|\mathbf{v} - \mathbf{u}\|^2 &= \|\mathbf{v} - (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} + (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} - \mathbf{u}\|^2\\ &= \|\mathbf{v} - (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v}\|^2 + \|(\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} - \mathbf{u}\|^2\\ &\geq \|\mathbf{v} - (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v}\|^2. \end{align} $$

Furthermore, equality holds only if $\|(\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v} - \mathbf{u}\|^2 = 0$ which holds only if $\mathbf{u} = (\mathcal{P}_{\{\mathbf{q}_i\}_{i=1}^m})\, \mathbf{v}$ by the point-separating property of the $2$-norm. The rightmost equality in $(*\!*\!*)$ follows from swapping $\{\mathbf{q}'_i\}_{i=1}^m$ for $\{\mathbf{q}_i\}_{i=1}^m$. $\square$

The Orthogonal Projection Theorem implies that any $\mathbf{v} \in V$ can be decomposed into its orthonogonal projection onto $\mathcal{U}$ and a vector orthogonal to it.

Definition (Orthogonal Complement): Let $\mathcal{U} \subseteq V$ be a linear subspace. The orthogonal complement of $\mathcal{U}$, denoted $\mathcal{U}^\perp$, is defined as

$$ \mathcal{U}^\perp = \{\mathbf{w} \in V\,:\, \langle \mathbf{w}, \mathbf{u}\rangle = 0, \forall \mathbf{u} \in \mathcal{U}\}. $$


Exercise: Establish that $\mathcal{U}^\perp$ is a linear subspace. $\lhd$

Lemma (Orthogonal Decomposition): Let $\mathcal{U} \subseteq V$ be a linear subspace and let $\mathbf{v} \in V$. Then $\mathbf{v}$ can be decomposed as $(\mathbf{v} - \mathcal{P}_\mathcal{U} \mathbf{v}) + \mathcal{P}_\mathcal{U} \mathbf{v}$ where $(\mathbf{v} - \mathcal{P}_\mathcal{U} \mathbf{v}) \in \mathcal{U}^\perp$ and $\mathcal{P}_\mathcal{U} \mathbf{v} \in \mathcal{U}$.

Proof: Immediate consequence of the previous theorem. $\square$

The map $\mathcal{P}_\mathcal{U}$ is linear, that is, $\mathcal{P}_\mathcal{U} (\alpha \,\mathbf{x} + \mathbf{y}) = \alpha \,\mathcal{P}_\mathcal{U} \mathbf{x} + \mathcal{P}_\mathcal{U} \mathbf{y}$ for all $\alpha \in \mathbb{R}$ and $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. Indeed,

$$ \mathcal{P}_\mathcal{U} (\alpha \,\mathbf{x} + \mathbf{y}) = \sum_{j=1}^m \langle \alpha \,\mathbf{x} + \mathbf{y}, \mathbf{q}_j \rangle \,\mathbf{q}_j = \sum_{j=1}^m \left\{\alpha \, \langle \mathbf{x}, \mathbf{q}_j \rangle + \langle \mathbf{y}, \mathbf{q}_j \rangle\right\} \mathbf{q}_j = \alpha \,\mathcal{P}_\mathcal{U} \mathbf{x} + \mathcal{P}_\mathcal{U} \mathbf{y}. $$

Therefore it can be encoded as an $n \times n$ matrix $P$. Let

$$ Q = \begin{pmatrix} | & & | \\ \mathbf{q}_1 & \ldots & \mathbf{q}_m \\ | & & | \end{pmatrix} $$

and note that computing

$$ Q^T \mathbf{v} = \begin{pmatrix} \langle \mathbf{v}, \mathbf{q}_1 \rangle \\ \cdots \\ \langle \mathbf{v}, \mathbf{q}_m \rangle \end{pmatrix} $$

lists the coefficients in the expansion of $\mathcal{P}_\mathcal{U} \mathbf{v}$ over the basis $\mathbf{q}_1,\ldots,\mathbf{q}_m$.

Hence we see that

$$ P = Q Q^T. $$

This is not to be confused with

$$ Q^T Q = \begin{pmatrix} \langle \mathbf{q}_1, \mathbf{q}_1 \rangle & \cdots & \langle \mathbf{q}_1, \mathbf{q}_m \rangle \\ \langle \mathbf{q}_2, \mathbf{q}_1 \rangle & \cdots & \langle \mathbf{q}_2, \mathbf{q}_m \rangle \\ \vdots & \ddots & \vdots \\ \langle \mathbf{q}_m, \mathbf{q}_1 \rangle & \cdots & \langle \mathbf{q}_m, \mathbf{q}_m \rangle \end{pmatrix} = I_{m \times m} $$

where $I_{m \times m}$ denotes the $m \times m$ identity matrix.

Exercise: Let $\mathcal{U} \subseteq V$ be a linear subspace and let $\mathbf{v} \in \mathcal{U}$. Show that $\mathcal{P}_\mathcal{U} \mathbf{v} = \mathbf{v}$. $\lhd$

2.3 Gram-Schmidt

We have some business left over: constructing orthonormal bases. Let $\mathbf{a}_1,\ldots,\mathbf{a}_m$ be linearly independent. We use the Gram-Schmidt algorithm to obtain an orthonormal basis of $\mathrm{span}(\mathbf{a}_1,\ldots,\mathbf{a}_m)$. The process takes advantage of the properties of the orthogonal projection derived above. In essence we add the vectors $\mathbf{a}_i$ one by one, but only after taking out their orthogonal projection on the previously included vectors. The outcome spans the same subspace and the Orthogonal Decomposition Lemma ensures orthogonality.

Theorem (Gram-Schmidt): Let $\mathbf{a}_1,\ldots,\mathbf{a}_m$ be linearly independent. Then there exists an orthonormal basis $\mathbf{q}_1,\ldots,\mathbf{q}_m$ of $\mathrm{span}(\mathbf{a}_1,\ldots,\mathbf{a}_m)$.

Proof idea: Suppose first that $m=1$. In that case, all that needs to be done is to divide $\mathbf{a}_1$ by its norm to obtain a unit vector whose span is the same as $\mathbf{a}_1$, that is, we set $\mathbf{q}_1 = \frac{\mathbf{a}_1}{\|\mathbf{a}_1\|}$.

Suppose now that $m=2$. We first let $\mathbf{q}_1 = \frac{\mathbf{a}_1}{\|\mathbf{a}_1\|}$ as in the previous case. Then we subtract from $\mathbf{a}_2$ its projection on $\mathbf{q}_1$, that is, we set $\mathbf{v}_2 = \mathbf{a}_2 - \langle \mathbf{q}_1, \mathbf{a}_2 \rangle \,\mathbf{q}_1$. By the Orthogonal Projection Theorem, $\mathbf{v}_2$ is orthogonal to $\mathbf{q}_1$. Moreover, because $\mathbf{a}_2$ is a linear combination of $\mathbf{q}_1$ and $\mathbf{v}_2$, we have $\mathrm{span}(\mathbf{q}_1,\mathbf{v}_2) = \mathrm{span}(\mathbf{a}_1,\mathbf{a}_2)$. It remains to divide by the norm of the resulting vector: $\mathbf{q}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|}$.

For general $m$, we proceed similarly but project onto the subspace spanned by the previously added vectors at each step.

GS (Source)

Proof: The inductive step is the following. Assume that we have constructed orthonormal vectors $\mathbf{q}_1,\ldots,\mathbf{q}_{j-1}$ such that

$$ U_{j-1} := \mathrm{span}(\mathbf{q}_1,\ldots,\mathbf{q}_{j-1}) = \mathrm{span}(\mathbf{a}_1,\ldots,\mathbf{a}_{j-1}). $$

By the Properties of Orthonormal Lists, $\{\mathbf{q}\}_{i=1}^{j-1}$ forms an orthonormal basis for $U_{j-1}$, so we can compute the orthogonal projection of $\mathbf{a}_j$ on $U_{j-1}$ as

$$ \mathcal{P}_{U_{j-1}}\mathbf{a}_j = \sum_{i=1}^{j-1} r_{ij} \,\mathbf{q}_i. $$

where we defined $r_{ij} = \langle \mathbf{q}_i , \mathbf{a}_j\rangle$. And we set

$$ \mathbf{v}_j = \mathbf{a}_j - \mathcal{P}_{U_{j-1}}\mathbf{a}_j \quad \text{and} \quad \mathbf{q}_j = \frac{\mathbf{v}_j}{\|\mathbf{v}_j\|}. $$

Here we used that $\|\mathbf{v}_j\| > 0$: indeed otherwise $\mathbf{a}_j$ would be equal to its projection $\mathcal{P}_{U_{j-1}}\mathbf{a}_j \in \mathrm{span}(\mathbf{a}_1,\ldots,\mathbf{a}_{j-1})$ which would contradict linear independence of the $\mathbf{a}_k$'s.

By the Orthogonal Projection Theorem, $\mathbf{q}_j$ is orthogonal to $\mathrm{span}(\mathbf{q}_1,\ldots,\mathbf{q}_{j-1})$ and, unrolling the calculations above, $\mathbf{a}_j$ can be re-written as the following linear combination of $\mathbf{q}_1,\ldots,\mathbf{q}_j$

$$ \begin{align} \mathbf{a}_j &= \mathcal{P}_{U_{j-1}}\mathbf{a}_j + \mathbf{v}_j\\ &= \mathcal{P}_{U_{j-1}}\mathbf{a}_j + \|\mathbf{v}_j\| \mathbf{q}_j\\ &= \mathcal{P}_{U_{j-1}}\mathbf{a}_j + \|\mathbf{a}_j - \mathcal{P}_{U_{j-1}}\mathbf{a}_j\| \mathbf{q}_j\\ &= \sum_{i=1}^{j-1} r_{ij} \,\mathbf{q}_i + \left\|\mathbf{a}_j - \sum_{i=1}^{j-1} r_{ij}\,\mathbf{q}_i\right\| \,\mathbf{q}_j\\ &= \sum_{i=1}^{j-1} r_{ij} \,\mathbf{q}_i + r_{jj} \,\mathbf{q}_j, \end{align} $$

where we defined $r_{jj} = \left\|\mathbf{a}_j - \sum_{i=1}^{j-1} r_{ij}\,\mathbf{q}_i\right\| = \|\mathbf{v}_j\|$.

Hence $\mathbf{q}_1,\ldots,\mathbf{q}_j$ forms an orthonormal list with $\mathrm{span}(\mathbf{a}_1,\ldots,\mathbf{a}_{j}) \subseteq \mathrm{span}(\mathbf{q}_1,\ldots,\mathbf{q}_{j})$. The opposite inclusion holds by construction. Moreover, because $\mathbf{q}_1,\ldots,\mathbf{q}_j$ are orthonormal, they are linearly independent by the Properties of Orthonormal Lists so must form a basis of their span. So induction goes through. $\square$

NUMERICAL CORNER We implement the Gram-Schmidt algorithm in Julia. For reasons that will become clear in a future notebook, we output the $\mathbf{q}_j$'s and $r_{ij}$'s, each in matrix form.

In [1]:
# Julia version: 1.5.1
using LinearAlgebra
In [2]:
function mmids_gramschmidt(A)
    n,m = size(A)
    Q = zeros(Float64,n,m)
    R = zeros(Float64,m,m)
    for j = 1:m
        v = A[:,j]
        for i = 1:j-1
            R[i,j] = dot(Q[:,i],A[:,j])
            v -= R[i,j]*Q[:,i]
        R[j,j] = norm(v)
        Q[:,j] = v/R[j,j]
    return Q,R
mmids_gramschmidt (generic function with 1 method)

Let's try a simple example.

In [3]:
w1, w2 = [1,0,1], [0,1,1]
A = hcat(w1,w2)
3×2 Array{Int64,2}:
 1  0
 0  1
 1  1
In [4]:
Q,R = mmids_gramschmidt(A);
In [5]:
3×2 Array{Float64,2}:
 0.707107  -0.408248
 0.0        0.816497
 0.707107   0.408248
In [6]:
2×2 Array{Float64,2}:
 1.41421  0.707107
 0.0      1.22474

Exercise: Let $B \in \mathbb{R}^{n \times m}$ be a matrix. Show that there exist orthonormal bases of $\mathrm{col}(B)$ and $\mathrm{null}(B)$. $\lhd$

Exercise: Let $\mathcal{W}$ be a linear subspace of $\mathbb{R}^d$ and let $\mathbf{w}_1,\ldots,\mathbf{w}_k$ be an orthonormal basis of $\mathcal{W}$. Show that there exists an orthonormal basis of $\mathbb{R}^d$ that includes the $\mathbf{w}_i$'s. $\lhd$

Lemma (Nullity): Let $B \in \mathbb{R}^{n \times m}$ be a matrix of rank $= k$. Then $\mathrm{dim}(\mathrm{null}(B)) = n - k$. Put differently,

$$ \mathrm{dim}(\mathrm{col}(B)) + \mathrm{dim}(\mathrm{null}(B)) = n. $$

For a proof, see Wikipedia.