# Linear Algebra/Orthogonal Projection Onto a Line/Solutions

## Solutions

This exercise is recommended for all readers.
Problem 1

Project the first vector orthogonally onto the line spanned by the second vector.

1. $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$
2. $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$
3. $\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}$, $\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$
4. $\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}$, $\begin{pmatrix} 3 \\ 3 \\ 12 \end{pmatrix}$

Each is a straightforward application of the formula from Definition 1.1.

1. $\displaystyle \frac{\begin{pmatrix} 2 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ -2 \end{pmatrix}}{ \begin{pmatrix} 3 \\ -2 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ -2 \end{pmatrix}} \cdot\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\frac{4}{13} \cdot\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} 12/13 \\ -8/13 \end{pmatrix}$
2. $\displaystyle \frac{\begin{pmatrix} 2 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 0 \end{pmatrix}}{ \begin{pmatrix} 3 \\ 0 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 0 \end{pmatrix}} \cdot\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\frac{2}{3} \cdot\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} 2 \\ 0 \end{pmatrix}$
3. $\displaystyle \frac{\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}}{ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}} \cdot\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} =\frac{-1}{6} \cdot\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} =\begin{pmatrix} -1/6 \\ -1/3 \\ 1/6 \end{pmatrix}$
4. $\displaystyle \frac{\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 3 \\ 12 \end{pmatrix}}{ \begin{pmatrix} 3 \\ 3 \\ 12 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 3 \\ 12 \end{pmatrix}} \cdot\begin{pmatrix} 3 \\ 3 \\ 12 \end{pmatrix} =\frac{1}{3} \cdot\begin{pmatrix} 3 \\ 3 \\ 12 \end{pmatrix} =\begin{pmatrix} 1 \\ 1 \\ 4 \end{pmatrix}$
This exercise is recommended for all readers.
Problem 2

Project the vector orthogonally onto the line.

1. $\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}, \{c\begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}\,\big|\, c\in\mathbb{R}\}$
2. $\begin{pmatrix} -1 \\ -1 \end{pmatrix}$, the line $y=3x$
1. $\displaystyle \frac{\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\cdot\begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}}{ \begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}\cdot\begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix}} \cdot\begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix} =\frac{-19}{19}\cdot\begin{pmatrix} -3 \\ 1 \\ -3 \end{pmatrix} =\begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}$
2. Writing the line as $\{c\cdot \begin{pmatrix} 1 \\ 3 \end{pmatrix}\,\big|\, c\in\mathbb{R}\}$ gives this projection.
$\frac{\begin{pmatrix} -1 \\ -1 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix}}{ \begin{pmatrix} 1 \\ 3 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix}} \cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix} =\frac{-4}{10}\cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix} =\begin{pmatrix} -2/5 \\ -6/5 \end{pmatrix}$
Problem 3

Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In $\mathbb{R}^4$ project this vector onto this line.

$\vec{v}=\begin{pmatrix} 1 \\ 2 \\ 1 \\ 3 \end{pmatrix} \qquad \ell=\{c\cdot\begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix}\,\big|\, c\in\mathbb{R}\}$

$\displaystyle \frac{\begin{pmatrix} 1 \\ 2 \\ 1 \\ 3 \end{pmatrix}\cdot\begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix}}{ \begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix}} \cdot\begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix} =\frac{3}{4} \cdot\begin{pmatrix} -1 \\ 1 \\ -1 \\ 1 \end{pmatrix} =\begin{pmatrix} -3/4 \\ 3/4 \\ -3/4 \\ 3/4 \end{pmatrix}$

This exercise is recommended for all readers.
Problem 4

Definition 1.1 uses two vectors $\vec{s}$ and $\vec{v}$. Consider the transformation of $\mathbb{R}^2$ resulting from fixing

$\vec{s}=\begin{pmatrix} 3 \\ 1 \end{pmatrix}$

and projecting $\vec{v}$ onto the line that is the span of $\vec{s}$. Apply it to these vectors.

1. $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$
2. $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$

Show that in general the projection tranformation is this.

$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} (x_1+3x_2)/10 \\ (3x_1+9x_2)/10 \end{pmatrix}$

Express the action of this transformation with a matrix.

1. $\displaystyle \frac{\begin{pmatrix} 1 \\ 2 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix}}{ \begin{pmatrix} 3 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix}} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} =\frac{1}{2}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix} =\begin{pmatrix} 3/2 \\ 1/2 \end{pmatrix}$
2. $\displaystyle \frac{\begin{pmatrix} 0 \\ 4 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix}}{ \begin{pmatrix} 3 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix}} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} =\frac{2}{5}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix} =\begin{pmatrix} 6/5 \\ 2/5 \end{pmatrix}$

In general the projection is this.

$\frac{\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix}}{ \begin{pmatrix} 3 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix}} \cdot \begin{pmatrix} 3 \\ 1 \end{pmatrix} =\frac{3x_1+x_2}{10}\cdot\begin{pmatrix} 3 \\ 1 \end{pmatrix} =\begin{pmatrix} (9x_1+3x_2)/10 \\ (3x_1+x_2)/10 \end{pmatrix}$

The appropriate matrix is this.

$\begin{pmatrix} 9/10 &3/10 \\ 3/10 &1/10 \end{pmatrix}$
Problem 5

Example 1.5 suggests that projection breaks $\vec{v}$ into two parts, $\mbox{proj}_{[\vec{s}\,]}({\vec{v}\,})$ and $\vec{v}-\mbox{proj}_{[\vec{s}\,]}({\vec{v}\,})$, that are "not interacting". Recall that the two are orthogonal. Show that any two nonzero orthogonal vectors make up a linearly independent set.

Suppose that $\vec{v}_1$ and $\vec{v}_2$ are nonzero and orthogonal. Consider the linear relationship $c_1\vec{v}_1+c_2\vec{v}_2=\vec{0}$. Take the dot product of both sides of the equation with $\vec{v}_1$ to get that

$\vec{v}_1\cdot(c_1\vec{v}_1+c_2\vec{v}_2) =c_1\cdot(\vec{v}_1\cdot\vec{v}_1) +c_2\cdot(\vec{v}_1\cdot\vec{v}_2) =c_1\cdot (\vec{v}_1\cdot\vec{v}_1)+c_2\cdot 0 =c_1\cdot (\vec{v}_1\cdot\vec{v}_1)$

is equal to $\vec{v}_1\cdot\vec{0}=\vec{0}$. With the assumption that $\vec{v}_1$ is nonzero, this gives that $c_1$ is zero. Showing that $c_2$ is zero is similar.

Problem 6
1. What is the orthogonal projection of $\vec{v}$ onto a line if $\vec{v}$ is a member of that line?
2. Show that if $\vec{v}$ is not a member of the line then the set $\{\vec{v},\vec{v}-\mbox{proj}_{[\vec{s}\,]}({\vec{v}\,})\}$ is linearly independent.
1. If the vector $\vec{v}\,$ is in the line then the orthogonal projection is $\vec{v}$. To verify this by calculation, note that since $\vec{v}\,$ is in the line we have that $\vec{v}=c_{\vec{v}}\cdot\vec{s}\,$ for some scalar $c_{\vec{v}}$.
$\frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\cdot\vec{s} =\frac{c_{\vec{v}}\cdot \vec{s}\cdot\vec{s}}{ \vec{s}\cdot\vec{s}}\cdot\vec{s} =c_{\vec{v}}\cdot\frac{\vec{s}\cdot\vec{s}}{ \vec{s}\cdot\vec{s}}\cdot\vec{s} =c_{\vec{v}}\cdot 1 \cdot\vec{s} =\vec{v}$
(Remark. If we assume that $\vec{v}\,$ is nonzero then the above is simplified on taking $\vec{s}\,$ to be $\vec{v}$.)
2. Write $c_{\vec{p}}\vec{s}\,$ for the projection $\mbox{proj}_{[\vec{s}\,]}({\vec{v}})$. Note that, by the assumption that $\vec{v}$ is not in the line, both $\vec{v}$ and $\vec{v}-c_{\vec{p}}\vec{s}\,$ are nonzero. Note also that if $c_{\vec{p}}\,$ is zero then we are actually considering the one-element set $\{\vec{v}\,\}$, and with $\vec{v}$ nonzero, this set is necessarily linearly independent. Therefore, we are left considering the case that $c_{\vec{p}}$ is nonzero. Setting up a linear relationship
$a_1(\vec{v})+a_2(\vec{v}-c_{\vec{p}}\vec{s})=\vec{0}$
leads to the equation $(a_1+a_2)\cdot\vec{v}=a_2c_{\vec{p}}\cdot\vec{s}$. Because $\vec{v}\,$ isn't in the line, the scalars $a_1+a_2$ and $a_2 c_{\vec{p}}$ must both be zero. The $c_{\vec{p}}=0$ case is handled above, so the remaining case is that $a_2=0$, and this gives that $a_1=0$ also. Hence the set is linearly independent.
Problem 7

Definition 1.1 requires that $\vec{s}\,$ be nonzero. Why? What is the right definition of the orthogonal projection of a vector onto the (degenerate) line spanned by the zero vector?

If $\vec{s}\,$ is the zero vector then the expression

$\mbox{proj}_{[\vec{s}\,]}({\vec{v}})= \frac{ \vec{v}\cdot\vec{s} }{ \vec{s}\cdot\vec{s} }\cdot\vec{s}$

contains a division by zero, and so is undefined. As for the right definition, for the projection to lie in the span of the zero vector, it must be defined to be $\vec{0}$.

Problem 8

Are all vectors the projection of some other vector onto some line?

Any vector in $\mathbb{R}^n$ is the projection of some other onto a line, provided that the dimension $n$ is greater than one. (Clearly, any vector is the projection of itself into a line containing itself; the question is to produce some vector other than $\vec{v}$ that projects to $\vec{v}$.)

Suppose that $\vec{v}\in\mathbb{R}^n$ with $n>1$. If $\vec{v}\neq\vec{0}$ then we consider the line $\ell=\{c\vec{v}\,\big|\, c\in\mathbb{R}\}$ and if $\vec{v}=\vec{0}$ we take $\ell$ to be any (nondegenerate) line at all (actually, we needn't distinguish between these two cases— see the prior exercise). Let $v_1,\dots,v_n$ be the components of $\vec{v}$; since $n>1$, there are at least two. If some $v_i$ is zero then the vector $\vec{w}=\vec{e}_i$ is perpendicular to $\vec{v}$. If none of the components is zero then the vector $\vec{w}\,$ whose components are $v_2,-v_1,0,\dots ,0$ is perpendicular to $\vec{v}$. In either case, observe that $\vec{v}+\vec{w}$ does not equal $\vec{v}$, and that $\vec{v}$ is the projection of $\vec{v}+\vec{w}$ onto $\ell$.

$\frac{(\vec{v}+\vec{w})\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\cdot\vec{v}=\bigl(\frac{\vec{v}\cdot\vec{v}}{\vec{v}\cdot\vec{v}} +\frac{\vec{w}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\bigr)\cdot\vec{v}=\frac{\vec{v}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\cdot\vec{v} =\vec{v}$

We can dispose of the remaining $n=0$ and $n=1$ cases. The dimension $n=0$ case is the trivial vector space, here there is only one vector and so it cannot be expressed as the projection of a different vector. In the dimension $n=1$ case there is only one (nondegenerate) line, and every vector is in it, hence every vector is the projection only of itself.

This exercise is recommended for all readers.
Problem 9

Show that the projection of $\vec{v}$ onto the line spanned by $\vec{s}$ has length equal to the absolute value of the number $\vec{v}\cdot\vec{s}$ divided by the length of the vector $\vec{s}$.

The proof is simply a calculation.

$|\frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}} \cdot\vec{s}\, | = |\frac{\vec{v}\cdot\vec{s}}{ \vec{s}\cdot\vec{s}}|\cdot |\vec{s}\, | = \frac{|\vec{v}\cdot\vec{s}\,|}{ |\vec{s}\,|^2}\cdot|\vec{s}\,| = \frac{|\vec{v}\cdot\vec{s}\,|}{|\vec{s}\,|}$
Problem 10

Find the formula for the distance from a point to a line.

Because the projection of $\vec{v}$ onto the line spanned by $\vec{s}$ is

$\frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\cdot\vec{s}$

the distance squared from the point to the line is this (a vector dotted with itself $\vec{w}\cdot\vec{w}$ is written $\vec{w}^2$).

$\begin{array}{rl} |\vec{v}- \frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}} \cdot\vec{s}\, |^2 &=\vec{v}\cdot\vec{v} -\vec{v}\cdot( \frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}\cdot\vec{s} ) -( \frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}} \cdot\vec{s}\, )\cdot\vec{v} +(\frac{\vec{v}\cdot\vec{s}}{ \vec{s}\cdot\vec{s}}\cdot\vec{s}\,)^2 \\ &=\vec{v}\cdot\vec{v} -2\cdot(\frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}) \cdot\vec{v}\cdot\vec{s} +(\frac{\vec{v}\cdot\vec{s}}{ \vec{s}\cdot\vec{s}})\cdot\vec{s}\cdot\vec{s} \\ &=\frac{(\vec{v}\cdot\vec{v}\,)\cdot(\vec{s}\cdot\vec{s}\,) -2\cdot(\vec{v}\cdot\vec{s}\,)^2 +(\vec{v}\cdot\vec{s}\,)^2}{ \vec{s}\cdot\vec{s}} \\ &=\frac{(\vec{v}\cdot\vec{v}\,)(\vec{s}\cdot\vec{s}\,) -(\vec{v}\cdot\vec{s}\,)^2}{\vec{s}\cdot\vec{s}} \end{array}$
Problem 11

Find the scalar $c$ such that $(cs_1,cs_2)$ is a minimum distance from the point $(v_1,v_2)$ by using calculus (i.e., consider the distance function, set the first derivative equal to zero, and solve). Generalize to $\mathbb{R}^n$.

Because square root is a strictly increasing function, we can minimize $d(c)=(cs_1-v_1)^2+(cs_2-v_2)^2$ instead of the square root of $d$. The derivative is $dd/dc=2(cs_1-v_1)\cdot s_1 +2(cs_2-v_2)\cdot s_2$. Setting it equal to zero $2(cs_1-v_1)\cdot s_1 +2(cs_2-v_2)\cdot s_2 =c\cdot(2s_1^2+2s_2^2)-(v_1s_1+v_2s_2)=0$ gives the only critical point.

$c=\frac{v_1s_1+v_2s_2}{{s_1}^2+{s_2}^2} =\frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}}$

Now the second derivative with respect to $c$

$\frac{d^2\,d}{dc^2}=2{s_1}^2+2{s_2}^2$

is strictly positive (as long as neither $s_1$ nor $s_2$ is zero, in which case the question is trivial) and so the critical point is a minimum.

The generalization to $\mathbb{R}^n$ is straightforward. Consider $d_n(c)=(cs_1-v_1)^2+\dots+(cs_n-v_n)^2$, take the derivative, etc.

This exercise is recommended for all readers.
Problem 12

Prove that the orthogonal projection of a vector onto a line is shorter than the vector.

The Cauchy-Schwarz inequality $|\vec{v}\cdot\vec{s}\,| \leq|\vec{v}\,|\cdot|\vec{s}\,|$ gives that this fraction

$|\frac{\vec{v}\cdot\vec{s}}{\vec{s}\cdot\vec{s}} \cdot\vec{s}\, | =|\frac{\vec{v}\cdot\vec{s}}{ \vec{s}\cdot\vec{s}}| \cdot|\vec{s}\, | = \frac{|\vec{v}\cdot\vec{s}\,|}{ |\vec{s}\,|^2}\cdot|\vec{s}\,| = \frac{|\vec{v}\cdot\vec{s}\,|}{|\vec{s}\,|}$

when divided by $|\vec{v}\,|$ is less than or equal to one. That is, $|\vec{v}\,|$ is larger than or equal to the fraction.

This exercise is recommended for all readers.
Problem 13

Show that the definition of orthogonal projection onto a line does not depend on the spanning vector: if $\vec{s}$ is a nonzero multiple of $\vec{q}$ then $(\vec{v}\cdot\vec{s}/\vec{s}\cdot\vec{s}\,)\cdot\vec{s}$ equals $(\vec{v}\cdot\vec{q}/\vec{q}\cdot\vec{q}\,)\cdot\vec{q}$.

Write $c\vec{s}$ for $\vec{q}$, and calculate: $(\vec{v}\cdot c\vec{s}/c\vec{s}\cdot c\vec{s}\,)\cdot c\vec{s}=(\vec{v}\cdot \vec{s}/\vec{s}\cdot \vec{s}\,)\cdot \vec{s}$.

This exercise is recommended for all readers.
Problem 14

Consider the function mapping to plane to itself that takes a vector to its projection onto the line $y=x$. These two each show that the map is linear, the first one in a way that is bound to the coordinates (that is, it fixes a basis and then computes) and the second in a way that is more conceptual.

1. Produce a matrix that describes the function's action.
2. Show also that this map can be obtained by first rotating everything in the plane $\pi/4$ radians clockwise, then projecting onto the $x$-axis, and then rotating $\pi/4$ radians counterclockwise.
1. Fixing
$\vec{s}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
as the vector whose span is the line, the formula gives this action,
$\begin{pmatrix} x \\ y \end{pmatrix} \mapsto \frac{\begin{pmatrix} x \\ y \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}}{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix} =\frac{x+y}{2}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix} =\begin{pmatrix} (x+y)/2 \\ (x+y)/2 \end{pmatrix}$
which is the effect of this matrix.
$\begin{pmatrix} 1/2 &1/2 \\ 1/2 &1/2 \end{pmatrix}$
2. Rotating the entire plane $\pi/4$ radians clockwise brings the $y=x$ line to lie on the $x$-axis. Now projecting and then rotating back has the desired effect.
Problem 15

For $\vec{a},\vec{b}\in\mathbb{R}^n$ let $\vec{v}_1$ be the projection of $\vec{a}$ onto the line spanned by $\vec{b}$, let $\vec{v}_2$ be the projection of $\vec{v}_1$ onto the line spanned by $\vec{a}$, let $\vec{v}_3$ be the projection of $\vec{v}_2$ onto the line spanned by $\vec{b}$, etc., back and forth between the spans of $\vec{a}$ and $\vec{b}$. That is, $\vec{v}_{i+1}$ is the projection of $\vec{v}_i$ onto the span of $\vec{a}$ if $i+1$ is even, and onto the span of $\vec{b}$ if $i+1$ is odd. Must that sequence of vectors eventually settle down— must there be a sufficiently large $i$ such that $\vec{v}_{i+2}$ equals $\vec{v}_{i}$ and $\vec{v}_{i+3}$ equals $\vec{v}_{i+1}$? If so, what is the earliest such $i$?

$\vec{a}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad \vec{b}=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
$\vec{v}_1=\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}, \quad \vec{v}_2=\begin{pmatrix} 1/2 \\ 0 \end{pmatrix}, \quad \vec{v}_3=\begin{pmatrix} 1/4 \\ 1/4 \end{pmatrix}, \quad \ldots$