Linear Algebra/Orthogonal Projection Onto a Line/Solutions

Solutions[edit | edit source]

This exercise is recommended for all readers.

Problem 1

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

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

Answer

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

$\displaystyle {\frac {{\begin{pmatrix}2\\1\end{pmatrix}}\cdot {\begin{pmatrix}3\\-2\end{pmatrix}}}{{\begin{pmatrix}3\\-2\end{pmatrix}}\cdot {\begin{pmatrix}3\\-2\end{pmatrix}}}}\cdot {\begin{pmatrix}3\\-2\end{pmatrix}}={\frac {4}{13}}\cdot {\begin{pmatrix}3\\-2\end{pmatrix}}={\begin{pmatrix}12/13\\-8/13\end{pmatrix}}$
$\displaystyle {\frac {{\begin{pmatrix}2\\1\end{pmatrix}}\cdot {\begin{pmatrix}3\\0\end{pmatrix}}}{{\begin{pmatrix}3\\0\end{pmatrix}}\cdot {\begin{pmatrix}3\\0\end{pmatrix}}}}\cdot {\begin{pmatrix}3\\0\end{pmatrix}}={\frac {2}{3}}\cdot {\begin{pmatrix}3\\0\end{pmatrix}}={\begin{pmatrix}2\\0\end{pmatrix}}$
$\displaystyle {\frac {{\begin{pmatrix}1\\1\\4\end{pmatrix}}\cdot {\begin{pmatrix}1\\2\\-1\end{pmatrix}}}{{\begin{pmatrix}1\\2\\-1\end{pmatrix}}\cdot {\begin{pmatrix}1\\2\\-1\end{pmatrix}}}}\cdot {\begin{pmatrix}1\\2\\-1\end{pmatrix}}={\frac {-1}{6}}\cdot {\begin{pmatrix}1\\2\\-1\end{pmatrix}}={\begin{pmatrix}-1/6\\-1/3\\1/6\end{pmatrix}}$
$\displaystyle {\frac {{\begin{pmatrix}1\\1\\4\end{pmatrix}}\cdot {\begin{pmatrix}3\\3\\12\end{pmatrix}}}{{\begin{pmatrix}3\\3\\12\end{pmatrix}}\cdot {\begin{pmatrix}3\\3\\12\end{pmatrix}}}}\cdot {\begin{pmatrix}3\\3\\12\end{pmatrix}}={\frac {1}{3}}\cdot {\begin{pmatrix}3\\3\\12\end{pmatrix}}={\begin{pmatrix}1\\1\\4\end{pmatrix}}$

This exercise is recommended for all readers.

Problem 2

Project the vector orthogonally onto the line.

${\begin{pmatrix}2\\-1\\4\end{pmatrix}},\{c{\begin{pmatrix}-3\\1\\-3\end{pmatrix}}\,{\big |}\,c\in \mathbb {R} \}$
${\begin{pmatrix}-1\\-1\end{pmatrix}}$ , the line $y=3x$

Answer

$\displaystyle {\frac {{\begin{pmatrix}2\\-1\\4\end{pmatrix}}\cdot {\begin{pmatrix}-3\\1\\-3\end{pmatrix}}}{{\begin{pmatrix}-3\\1\\-3\end{pmatrix}}\cdot {\begin{pmatrix}-3\\1\\-3\end{pmatrix}}}}\cdot {\begin{pmatrix}-3\\1\\-3\end{pmatrix}}={\frac {-19}{19}}\cdot {\begin{pmatrix}-3\\1\\-3\end{pmatrix}}={\begin{pmatrix}3\\-1\\3\end{pmatrix}}$
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} \}

Answer

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

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.

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

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.

Answer

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

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.

Answer

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

What is the orthogonal projection of ${\vec {v}}$ onto a line if ${\vec {v}}$ is a member of that line?
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.

Answer

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 ${\vec {v}}$ .)
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 $(a_{1}+a_{2})\cdot {\vec {v}}=a_{2}c_{\vec {p}}\cdot {\vec {s}}$ . 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?

Answer

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?

Answer

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 ${\vec {s}}$ .

Answer

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.

Answer

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 $\mathbb {R} ^{n}$ .

Answer

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 $2(cs_{1}-v_{1})\cdot s_{1}+2(cs_{2}-v_{2})\cdot s_{2}=c\cdot (2s_{1}^{2}+2s_{2}^{2})-(2v_{1}s_{1}+2v_{2}s_{2})=0$ gives the only critical point.

c={\frac {v_{1}s_{1}+v_{2}s_{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.

Answer

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 $({\vec {v}}\cdot {\vec {q}}/{\vec {q}}\cdot {\vec {q}}\,)\cdot {\vec {q}}$ .

Answer

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.

Produce a matrix that describes the function's action.
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.

Answer

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.
${\begin{pmatrix}1/2&1/2\\1/2&1/2\end{pmatrix}}$
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 $i$ ?