| Linear Algebra | ||
| ←Projection | Orthogonal Projection Onto a Line | Gram-Schmidt Orthogonalization→ |
We first consider orthogonal projection onto a line.To orthogonally project a vector 
onto a line 
,mark the point on the line at which someone standing on that point could see by looking straight up or down (from that person's point of view).
The picture shows someone who has walked out on the lineuntil the tip of is straight overhead.That is, where the line is described as the span of some nonzerovector 
,the person has walked out to find the coefficient 
with the property that 
is orthogonalto 
.
We can solve for this coefficient by noting that because
is orthogonal to a scalar multipleof 
it must be orthogonal to itself,and then the consequent fact that the dot product
is zero gives that
.
- Definition 1.1
The orthogonal projection of onto the line spanned by a nonzero is this vector.
Problem 13 checks that theoutcome of the calculation depends only on the line and not on which vector happens to be used to describe that line.
- Remark 1.2
The wording of that definition says "spanned by " instead the more formal "the span of the set 
". This casual first phrase is common.
- Example 1.3
To orthogonally projectthe vector 
onto the line 
,we first pick a direction vector for the line.For instance,
will do.Then the calculation is routine.
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- Example 1.4
In 
, the orthogonal projection of a general vector
onto the 
-axis is
which matches our intuitive expectation.
The picture above with the stick figure walking out on the line until's tip is overhead is one way to think of the orthogonal projection ofa vector onto a line.We finish this subsection with two other ways.
- Example 1.5
A railroad car left on an east-west track without its brake is pushed bya wind blowing toward the northeast at fifteen miles per hour;what speed will the car reach?
For the wind we use a vector of length 
that points towardthe northeast.
The car can only be affected by the part of the wind blowing in theeast-west direction— the part of in the directionof the 
-axis is this (the picture has the same perspective as therailroad car picture above).
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So the car will reach a velocity of 
miles per hour toward the east.
Thus, another way to thinkof the picture that precedes the definition is that it shows as decomposed into two parts, the part with the line (here, the partwith the tracks, 
),and the part that is orthogonal to the line(shown here lying on the north-south axis).These two are "not interacting" or "independent",in the sense that the east-west car is not at all affected by thenorth-south part of the wind (see Problem 5).So the orthogonal projection of onto the line spanned by can be thought of asthe part of 
that lies in the direction of 
.
Finally, another useful way to think of the orthogonal projectionis to have the person stand not on the line, but on the vector that is to beprojected to the line.This person has a rope over the line and pulls it tight,naturally making the rope orthogonal to the line.
That is, we can think of the projection 
as being the vectorin the line that is closest to (see Problem 11).
- Example 1.6
A submarine is tracking a ship moving along the line 
.Torpedo range is one-half mile.Can the sub stay where it is, at the origin on the chart below,or must it move to reach a place where the ship will pass within range?
The formula for projectiononto a line does not immediately apply because the line doesn't pass throughthe origin, and so isn't the span of any .To adjust for this, we start by shifting the entire map down two units.Now the line is 
, which is a subspace, and we can project to getthe point of closest approach, the point on theline through the origin closest to
the sub's shifted position.
The distance between and is approximately 
miles and so the sub must move to get in range.
This subsection has developed a natural projection map: orthogonal projectiononto a line.As suggested by the examples, it is often called for in applications.The next subsection shows how the definition of orthogonalprojection onto a line gives us a way to calculate especially convienent basesfor vector spaces, again something that is common in applications.The final subsection completely generalizes projection, orthogonal or not,onto any subspace at all.
Exercises
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- This exercise is recommended for all readers.
- Problem 1
Project the first vector orthogonallyonto the line spanned by the second vector.
-
, - ,
- ,
- ,
- This exercise is recommended for all readers.
- Problem 2
Project the vector orthogonally onto the line.
- , the line
- Problem 3
Although the development of Definition 1.1is guided by the pictures, we are not restricted to spaces thatwe can draw.In 
project this vector onto this line.
- This exercise is recommended for all readers.
- Problem 4
Definition 1.1 uses two vectors and.Consider the transformation of 
resulting from fixing
and projecting onto the line that is the span of .Apply it to these vectors.
Show that in general the projection tranformation is this.
Express the action of this transformation with a matrix.
- Problem 5
Example 1.5 suggests that projection breaks intotwo parts, 
and
, that are"not interacting".Recall that the two are orthogonal.Show that any two nonzero orthogonal vectors make up a linearlyindependent set.
- Problem 6
- What is the orthogonal projection of onto a line if is a member of that line?
- Show thatif is not a member of the linethen the setis linearly independent.
- Problem 7
Definition 1.1 requires that benonzero.Why?What is the right definition of the orthogonal projectionof a vector onto the (degenerate) line spanned by the zero vector?
- Problem 8
Are all vectors the projection of some other vector onto some line?
- This exercise is recommended for all readers.
- Problem 9
Show that the projection of onto the line spanned by has length equal to the absolute value of the number
divided by the length of the vector.
- Problem 10
Find the formula for the distance from a point to a line.
- Problem 11
Find the scalar 
such that 
is a minimum distance from the point 
by using calculus (i.e., consider the distance function, set thefirst derivative equal to zero, and solve).Generalize to 
.
- This exercise is recommended for all readers.
- Problem 12
Prove that the orthogonal projection of a vector onto a line is shorterthan the vector.
- This exercise is recommended for all readers.
- Problem 13
Show that the definition of orthogonal projection onto a linedoes not dependon the spanning vector: if is a nonzero multipleof 
then
equals
.
- This exercise is recommended for all readers.
- Problem 14
Consider the function mapping to plane to itself that takesa vector to its projection onto the line 
.These two each show that the map is linear, the first one in a way thatis 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 rotatingeverything in the plane radians clockwise,then projecting onto the -axis,and then rotating radians counterclockwise.
- Problem 15
For 
let 
be theprojection of 
onto the line spanned by 
, let
be the projection of onto the line spannedby , let 
be the projection of onto the line spanned by , etc.,back and forth between the spans of and .That is, 
is the projection of
onto the span of if 
is even, andonto the span of if is odd.Must that sequence of vectors eventually settle down— mustthere be a sufficiently large 
such that
equals and 
equals ?If so, what is the earliest such ?
Solutions
| Linear Algebra | ||
| ←Projection | Orthogonal Projection Onto a Line | Gram-Schmidt Orthogonalization→ |
