Math 2270 - Lecture 22 : Projections

Dylan Zwick

Fall 2012

This lecture coverssection 4.2from the textbook.

In our last lecture we learned that ifAis anm×nmatrix then every vectorx?Rncan be written as x=xrow+xnull wherexrowis in the rowspace ofA, writtenxrow?C(AT), andxnullis in the nullspace ofA, writtenxnull?N(A). We left relatively unexplored how, given the rowspace or the nullspace, we would actually find these vectorsxrowandxnull. In order to find these vectors, we needprojections, and projections are the subject of this lecture. The exercises for section 4.2 will be:

Section 4.2- 1, 11, 12, 13, 17

1 Projections Onto Lines

When we project a vectorbonto a line1we want to find the point on the line closest to the vectorb. Call the vector that points from the origin to this closest pointp, and note thatpis a vector on this line. The vector that points fromptobis called the error vector", and it will be perpendicular top, and indeed perpendicular to the line itself.

1We"ll assume we"re always dealing with lines through the origin, as these are the

one-dimensional subspaces ofRn. 1 We want to figure out how we can calculatepfrom the vectorband the line. Well, a line through the origin can be represented as all multiples of a direction vectora, and so we"ll describe the line using a direction vector aparallel to the line. Our vectorpwill be some constant multiple of this direction vectora, sop= ˆxa. We want to calculateˆx. Now, the error vectore=b-pwill be perpendicular toa, so a·(b-ˆxa) = 0.

We can rewrite this as

a·b= ˆxa·a, and solve forˆxto get:

ˆx=a·b

a·a=aTbaTb. I rewrote it using transposes because that"s an idea that will generalize beyond just lines. So, to recap, the projection,p, of a vectorbonto the line spanned by the direction vectorawill be: p=aTb aTaa. 2 Example- Calculate the projection of the vector?23? onto the line spanned by the vector ?11?

2 Projection Onto a Subspace

A line inRmis a one-dimensional subspace. Suppose we have a higher- dimensional subspaceV, and we want to project a vectorbonto it. First, we need a description ofV, and the best description is a set of basis vec- tors. So, supposeVis a subspace ofRmwith basisa1,...,an. The projec- tion ofbontoVis the vector inVclosest tob. This projection vector,p, will be by definition a linear combination of the basis vectors ofV: p= ˆx1a1+···+ ˆxnan. IfAis a matrix with column vectorsa1,...,anandˆxis a vector with componentsˆx1,...,ˆxnthenp=Aˆx. The error vector, the vector that points fromptob, will bee=b-p=b-Aˆx. This error vector will be perpen- dicular to the subspaceV, which is equivalent to being perpendicular to the basis vectorsa1,...,an. Stated mathematically, this is 3 aT1(b-Aˆx) = 0, a T n(b-Aˆx) = 0.

We can write all of the above in matrix form:

T(b-Aˆx) =0.

This in turn can be written as

TAˆx=ATb.

IfATAis invertible we can solve forˆxto get:

x= (ATA)-1ATb. Now, the projection vectorpis the vectorAˆx, so p=A(ATA)-1ATb. In general, to calculate the projection of any vector onto the spaceW we multiply the vector by theprojection matrixP=A(ATA)-1AT. Wow, that was a lot of work! Let"s see how we"d actually use these in an example. 4

Example- IfA=((

1 0 1 1 1 2)) andb=(( 6quotesdbs_dbs2.pdfusesText_3

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