Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:40 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Wednesday February 16 2.8 Assignment Page 175 Problems 5,6,31,32,33,34,35,36 Main Idea: The Forest is the important thing. The trees individually do not determine much. The vector space is the forest. The individual vectors do not determine much. Key Words: Subspace, closed under addition, closed under scalar multiplication, Null Space, Column Space, basis Goal: Express a vector space in terms of a basis. Understand that there will be many many choices of bases. Previous Assignment 1. Find a quaternion which rotates through 30 degrees about the axis vector <1,2,3>. 2. Find a 3x3 matrix which rotates through 30 degrees about the axis vector <1,2,3>. ans = { Cos[Pi/12], Sin[Pi/12]/Sqrt[14],Sin[Pi/12]*2/Sqrt[14],Sin[Pi/12]*3/Sqrt[14]}; n[{a_,b_,c_,d_}] := a^2+b^2+c^2+d^2; -------------------------------------------------------------------------- x[{a_,b_,c_,d_},{e_,f_,g_,h_}] := {a e-b f-c g-d h, +b e+a f-d g+c h, +c e+a g+d f-b h, +b g-c f+d e+a h}; conj[{a_,b_,c_,d_}]:= {a,-b,-c,-d}; inv[{a_,b_,c_,d_}] := 1/n[{a,b,c,d}]*conj[{a,b,c,d}]; p[{a_,b_,c_,d_}] := {b,c,d}; r[{a_,b_,c_,d_},{t_,xx_,y_,z_}] := x[x[{a,b,c,d},{t,xx,y,z}],inv[{a,b,c,d}]]; ---------------------------------------------------------------------------- This is the matrix which rotates elements 30 degrees about {1,2,3} | 0.875595 -0.381753 0.29597 | A = | 0.420031 0.904304 -0.0762129 | | -0.238552 0.191048 0.952152 | --------------------------------------------------------- A = {{ 0.875595, -0.381753, 0.29597 }, { 0.420031, 0.904304, -0.0762129}, {-0.238552, 0.191048, 0.952152 }}; f[{a_,b_,c_}] := ( H = Table[0,{i,1,36}]; H[[1]] = {a,b,c}; Do[H[[i+1]] = A.H[[i]],{i,1,35}]; K = Table[Point[H[[i]]],{i,1,35}]; Return[Show[Graphics3D[K]]]); W = Table[f[{2*i,2*i,2*i}],{i,-4,4}]; line = ParametricPlot3D[{x,x,x},{x,-8,8}]; p1 = Show[W,line,PlotLabel->"Rotate 30 degrees about {1,2,3}"]; Display["cone.ps",p1]; p2 = Show[W,ViewPoint->{6,8,10},PlotLabel->"View Point at {1,2,3}"]; Display["circle.ps",p2]; ---------------------------------------------------------- Discussion: We are working with stuff that has a lot of applications and possibilities. Every application has its own special terminology. So when you talk to somebody, expect them to be using the same basic properties that we are now learning. But they will call them by different names. So now I will give the "traditional" words. ------------------------------------------------------ A vector is a single column. That is, a nx1 matrix. A matrix has more than one column. We sometimes think of the matrix as consisting of columns. If V , V , ..., V are vectors and c , c , ..., c are numbers, 1 2 m 1 2 m then the vector v = c V + c V + ... + c V is called a 1 1 2 2 m m linear combination of V , V , ..., V with coefficients 1 2 m c , c , ..., c . 1 2 m ---------------------------------------------------------- *********************************************************** * AV really means a way of combining the columns of A * *********************************************************** Notice that if A is the matrix vith columns V , V , ..., V , 1 2 m | c | | 1 | A = [V V ... V ] and if V is the vector | c |, 1 2 m | 2 | | . | | . | | . | | c | | m | then AV is simply the linear combination c V + c V + ... c V . 1 1 2 2 m m Thus matrix multiplication by a matrix A is actually looking at linear combinations of the columns of A. We can solve AX = B if and only if B is a linear combination of the columns of A. The set of all vectors which are linear combinations of the columns of A is CS(A) = "the column space of A". This brings us to discuss the concept of a vector space. ************************************************************ * Keep your eye on the Subspace * ************************************************************ In linear algebra we have to describe things. Fortunately, there are good ways to describe the objects. Unfortunately, there are many good ways to describe objects. Since an object can have several distinctive names, we are often faced with the question of deciding if two given names refer to the same object. When ever we try to describe something in detail, we have to choose the way we present it. There is no good rule for making this choice. Everybody can have different looking presentations. The same correct answer can appear very different. What we have to do is try to focus on the item itself, not on how it was written. The item itself will be an object called a "vector space". n Definition of Vector Subspace of R . ====================================== n n A subset W of R is called a subspace of R if it has the following properties: n (a) W contains the zero vector in R (b) W is closed under addition. (* That is, if W and W are in W, then W +W is in W *) 1 2 1 2 (c) W is closed under scalar multiplication. (* If W is in W and k is any scalar, than kW is in W *) ----------------------------------------------------------- (* Common subspaces are the kernel and the image *) Theorem: If Arxs is any matrix, then s (1) ker(A) = {W in R | AW = 0} is a subspace. r (2) im(A) = {AW | W is in R } is a subspace. -------------------------------------------------- Proof that the kernel of A is a subspace. Part (a). A0 = 0 so 0 is in ker(A). Part (b). If W and W are in ker(A), 1 2 A(W +W ) = AW +AW = 0+0 = 0. 1 2 1 2 then AW = 0 and AW = 0. 1 2 Part (c). If W is in ker(A) and k is any scalar, then A(kW) = k(AW) = k0 = 0. Since ker(A) has 0, is closed under addition, and closed under scalar multiplication, ker(A) is a subspace. --------------------------------------------------- Proof of (2). Part (a). A0 = 0 so 0 is in im(A). Part (b). If AW and AW are elements in im(A), then 1 2 AW +AW = A(W +W ) is also in the image of A. 1 2 1 2 Part (c) If AW is in im(A) and k is any scalar, then k(AW) = A(kW) is also in the image of A. -------------------------------------------------------------- The span of a set of vectors is everything that they generate. Definition of Span: =================== n Consider the vectors V ,V , ..., V in R . The set of all 1 2 m linear combinations of the vectors V , V , ..., V is called 1 2 m their span; < V ,V , ..., V > = {c V +c V +...+c V | c are scalars} 1 2 m 1 1 2 2 m m i | 1 | | 0 | Example: What is the span of | 1 | and | 0 | ? | 0 | | 1 | | 1 | | 0 | | 0 | Example; What is the span of | 0 | | 1 | | 0 | ? | 0 |,| 0 |,| 1 | | 1 | | 1 | | 0 | Example: What is the span of | 0 | | 1 | | 1 | ? | 1 |, | 2 |, | 1 | | 1 1 0 | Equivalently, we can ask, what is the image of | 0 1 1 |? | 1 2 1 | | a | The span simply asks, For which | b | do there exist | c | x ,x ,x such that 1 2 3 | 1 | | 1 | | 0 | | a | x | 0 | + x | 1 | + x | 1 | = | b | 1 | 1 | 2 | 2 | 3 | 1 | | c | Solve this system by reducing it to Row Canonical Form. | 1 1 0 a | | 1 1 0 a | | 1 0 -1 a-b | | 0 1 1 b | | 0 1 1 b | | 0 1 1 b | | 1 2 1 c | | 0 1 1 -a+c | | 0 0 0 -a-b+c| The system is solvable when there is no stairstep one in the last column. This means it is solvable when a+b=c. So the span of these three vectors | a | is all vectors of the form | b |. Check that the |a+b| original vectors were of this form. To write this space in terms of a basis, separate on the coefficients. | a | | 1 | | 0 | | b | = a| 0 | + b | 1 | |a+b| | 1 | | 1 | | 1 | | 0 | The basis is: { | 0 | | 1 | }. | 1 |, | 1 | Spanning sets without unnecessary elements are called bases. Definition of Linear Independence; ================================== Consider a sequence v , v , ..., v of vectors in a subspace V 1 2 m n of R . The vectors v , v , ..., v are called linearly 1 2 m independent if none of them is a linear combination of the others. Otherwise, they are called linearly dependent. Definition of Basis =================== We say that the vectors v , v , ..., v form a basis of V if 1 2 m they span V and are linearly independent. Linear Relation =============== n Consider the vectors v ,v , ..., v in R . An equation of the 1 2 m form c V + c V + ... + c V = 0 1 1 2 2 m m is called a linear relation among the vectors V , V , ..., V . 1 2 m There is always the trivial relation, with c = c = ... = c = 0. 1 2 m Non trivial relations may or may not exist among the V , V , ..., V . 1 2 m Find a linear dependence relation among the vectors | 1 | | 1 | | 5 | | 1 | | 1 | | 2 | | 5 | | 4 | | 8 | | 1 | | 9 | | 1 | | 9 | | 1 | | 1 | | 1 |, | 5 |, | 1 |, | 5 |, | 1 | Solution: We solve for the coefficients x , x , x , x , x 1 2 3 4 5 such that | 1 | | 1 | | 5 | | 1 | | 1 | | 0 | x | 2 |+x | 5 |+x | 4 |+x | 8 |+x | 1 | = | 0 | 1| 9 | 1| 1 | 3| 9 | 4| 1 | 5| 1 | | 0 | | 1 | | 5 | | 1 | | 5 | | 1 | | 0 | | 1 1 5 1 1 0 | | 1 1 5 1 1 0 | | 1 1 5 1 1 0 | | 2 5 4 8 1 0 | | 0 3 -6 6 -1 0 | | 0 3 -6 6 -1 0 | | 9 1 9 1 1 0 | | 0 -8-36 -8 -8 0 | | 0 -8-36 -8 -8 0 | | 1 5 1 5 1 0 | | 0 4 -4 4 0 0 | | 0 1 -1 1 0 0 | | 1 1 5 1 1 0 | | 1 0 6 0 1 0 | | 1 0 6 0 1 0 | | 0 1 -1 1 0 0 | | 0 1 -1 1 0 0 | | 0 1 -1 1 0 0 | | 0 3 -6 6 -1 0 | | 0 0 -3 3 -1 0 | | 0 0 1 -1 1/3 0 | | 0 -8-36 -8 -8 0 | | 0 0-44 0 -8 0 | | 0 0-44 0 -8 0 | | 1 0 0 6 -1 0| |1 0 0 6 -1 0 | | 1 0 0 6-3/33 0 | | 0 1 0 0 1/3 0| |0 1 0 0 1/3 0 | | 0 1 0 0 1/3 0 | | 0 0 1 -1 1/3 0| |0 0 1 -1 1/3 0 | | 0 0 1 -1 1/3 0 | | 0 0 0-44 20/3 0| |0 0 0 11 -5/3 0 | | 0 0 0 1-5/33 0 | x x x x x =a RHS 1 2 3 4 5 | 1 0 0 6 -1 0 ||1 0 0 0-3/33 0||1 0 0 0 -3/33 0 | | 0 1 0 0 1/3 0 ||0 1 0 0 1/3 0||0 1 0 0 11/33 0 | | 0 0 1 0 6/33 0 ||0 0 1 0 6/33 0||0 0 1 0 6/33 0 | | 0 0 0 1-5/33 0 ||0 0 0 1-5/33 0||0 0 0 1 -5/33 0 | | x1 | | 3 | | x2 | |-11 | | x3 | = a/33 | -6 | | x4 | | 5 | | x5 | | 33 | | 1 | | 1 | | 5 | | 1 | | 1 | | 0 | 3| 2 |-11| 5 |- 6| 4 |+5 | 8 |+33| 1 | = | 0 | | 9 | | 1 | | 9 | | 1 | | 1 | | 0 | | 1 | | 5 | | 1 | | 5 | | 1 | | 0 | | 3 -11 - 30 + 5 + 33 | | 0 | | 6 -55 - 24 +40 + 33 | = | 0 | | 27 -11 - 54 + 5 + 33 | | 0 | | 3 -55 - 6 +25 + 33 | | 0 | A dependence between the columns of a matrix A is found by solving AX = 0. A basis for the column space of A will be those columns of A which become stair step one columns in the RCF of A.