Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:50 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, March 9 Section 4.4 Main Idea: If you are not confused, you are not paying attention. Key Words: V = B V = C V B C If T(V) = W then T V = W BB B B Goal: Learn to change V to V and T to T B C BB CC =============================================================== Previous Assignment Page 243 Problem 11, 21, 22, 38 Page 243 Problem 11 3 Find a basis for the set of vectors in R in the plane x+2y+z = 0. x y=a z=b | 1 2 1 | | x | | -2 | | -1 | | y | = a | 1 | + b | 0 | | z | | 0 | | 1 | <-----------------> Basis of the space. -------------------------------------------------------- a = Plot3D[x+2y,{x,-5,5},{y,-5,5},Shading->False]; b = Graphics3D[{Thickness[0.02 ],Line[{{ 0, 0, 0},{-2,1,0}}]}]; c = Graphics3D[{Thickness[0.009],Line[{{ 4,-2, 0},{-4,2,0}}]}]; d = Graphics3D[{Thickness[0.02 ],Line[{{ 0, 0, 0},{-1,0,1}}]}]; e = Graphics3D[{Thickness[0.009],Line[{{ 3, 0,-3},{-3,0,3}}]}]; f = Show[a,b,c,d,e,PlotLabel->"P243 P11 x+2y+z = 0"]; Display["11.ps",f]; -------------------------------------------------------------- Page 243 Problem 21 (a) A single vector by itself is linearly dependent. Yes if the vector is zero. No if the vector is not zero. (b) If H = , then { B1,B2, ..., Bp} is a basis for H. True only if B1, B2, ..., Bp are also linearly independent. n (c) The columns of an invertible nxn matrix form a basis for R . True. (d) A basis is a spanning set that is as large as possible. False: A basis is actually a spanning set that is as SMALL as possible. (e) In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False. ------------------------------------------------------------ Page 243 Problem 22 (a) A linearly independent set in a subspace H is a basis for H. Not true unless the set is a spanning set of H. (b) If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V. True. (c) A basis is a linearly independent set that is as large as possible. True. (d) The standard method for producing a spanning set for the Null Space of A, described in section 4.2 sometimes fails to produce a basis for the Null space of A. When the Null Space of A is the zero space, it has no basis. When the Null Space is not zero, the method works. (e) If B is an echelon form of a matrix A, then the pivot columns of B form a basis for the Column Space of A. True. --------------------------------------------------------- Page 243 Problem 38 Show that {1, Cos [t], Cos^2 [t], ... Cos^6 [t] } is a linear independent set of functions defined on R. If a,b,c,d,e,f,g are distinct then | 1 a a^2 a^3 a^4 a^5 a^6 | | 1 b b^2 b^3 b^4 b^5 b^6 | | 1 c c^2 c^3 c^4 c^5 c^6 | Det | 1 d d^2 d^3 d^4 d^5 d^6 | =/= 0. | 1 e e^2 e^3 e^4 e^5 e^6 | | 1 f f^2 f^3 f^4 f^5 f^6 | | 1 g g^2 g^3 g^4 g^5 g^6 | Proof: | 1 x x^2 x^3 x^4 x^5 x^6 | | 1 b b^2 b^3 b^4 b^5 b^6 | Det | 1 c c^2 c^3 c^4 c^5 c^6 | | 1 d d^2 d^3 d^4 d^5 d^6 | | 1 e e^2 e^3 e^4 e^5 e^6 | | 1 f f^2 f^3 f^4 f^5 f^6 | | 1 g g^2 g^3 g^4 g^5 g^6 | is a polynomial of degree 6. It has at most six roots. But b,c,d,e,f,g are obviously roots. Thus if a is distinct from the rest, the determinant does not evaluate to zero when x = a. Replacing a,b,c,d,e,f,g by distinct values of the cosine, we find that no linear combination of the powers of cosines can be zero because such a linear combination would be in the null space of an invertible matrix. ------------------------------------------------------- New Material: I really doubt that anybody can read the material in the book and even remotely understand what the author is talking about. It is the same for all books. It does not get any easier. I am presenting the material as I think it is the easiest to learn. This is essentially what I think the book is attempting to teach. Please give the material your best shot. It really is important. Hopefully, you will be able to understand what is going on. Then try explaining it to someone else. ============================================================ Theory: Suppose we have a basis B1 B2 ... Bn Then every vector can be expressed as a linear combination of that basis as V = x1 B1 + x2 B2 + ... + xn Bn. We write |x1| |x2| |x3| V = [B1 B2 ... Bn] | .| | .| |xn| ---------------------------------------------------- | 1 | |1| |1| |0| Example. Express | 2 | in the basis |0| |0| |1|. | 3 | |0|,|1|,|0| That is, find the c1, c2, and c3 such that the following holds. | 1 | | 1 1 0 | | x1 | | 2 | = | 0 0 1 | | x2 | | 3 | | 0 1 0 | | x3 | answer | x1 | = |-2| | x2 | = | 3| | x3 | = | 2| --------------------------------------------------------- | 1 | |0| |0| |1| Example. Express | 2 | in the basis |1| |1| |0|. | 3 | |1|,|0|,|1| That is, find the x1, x2, and x3 such that the following holds. | 1 | | 0 0 1 | | x1 | | 2 | = | 1 1 0 | | x2 | | 3 | | 1 0 1 | | x3 | answer | x1 | = | 2| | x2 | = | 0| | x3 | = | 1| --------------------------------------------------------- If V = [B1 B2 B3] V = [ C1 C2 C3] V B C <---B----> <----C----> We say that the representation of V in the basis B is V B We say that the representation of V in the basis C is V C | 1 | | -2 | | 2 | V = | 2 | V = | 3 | V = | 0 | | 3 | B | 2 | C | 1 | ----------------------------------------------------- The basic thing to remember is [B] V = [C] V B C So if you want to convert from V to V B C -1 You can solve for V = [B] [C] V B C This means we can convert from the C representation of a vector to the B representation of a vector by multiplying by the matrix -1 [B] [C]. We call this matrix the change of basis matrix P BC Notice V = P V B BC C ------------------------------------------------------------ Example. Find the matrix P which changes the representation BC | 1 0 1 | | 0 1 0 | in the basis B = | 0 0 1 | to C = | 1 0 0 | | 1 1 0 | | 0 0 1 | | 2 | If V = | 3 | what is V ? C | 4 | B Always start [C] V = [B] V C B -1 V = [B] [C] V B C -1 | 1 0 1 | | 0 1 0 | P = | 0 0 1 | | 1 0 0 | BC | 1 1 0 | | 0 0 1 | | 1-1 0| | 0 1 0 | = |-1 1 1| | 1 0 0 | | 0 1 0| | 0 0 1 | |-1 1 0 | = | 1 -1 1 | | 1 0 0 | V = P V B BC C |-1 1 0 || 2 | | 1 | V = | 1 -1 1 || 3 | = | 3 | B | 1 0 0 || 4 | | 2 | C B Check V = [B] V = [C] V B C | 1 0 1 ||1| | 3| | 0 1 0 ||2| |3| [B] V = | 0 0 1 ||3| = | 2| [C] V = | 1 0 0 ||3| =|2| B | 1 1 0 ||2| | 4| C | 0 0 1 ||4| |4| same...........................same so it checks. ================================================================== Matrix of a linear transformation with respect to a basis B. put the basis B across the top and down the left hand side. B1 B2 ... Bn -------------------------------------- B1 | | | | B2 | | | | . | | . | | . | | Bn | | --------------------------------------- Now apply the transformation to the left hand column and write the output in terms of the basis Bi across the top. The transpose of the matrix is the matrix of the linear transformation with respect to the B basis. Example: for the linear transformation | 1 2 3 | T = | 0 1 2 | | 1 0 0 | | 1 0 1 | write the matrix of T with respect to the basis | 0 0 1 | | 0 1 0 | 1 0 1 0 0 1 0 1 0 ----------------------------------- 1 1 | | 0 --> 0 | 1 1 0 | 0 1 | | | | 0 3 | | 0 --> 2 | 1 0 2 | 1 0 | | | | 1 3 | | 1 --> 1 | 2 1 1 | 0 1 | | ----------------------------------- | 1 1 2 | The matrix of T with respect to the basis B is | 1 0 1 |. BB | 0 2 1 | ////////////////////////////////////////////////////////////////////// The following should make sense. It is non standard, not because the ideas are wrong, but because it is done from the basic level of understanding. (* If you know where a basis goes, you know where everything else goes *) (* If you know the prices of the individual items, you can find the price *) (* of a whole bag of groceries. *) (* If you know the calories in one ounce of the prepared foods, you can *) (* figure the total calories on the meal. *) Suppose a linear transformation T sends a basis [B1 B2 ... Bn] to [C1 C2 ... Cn]. Then linear combinations will follow along in the same manner. Thus for any X, the linear combination [B1 B2 ... Bn] X will be sent to [C1 C2 ... Cn ] X. [B1 B2 ... Bn] X --------------> [C1 C2 ... Cn] X The idea is that if you know where the elements of a basis are sent, you can figure where all the elements must be sent. Suppose that I know | 1 | ------------> | 2 | | 1 | | 1 | and | 1 | ------------> | 0 | |-1 | | 1 | Where does | 3 | go? | 4 | Where does | 1 | go? | 0 | Where does | 0 | go? | 1 | To find where an element goes, I have to first express it in terms of the elements whose destination I know. | 3 | = | 1 1 | | 7/2 | | 4 | | 1-1 | | -1/2 | V B V B Now use V to get the image of | 3 | . B | 4 | | 3 | = | 1 1 || 7/2 | -----T-----> | 2 0 | | 7/2 | = | 7 | | 4 | | 1 -1 ||-1/2 | | 1 1 | |-1/2 | | 3 | V [B] X [C] X image | 1 | = |1 1| | 1/2 | ---------> | 2 0 | |1/2| = |1| | 0 | |1 -1| | 1/2 | | 1 1 | |1/2| |1| V [B] X [C] X image | 0 | = |1 1| | 1/2 | ----------> | 2 0 | | 1/2| = | 1 | | 1 | |1 -1| |-1/2 | | 1 1 | |-1/2| | 0 | V [B] X [C] X image --------------------------------------------------------- Suppose we know T How do we get T ? BB CC We know T V = W BB B B We know T V = W CC C C We assume we have a linear transformation T. But it can have various matrix representations, depending on what basis you use to represent the vectors. P T P V = W CB BB BC C C Translation, Start with V . Change it to V . C B Find where V goes in the B basis. Call the image B W . Now change the vector W back to the C basis. B B We can mechanize this process using a change of basis matrix. So T = P T P CC CB BB BC ===========================================================