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iastate.eduhttp://carolannpeacock.com/hentzel/class.307.05Text: direct Algebra and also its Applications, 3rd Edition David C. LayWednesday, march 9 section 4.4Main Idea: If you space not confused, you are not paying attention. An essential Words: V = B V = C V B C If T(V) = W then T V = W BB B B Goal: learn to adjust V to V and T come T B C BB CC===============================================================Previous AssignmentPage 243 trouble 11, 21, 22, 38Page 243 problem 11 3Find a basis for the set of vectors in R in the planex+2y+z = 0. X y=a z=b | 1 2 1 | | x | | -2 | | -1 | | y | = a | 1 | + b | 0 | | z | | 0 | | 1 | communication of the space.--------------------------------------------------------a = Plot3DFalse>;b = Graphics3D,Line< 0, 0, 0,-2,1,0>>; c = Graphics3D,Line< 4,-2, 0,-4,2,0>>; d = Graphics3D,Line< 0, 0, 0,-1,0,1>>; e = Graphics3D,Line< 3, 0,-3,-3,0,3>>; f = Show"P243 P11 x+2y+z = 0">;Display<"11.ps",f>;--------------------------------------------------------------Page 243 difficulty 21(a) A single vector by itself is linearly dependent.Yes if the vector is zero. No if the vector is no zero. (b) If H = , climate B1,B2, ..., Bp isa basis because that H.True only if B1, B2, ..., Bp are additionally linearly independent. N(c) The columns of an invertible nxn matrix kind a basis because that R .True.(d) A basis is a spanning set that is as huge as possible.False: A communication is in reality a spanning collection that is as small as possible.(e) In part cases, the linear dependence relations among the columns that a matrix have the right to be influenced by certain elementary row operations ~ above the matrix.False. ------------------------------------------------------------Page 243 trouble 22(a) A linearly independent set in a subspace H is a basis for H.Not true unless the set is a spanning collection of H.(b) If a finite set S that nonzero vectors spans a vector an are V,then some subset of S is a basis for V.True.(c) A basis is a linearly independent set that is as big as possible.True.(d) The standard technique for producing a spanning set for the Null space of A, defined in section 4.2 sometimes stops working to producea basis for the Null space of A.When the Null room of A is the zero space, it has actually no basis.When the Null room is no zero, the technique works.(e) If B is one echelon kind of a matrix A, climate the pivotcolumns that B form a basis because that the Column space of A.True.---------------------------------------------------------Page 243 trouble 38Show the 1, Cos , Cos^2 , ... Cos^6 is a linear independent collection of functions characterized 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 contends most 6 roots. But b,c,d,e,f,g space obviously roots. Hence if a isdistinct from the rest, the determinant does notevaluate to zero as soon as x = a.Replacing a,b,c,d,e,f,g by unique values of thecosine, we uncover that no linear mix of thepowers that cosines have the right to be zero because such a linearcombination would be in the null space of one invertiblematrix.-------------------------------------------------------New Material: ns really doubt that anybody deserve to read the product in thebook and even remotely recognize what the author is talkingabout. It is the very same for all books. It does not get anyeasier. I am presenting the product as i think the is the most basic to learn. This is essentially what ns think the bookis attempting come teach. Please offer the material your bestshot. It really is important. Hopefully, friend will have the ability to understand what is walking on. Then shot explaining the tosomeone else.============================================================Theory: suppose we have actually a basis B1 B2 ... BnThen every vector can be expressed as a linear mix ofthat basis as V = x1 B1 + x2 B2 + ... + xn Bn.We write |x1| |x2| |x3| V = | .| | .| |xn|---------------------------------------------------- | 1 | |1| |1| |0| Example. Refer | 2 | in the communication |0| |0| |1|. | 3 | |0|,|1|,|0| the is, uncover the c1, c2, and also c3 such the the complying with 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. Refer | 2 | in the basis |1| |1| |0|. | 3 | |1|,|0|,|1| the is, find the x1, x2, and x3 such that the complying with 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 = V = < C1 C2 C3> V B C us say the the depiction of V in the basis B is V BWe say that the representation of V in the communication C is V C | 1 | | -2 | | 2 |V = | 2 | V = | 3 | V = | 0 | | 3 | B | 2 | C | 1 |-----------------------------------------------------The simple thing come remember is V = V B CSo if you desire to transform from V come V B C -1You can solve for V = V B CThis method we can convert from the C representation of a vectorto the B representation of a vector by multiply by the procession -1 . We call this matrix the readjust of basis matrix P BC notification V = p V B BC C ------------------------------------------------------------Example. Uncover the procession P which changes the depiction BC | 1 0 1 | | 0 1 0 |in the communication B = | 0 0 1 | to C = | 1 0 0 | | 1 1 0 | | 0 0 1 | | 2 |If V = | 3 | what is V ? C | 4 | BAlways start V = V C B -1V = 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 = ns V B BC C |-1 1 0 || 2 | | 1 |V = | 1 -1 1 || 3 | = | 3 | B | 1 0 0 || 4 | | 2 | C B inspect V = V = V B C | 1 0 1 ||1| | 3| | 0 1 0 ||2| |3| V = | 0 0 1 ||3| = | 2| V = | 1 0 0 ||3| =|2| B | 1 1 0 ||2| | 4| C | 0 0 1 ||4| |4| same...........................same so the checks.==================================================================Matrix that a linear revolution with respect to a communication B. Placed the communication B across the top and also down the left hand side. B1 B2 ... Bn --------------------------------------B1 | | | |B2 | | | |. | |. | |. | |Bn | | --------------------------------------- Now apply the change to the left hand column and also write the calculation in terms of the communication Bi throughout the top.The transpose the the matrix is the procession of the lineartransformation through respect come the B basis.Example: because that the linear revolution | 1 2 3 | T = | 0 1 2 | | 1 0 0 | | 1 0 1 |write the procession of T through 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 come the basis B is | 1 0 1 |. BB | 0 2 1 |//////////////////////////////////////////////////////////////////////The following should do sense. That is no standard, not since theideas are wrong, but since it is excellent from the an easy level of understanding. (* If you know where a basis goes, you recognize where whatever else goes *)(* If you recognize the price of the separation, personal, instance items, girlfriend can find the price *)(* the a totality bag that groceries. *)(* If you recognize the calorie in one ounce of the all set foods, you can *)(* number the full calories ~ above the meal. *) expect a linear revolution T sends a communication come .Then linear combinations will follow along in the same manner. Hence for any X,the linear combination X will be sent out to X. X --------------> X The idea is the if you understand where the elements of a basis are sent, youcan figure where all the elements must be sent. Suppose that I understand | 1 | ------------> | 2 | | 1 | | 1 | and also | 1 | ------------> | 0 | |-1 | | 1 |Where does | 3 | go? | 4 |Where walk | 1 | go? | 0 |Where walk | 0 | go? | 1 |To find where an element goes, I have actually to very first express that in termsof the elements whose location I know. | 3 | = | 1 1 | | 7/2 | | 4 | | 1-1 | | -1/2 | V B V BNow use V to obtain the photo 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 X X image | 1 | = |1 1| | 1/2 | ---------> | 2 0 | |1/2| = |1|| 0 | |1 -1| | 1/2 | | 1 1 | |1/2| |1| V X X picture | 0 | = |1 1| | 1/2 | ----------> | 2 0 | | 1/2| = | 1 || 1 | |1 -1| |-1/2 | | 1 1 | |-1/2| | 0 | V X X picture ---------------------------------------------------------Suppose we know T just how do we get T ? BB CC We know T V = W BB B B We know T V = W CC C CWe i think we have actually a linear change T. However it deserve to havevarious matrix representations, relying on what basis you useto stand for the vectors. Ns T ns V = W CB BB BC C C Translation, start with V . Adjust it come V . C BFind whereby V go in the B basis. Contact the picture BW . Now change the vector W ago to the C basis.

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B BWe have the right to mechanize this process using a readjust of communication matrix. For this reason T = ns T p CC CB BB BC==========================================================