Preparation problems for the discussion sections on October 7th and 9th1. Determine a basis for each of the following subspaces:(i) H = {4s−3s−t: s, t ∈ R},(ii) K = {abcd: a − 3b + c = 0},(iii) Col (1 2 3 0 00 0 1 0 10 0 0 1 0),(iv) Nul(1 2 3 0 00 0 1 0 10 0 0 1 0).Solution:(i): Every vector in H is of them form4s−3st= s4−30+ t001.SoH = span{4−30,001}.Since these two vectors are linearly independent (they are not multiples of each other),{4−30,001} is a basis of H.(ii) Every vector in K is of them formabcd=3b − cbcd= b3100+ c−1010+ d0001.Thus,K = span({3100,−1010,0001}).I will leave it to you to check that these vectors are linearly independent and hence forma basis for K.(iii) The matrixA :=1 2 3 0 00 0 1 0 10 0 0 1 012is already of echelon form. Hence its pivot columns form a basis of Col(A). Hence{100,310,001}is basis of Col(A).(iv) Let A be as above. We bring A to reduced echelon form:1 2 3 0 00 0 1 0 10 0 0 1 0R1→R1−3R2−−−−−−−→1 2 0 0 −30 0 1 0 10 0 0 1 0Hence the free variables of Ax = 0 are x2, x5. We have Ax = 0 iff (that is, if and only if)x1= −2x2+ 3x5x3= −x5x4= 0.Hence every vector v ∈ Nul(A) is of the form−2x2+ 3x5x2−x50x5= x2−21000+ x530−101.Hence{−21000,−30−101}form a basis of Nul(A).[Note: if you transform A into reduced echelon form and write Nul(A) as span of coef-ficients of free variables (after substituting dependent variables in terms of free variables)then those vectors always will be linearly independent. So here there is no need to checklinear independence, since we know they are in fact linearly independent.]2. Determine the dimension of N ul(A) and Col(A), whereA :=1 2 3 −4 81 2 0 2 82 4 −3 10 93 6 0 6 9.Solution: In order to determine the dimension of the two subspaces, we just have todetermine the number of pivot columns and free variables of A. So we bring A to echelon3form:1 2 3 −4 81 2 0 2 82 4 −3 10 93 6 0 6 9R2→R2−R1,R3→R3−2R1,R4→R4−3R1−−−−−−−−−−−−−−−−−−−−−−−→1 2 3 −4 80 0 −3 6 00 0 −9 18 −70 0 −9 18 −15R3→R3−3R2,R4→R4−3R2−−−−−−−−−−−−−−−→1 2 3 −4 80 0 −3 6 00 0 0 0 −70 0 0 0 −15R4→R4−(15/7)R2−−−−−−−−−−→1 2 3 −4 80 0 −3 6 00 0 0 0 −70 0 0 0 0Hence the echelon form of A has three pivots columns and two non-pivot columns. Hencedim Col(A) =number of pivot columns = 3 and dim Nul(A) =number of free variables= 2.3. Let A, B be two 4 × 3 matrices. Let a1, a2, a3be the columns of A and let b1, b2, b3bethe columns of B.(i) Suppose that {a1, a2, a3} is linearly independent. Find a basis for Col(A) anddescribe Nul(A).(ii) Suppose that {b1, b2} is linearly independent and b3= 2b1+ 7b2. Find a basis forCol (B) and a basis for Nul(B).Solution: (i) {a1, a2, a3} is a basis of Col(A), because it spans Col(A) (they are thecolumns of A!) and they are linearly independent (by assumption). Therefore, the rankof A is 3 and Ax = 0 has only the trivial solution. Hence Nul(A) = {000}.(ii) {b1, b2} is a basis of Col(B): by assumption it is linearly independent and b3=2b1+ 7b2, so we can show each column of B as a linear combination of b1and b2.Hence,we have span({b1, b2}) = Col(B). Now letx1x2x3be in Nul(B). Then000= Bx1x2x3= [b1b2b3]x1x2x3= x1b1+ x2b2+ x3b3.Since b3= 2b1+ 7b2, this happens iff000= x1b1+ x2b2+ x3(2b1+ 7b2) = (x1+ 2x3)b1+ (x2+ 7x3)b2.Since {b1, b2} is linearly independent, the equation y1b1+ y2b2= 0 has only the solutiony1= y2= 0. Hence the above equation gives0 = x1+ 2x3= x2+ 7x3.4Hence every vector in Nul(B) is of the form−2x3−7x3x3= x3−2−71.Hence {−2−71} is a basis of Nul(B).[Here is a quicker way to do that: since the dimension of the column space and the nullspace add up to 3 (each column of B has to correspond to a free variable or contain apivot), we know that dim N ul(B) = 1. Elements of Nul(B) correspond to linear relationsbetween the columns of B. The fact that 2b1+ 7b2− b3= 0, means that [2, 7, −1]Tis inNul(B). Since the dimension is 1, this vector is a basis.]4. Let u1=11, u2=1−1and let B = {u1, u2}.(i) Let v =23. Express v in terms of the basis B.(ii) Let w =11. Express w in terms of the basis B.(iii) Let T : R2→ R2be defined such that T (v) is expressing v in terms of the basisB. (Convince yourself that this is a linear transformation.) Determine the matrixthat represents T with respect to the standard basis of R2.Solution: (i) We need to solve1 1 21 −1 3R2→R2−R1−−−−−−−→1 1 20 −2 1R2→R2/(−2)−−−−−−−→1 1 20 1 −.5R1→R1−R2−−−−−−−→1 0 2.50 1 −.5.Hence23= 2.511− .51−1.Hence v in terms of the basis B is2.5−.5.(ii) In contrast to the first part (but you can certainly proceed that way, if you don’tsee it), it is obvious that11= 111+ 01−1.Hence w in terms of the basis B is10.(iii) Since10= .511+ .51−1,5we haveT (10) =.5.5.Since01= .511− .51−1,we haveT (01) =.5−.5.Since.5.5and.5−.5in terms of the standard basis are.5.5and.5−.5, respectively,the matrix that represents T with respect to the standard basis of R2isA =.5 .5.5 −.5.A remark. Since A is the matrix that represents T with respect to the standard basisof R2, we haveT (v) = Av.For example,T (23) =.5 .5.5 −.523=2.5−.5.Hence once you have calculated the matrix A, you have a simple way to express any vectorv in terms of the basis B.5. Let L: R2→ R3be a
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