Linear Algebra with Applications (3rd Edition)

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Unformatted text preview: Math 2.53, Fall 2W1, Exam #2 sense E“ 3. 4. 5. Exam # 2 True or False? If four vectors span ill4 , then they must form a basis of 111‘ There exists a matrix A such that imiA) = keriA}. There exists a noninvertible 3 x3 matrix A that is similar to the identity matrix I, . IfA isa 9x9 matrix suchthat A: =fl,thentheinequaiitv rank{A)54 must hold. If hvo 2x 2 matrices have the same kernel, then they must have the same reduced row echelon form (rref‘j. True or False? if Vand W are both n-dimensional linear spaces and T is a linear transfonnation from Vto W, then Tmust be an isomorphism. 1 If Vis the set of all 2 x 2 matrioesA such that the vector 2 is in the kernel of A, then Visasubspaoe of Rm. The function T{f(x)}=f{5) isalineartransformationfi'om P to ill. Asusuai,P denoted the space of all poiynomials. a+c b+c Ihelineartransformation T(a+hx+cxi)= +b +b+ a a c from P; to Rm is an isomombism. The exists a two-dimensional subspace of Rm whose nonzero elements are invertible. Let T be the orthogonal projection onto the plane 2x+2y+ s = t] in llitJ . Find the matrix B of T with respect to a basis 53 of litJ of your choice. {Your answer will involve basis 53 and matrix B). Let Vbe the set of‘all polynomials fix) in P; such that ffl): f’fl) = i}. We are told that Via a subspace of P‘3 . Find a basis of V and determine the dimension of V. i] 1 Consider the fimction T (M) = in {1 i] U Rid—ML fllfi'omllim to km. a. Find the matrixBof‘Twith respectto the standard basis 93 of Rm , b. Find a basis of the kernel of T. Math 253. Fall lflfl 1, Exam it 2, Solutions 2 4. Step 1 :Write the general element of the ambient space, F; , involving some arbitrary,r constants: f{1]= or—i~.r5.:c+cr2 +0313. Step 1: Plug in the conditiouts) that define the subspace { f {1) = I], fail} = D in our case} and solve the ensuing system of linear equations: f"{x}=s+2cx+3ar’ , f(l)=o+o+c+d=i} f’fl}: b+2c+3d =3. e — c — 2d = fl This s stem reduces to , F a+k+M=fl withsolution o=c+2d, b =—2c—3d. Step 3: Use your answers in Steps 1 and 2 to write the general element of the su hspace: fix} = [c -+- 2d} + [-2c — 3d): + of + {11.13 . Step 4: Write your answer in Step 3 as a linear combination; use the arbitrary constants {c and d in our case) as the coefficients. f(x)=c[:-2x+x1)+d(2—3x+x3]. The fimotions l—2x+.r1 and 2—3x+x3 form a basis of V {check linear independencei). Answer: l—Ex—l—xz, 2—3x—l—x], so that dim[V} = 2. There are other hases. Your answer is right as long as {a} you get two fimotions, (b) they are linearly,r hidependem, a11r;i{c}1:he§«r satisf}r the tart: defining properties of I”, ffl) = l} and fl{i}=fl. D -2 l D 5. a. Use the familiar commutative diagram to find 3: ii 2 E j: . iii 0 ll i1 1 D . . U 1 . . b. Looking for relations among the columns of B we see that u , 2 is a basis of the {1 CI 1 {i i] 1 , _ , kernel of B, so that , 1s a bases of the kernel of T. 0 fl 2 U i} 1 1 G c. The image, or column space, of H has the basis [J , so that the matrices ‘—2 ...
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