This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ndimensional 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 twodimensional 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 ...
View
Full Document
 Spring '07
 GHITZA
 Linear Algebra, Algebra

Click to edit the document details