# Math 2270 Spring 2004 Homework 11 Solutions ```Math 2270 Spring 2004
Homework 11 Solutions
Section 4.2) 2, 5, 6, 10, 13, 14, 26, 27
(2) T (A) = 7A from ℜ2x2 to ℜ2x2 . Let A and B be 2 x 2 matrices and let k be a scalar.
Then, by the rules of matrix multiplication outlined in chapters 1 and 2 it follows that
T (A + B) = 7(A + B) = 7A + 7B = T (A) + T (B)
T (kA) = 7(kA) = k(7A) = kT (A)
so T is linear. To determine if T is an isomorphism, we look for its inverse. We see that
if T (A) = 7A = B, then T −1 ( 71 B) = A, so T is invertible with inverse transformation
T −1 (B) = 17 B. Thus, T is an isomorphism.
(5) T (A) = A2 from ℜ2x2 to ℜ2x2 . Let A be a 2 x 2 matrix and let k be a scalar. Then, by
the rules of matrix multiplication outlined in chapters 1 and 2 it follows that
T (kA) = (kA)2 = kAkA = k(kA2 ) = k(kT (A)) 6= kT (A)
so T is not linear.
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1 2
1 2
(6, 26) T (A) =
A from ℜ2x2 to ℜ2x2 . Define C =
, so T (A) = CA. Let A
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and B be 2 x 2 matrices and let k be a scalar. Then, by the rules of matrix multiplication
outlined in chapters 1 and 2 it follows that
T (A + B) = C(A + B) = CA + CB = T (A) + T (B)
T (kA) = C(kA) = k(CA) = kT (A)
so T is linear. We must determine if T is an isomorphism. Looking at the matrix C we
see that column two is two times column one, so the columns are not linearly independent.
Therefore, C is not invertible and there are infinitely many solutions, ~x ∈ ℜ2 , to the equation
C~x = ~0. Thus, there are infinitely many 2 x 2 matrices A such that CA = 0. (For example,
let A = [~x ~x] have columns in the kernel of C.) Therefore, the kernel of T contains more
than the zero matrix violating Fact 4.2.3 for isomorphisms. Thus, T is not an isomorphism.
To find the exact kernel of T we must solve
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1 2
3 6
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a b
c d
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=
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a + 2c b + 2d
3a + 6c 3b + 6d
1
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=
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0 0
0 0
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Math 2270 Spring 2004
Homework 11 Solutions
This reduces to two equations and four unknowns: a = −2c and b = −2d. Therefore,
matrices in the kernel have the general form
c
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−2 0
1 0
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+d
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0 −2
0 1
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and the nullity is two.
(10) T (x + iy) = x − iy from C to C. Let f = a + bi and g = c + di be two complex numbers
and let k be a scalar. Then,
T (a + bi + c + di) = T ((a + c) + i(b + d)) = (a + c) − i(b + d) =
a + c − ib − id = (a − ib) + (c − id) = T (a + ib) + T (c + id)
T (k(a + bi)) = T (ak + bki) = ak − bki = k(a − bi) = kT (a + bi)
showing that T is linear. Notice that T (T (a + ib)) = T (a − ib) = a + ib, so T is its own
inverse. Therefore, T is an isomorphism.
(13, 27) T (f ) = f ′′ + 4f ′ from P2 to P2 . Let f and g be polynomials in P2 and let k be a
scalar. Then, by the properties of derivatives
T (f + g) = (f + g)′′ + 4(f + g)′ = f ′′ + g ′′ + 4f ′ + 4g ′ = T (f ) + T (g)
T (kf ) = (kf )′′ + 4(kf )′ = kf ′′ + 4kf ′ = k(f ′′ + 4f ′) = kT (f )
so T is linear. We must determine if T is invertible. We first check to see that the image
of T is all of P2 . Take the function g(t) = t2 . We must determine if there is a polynomial
f (t) = a + bt + ct2 that solves the equation f ′′ + 4f ′ = g, but f ′′ + 4f ′ = 2c + 4(b + 2ct) =
2c + 4b + 8ct. We see that the terms of power 2 always disappear, so the image of T will not
be all of P2 . Therefore, by Fact 4.2.3, T is not an isomorphism.
We must find the image of T , or all polynomials g in P2 such that there is a solution
polynomial f (also in P2 ) to the equation f ′′ + 4f ′ = g. For general f = a + bt + ct2 , we
have f ′′ + 4f ′ = 2c + 4b + 8ct = 2c(1 + 4t) + 4b. Therefore, all elements of the image of T
are linear combinations of {1 + 4t, 1}, which is a basis for the image. Notice that this space
is just P1 , the set of all polynomials of degree less than or equal to one. It follows that the
rank of T is 2.
We must find the kernel of T , or all solutions to the equation f ′′ + 4f = 0. For general
f = a + bt + ct2 , we have f ′′ + 4f ′ = 2c + 4b + 8ct = 0. Looking at the coefficients, we see
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Math 2270 Spring 2004
Homework 11 Solutions
that 8c = 0 and 2c + 4b = 0. By the first equation c = 0. It follows that b = 0. However,
there are no restrictions on a. Therefore, elements of the kernel are linear combinations of
{1}. It follows that the nullity of T is 1.
(14) T (f (t)) = f (−t) from P2 to P2 , or T (a + bt + ct2 ) = a − bt + ct2 . Let a + bt + ct2 and
d + et + f t2 be general polynomials in P2 and let k be a scalar. Then,
T (a + bt + ct2 + d + et + f t2 ) = a − bt + ct2 + d − et + f t2 = T (a + bt + ct2 ) + T (d + et + f t2 )
T (k(a + bt + ct2 )) = T (ak + bkt + ckt2 ) = ak − bkt + ckt2 = k(a − bt + ct2 ) = kT (a + bt + ct2 )
so T is linear. Notice that T (T (a + bt + ct2 )) = T (a − bt + ct2 ) = a + bt + ct2 , so T is its own
inverse. Therefore, T is an isomorphism.
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