Download 10. Is the set of 3 × 3 matrices whose kernel contains 1 2 3 a

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10. Is the set of 3 × 3 matrices whose kernel contains 2 a subspace of
3
R3×3 ?
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We will denote 2 by w. The set of 3 × 3 matrices whose kernel contains
3
w will be denoted V . Since 0w = 0, we have 0 ∈ V .
Suppose A ∈ V and r ∈ R. Then w ∈ ker A so Aw = 0. Hence (rA)w =
r(Aw) = 0 so w ∈ ker rA and rA ∈ V . Thus V is closed under scalar multiplication.
Now suppose that A, B ∈ V . Then w ∈ ker A so Aw = 0 and similarly
Bw = 0. Then (A + B)w = Aw + Bw = 0 so w ∈ ker A + B. Hence A + B ∈ V ,
and V is closed under addition.
Since it is closed under addition and scalar multiplication and contains 0, V
is a subspace.
48. Find a basis for each of the following linear spaces and determine their
dimensions.
a. {f in P4 : f is even}
b.{f in P4 : f is odd}
a. Let V = {f in P4 : f is even}. Let f ∈ V . Write f (t) = at4 + bt3 +
ct + dt + e. Then f (−t) = at4 − bt3 + ct2 − dt + e. Since f (t) = f (−t), we
have b = d = 0 so f (t) = at4 + ct2 + e, so f is a linear combination of t4 , t2 ,
and 1. Since this holds for every f ∈ V , it follows that t4 , t2 , and 1 span V .
To show they are linearly independent, suppose there are x, y, z ∈ R such that
xt4 + yt2 + z = 0. Then by comparing coefficients we get x = y = z = 0, so
t4 , t2 , 1 form a basis. Since the basis has 3 elements, the dimension is 3.
2
b. Let V = {f in P4 : f is odd}. Let f ∈ V . Write f (t) = at4 + bt3 + ct2 +
dt + e. Then f (−t) = at4 − bt3 + ct2 − dt + e. Since f (t) = −f (−t), we have
a = c = e = 0 so f (t) = bt3 + dt, so f is a linear combination of t3 and t. Since
this holds for every f ∈ V , it follows that t3 and t span V . To show they are
linearly independent, suppose there are x, y ∈ R such that xt3 + yt = 0. Then
by comparing coefficients we get x = y = 0, so t3 , t form a basis. Since the basis
has 2 elements, the dimension is 2.
6. Define T : R2×2 → R2×2 by T (M ) = M
1
3
2
. Is T linear? Is it an
6
isomorphism?
Let A =
1
3
2
2
, so T (M ) = M A. Let U, V ∈ R2 and r ∈ R. Then T (U +V ) =
6
1
(U + V )A = U A + V A = T (U ) + T (V ) and T (rU ) = rU A = rT (A). Hence T
is linear.
To show T is not an isomorphism, it suffices to find a nonzero matrix B
a b
such that T (B) = 0 or equivalently BA = 0. Let B =
. Then BA =
c d
a + 3b 2a + 6b
. So we want a, b, c, d such that a + 3b = 2a + 6b = c + 3d =
c + 3d 2c + 6d
2c + 6d = 0. One choice is a = 1, b = −3, c = 0 and d = 0. Since this
gives a matrix B with T (B) = BA = 0, T is not injective, and hence not an
isomorphism.
Z
3
22. Define T : P2 → R by T (f ) =
f (t)dt. Is T linear? Is T an
−2
isomorphism?
Z
3
Z
3
Let f, g ∈ P2 and r ∈ R. Then T (f + g) =
f (t) + g(t)dt =
f (t)dt +
−2
Z 3
Z 3 −2
Z 3
g(t)dt = T (f ) + T (g). Also T (rf ) =
rf (t)dt = r
f (t)dt = rT (f ).
−2
−2
−2
Hence T is linear.
T cannot be an isomorphism because P2 and R have different dimensions.
52. Define T : R2×2 → R2×2 by T (M ) = M
nullity of T . Let A =
1
3
1
3
2
. Find the kernel and
6
2
, so T (M ) = M A. A matrix B is in the kernel
6
of T if and only if T (B) = 0 or equivalently BA = 0. Let B =
a
c
b
. Then
d
a + 3b 2a + 6b
. In particular BA = 0 if and only if a + 3b = 2a + 6b =
c + 3d 2c + 6d
c + 3d = 2c + 6d = 0. This system of equations reduces to a + 3b = 0 and
c + 3d = 0, or equivalently a = −3b and c = −3d. Thus the kernel is the set of
−3b b
matrices of the form
.
−3d d
−3 1
0 0
By inspection,
and
are a basis for the kernel of T , so the nullity
0 0
−3 1
of T is 2.
BA =
2