Download Solutions

Midterm Exam I, Math 121 A
27 October, 2006
Name:
Solutions
ID:
Instructions: Put away any electronic devices, papers, notes, and books. Use only
a pencil and eraser (or a pen if you are daring). Write neatly and only on the paper
provided. Use the back of the page if necessary.
1. (25 pts.) Let T : R3 → R2 be defined via T ((a1 , a2 , a3 )) = (5a1 , a3 − 2a2 ). Show
that T is linear. Find bases for N(T) and R(T). Confirm the Dimension Theorem
for this example. Is T one-to-one? Is T onto?
Solution: To show that T is linear let x = (a1 , a2 , a3 ) and y = (b1 , b2 , b3 ) and let
c ∈ F . Then
T(x + cy) =
=
=
=
=
T ((a1 + cb1 , a2 + cb2 , a3 + cb3 ))
(5(a1 + cb1 ), a3 + cb3 − 2(a2 + cb2 ))
(5a1 + c5b1 , a3 − 2a2 + c(b3 − 2b2 ))
(5a1 , a3 − 2a2 ) + c (5b1 , b3 − 2b2 )
T(x) + cT(y),
(1)
which shows that T is linear.
N(T) is the set of all x = (a1 , a2 , a3 ) such that T(x) = (0, 0). Equivalently, x ∈ N(T)
if and only if
5a1 = 0
a3 − 2a2 = 0.
(2)
Thus N(T) = {(0, s, 2s) | s ∈ R}. The set SN = {(0, 1, 2)} spans N(T). Since it is
also linearly independent, SN is a basis for N(T).
R(T) = R2 . To see this we will show that R2 ⊆ R(T). Let y = (b1 , b2 ) ∈ R2 be
arbitrary. If we can find some x = (a1 , a2 , a3 ) ∈ R3 such that y = T(x), then
y ∈ R(T). Now, y = T(x) if and only if
b1 = 5a1
b2 = a3 − 2a2 .
1
(3)
A solution to this system exists; one solution is x = (a1 , a2 , a3 ) = (b1 /5, 0, b2 ). Thus
y ∈ R(T), which implies that R2 ⊆ R(T). Since R(T) ⊆ R2 always, it must be that
R(T) = R2 . A basis for R2 is SR = {(1, 0), (0, 1)}; thus SR is a basis for R(T).
By definition dim(N(T)) = 1 and dim(R(T)) = 2. The Dimension Theorem
states that dim(N(T))+dim(R(T)) = dim(V), where V = R3 . Thus the Dimension
Theorem is confirmed.
T is not one-to-one because N(T) 6= {(0, 0, 0)}. Alternatively, T ((1, 1, 1)) = (5, −1)
and T ((1, 0, 1/2)) = (5, −1), which shows directly that T is not one-to-one.
T is onto because R(T) = R2 .///
2
2. (25 pts.) Let V and W be vector spaces and let T : V → W be linear. If V1 is a
vector subspace of V show that T(V1 ) = {w ∈ R(T) | T(v) = w for some v ∈ V1 } is
a vector subspace of W.
Solution: We must show that (a) 0W ∈ T(V1 ), (b) w1 + w2 ∈ T(V1 ), for all
w1 , w2 ∈ T(V1 ), and (c) cw1 ∈ T(V1 ), for all w1 ∈ T(V1 ) and c ∈ F .
(a) Since T is linear T(0V ) = 0W . It must be true that 0V ∈ V1 , because V1 is a
vector subspace of V. Thus 0W ∈ T(V1 ), because there is some element v ∈ V1 ,
namely v = 0V , such that T(v) = 0W .
(b) Let w1 , w2 ∈ T(V1 ). Then there exist v1 , v2 ∈ V1 such that T(v1 ) = w1 and
T(v2 ) = w2 . By the linearity of T, w1 + w2 = T(v1 + v2 ). Now, it must be true
that v1 + v2 ∈ V1 , because v1 , v2 ∈ V1 and V1 is a vector subspace of V. Thus
w1 + w2 ∈ T(V1 ), because there is some element v ∈ V1 , namely v = v1 + v2 , such
that T(v) = w1 + w2 .
(c) Let w1 ∈ T(V1 ). Then there exists some v1 ∈ V1 such that T(v1 ) = w1 . Let
c ∈ F . Then cv1 ∈ V1 , because V1 is a vector subspace of V. By the linearity of T,
cw1 = T(cv1 ). Thus cw1 ∈ T(V1 ), because there is some element v ∈ V1 , namely
v = cv1 , such that T(v) = cw1 .///
3
3. (25 pts.) Prove that if {A1 , . . . , Ak } is a linearly independent subset of Mm×n (F ),
then {At1 , . . . , Atk } is a linearly independent subset of Mn×m (F ).
Solution: Suppose that S = {A1 , . . . , Ak } is linearly independent. Let Z be the
zero matrix of Mm×n (F ): Zi,j = 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Then Z = 0Mm×n (F )
and Z t = 0Mn×m (F ) . Let a1 , . . . , ak ∈ F be arbitrary. S is linearly independent if
and only if
k
X
al Al = Z = 0Mm×n (F )
(4)
l=1
implies that al = 0 for all l = 1, . . . , k. Equation (4) holds if and only if the
following holds
k
X
al (Al )i,j = 0, 1 ≤ i ≤ m, 1 ≤ j ≤ n.
(5)
l=1
Now, consider an arbitrary representation of zero as a linear combination of vectors
from S t = {At1 , . . . , Atk }:
k
X
al Atl = Z t = 0Mn×m (F ) .
(6)
l=1
Equation (6) holds if and only if the following holds
k
X
al (Atl )i,j = 0, 1 ≤ i ≤ n, 1 ≤ j ≤ m.
(7)
l=1
Using (A)i,j = (At )j,i, Eq. (7) holds if and only if the following holds
k
X
al (Al )j,i = 0, 1 ≤ i ≤ n, 1 ≤ j ≤ m.
(8)
l=1
Equations (8) and (5) are equivalent (only the indices are different) and Eq. (5)
implies that al = 0 for all l = 1, . . . , k. Thus Eq. (8) implies that al = 0 for all
l = 1, . . . , k. Hence, S t is linearly independent.///
4
4. (25 pts.) Let V and W be vector spaces such that dim(V) = dim(W) < ∞, and
let T : V → W be linear. Show that there exist ordered bases β and γ for V and
W, respectively, such that [T]γβ is diagonal.
Solution: By the Dimension Theorem dim(V) = dim(N(T)) + dim(R(T)). Let
dim(V) = n, dim(N(T)) = k, and dim(R(T)) = n − k, where 0 ≤ k ≤ n. Let
βN = {vi }ki=1 be an ordered basis for N(T). We may extend βN to β = {vi }ni=1 ,
where β is an ordered basis for V. In the proof of the Dimension Theorem we
showed that γR = {T(vi )}ni=k+1 is an ordered basis for R(T). We may extend γR to
γ = {wi }ni=1 , where wi = T(vi ) for i = k + 1, . . . , n and γ is an ordered basis for W.
Let A ∈ Mn×n (F ) denote the matrix representation of T in the order bases β and
γ: A = [T]γβ . By definition
T(vj ) =
n
X
Ai,j wi .
(9)
i=1
Since vj ∈ N(T) for 1 ≤ j ≤ k, we have T(vj ) = 0W for 1 ≤ j ≤ k. Thus
n
X
Ai,j wi =
i=1
(
0W
wj
if 1 ≤ j ≤ k
.
if k + 1 ≤ j ≤ n
(10)
The unique matrix that satisfies the last equation is given by
Ai,j =
(
0
1
if i 6= j
.
if i = j and k + 1 ≤ j ≤ n
Thus A = [T]γβ is diagonal.///
5
(11)