Download Solutions to HW 5

Math 110, Spring 2015: Homework 5 Solutions
Section 2.4
Exercise 2.4.4: Let A and B be n × n invertible matrices. Prove that AB is invertible and (AB)−1 =
B −1 A−1 .
Proof. We first note that the matrices AB and B −1 A−1 are n × n matrices. By the definition of an
invertible n × n matrix (on page 100 of the text), we need only prove that
(AB)(B −1 A−1 ) = (B −1 A−1 )(AB) = I,
where I denotes the n × n identity matrix. We can readily check this by using the associativity of matrix
multiplication (Theorem 2.16) and the fact that IC = CI = C for all C ∈ Mn×n (R) (Theorem 2.12(c)):
(AB)(B −1 A−1 ) = A(BB −1 )A−1 = AIA−1 = AA−1 = I.
Similarly,
(B −1 A−1 )(AB) = B −1 (A−1 A)B = B −1 IB = B −1 B = I.
Exercise 2.4.6: Prove that if A is invertible and AB = 0 , then B = 0 .
Proof. Multiplying both sides of the equation AB = 0 on the left by the inverse A−1 of A, we obtain
A−1 AB = A−1 0 =⇒ IB = B = 0
as desired; here I denotes the identity matrix of the same dimensions as A.
1
Exercise 2.4.9: Let A and B be n × n matrices such that AB is invertible. Prove that A and B are
invertible. Give an example to show that arbitrary matrices A and B need not be invertible if AB is
invertible.
Solution: We first prove that A and B are invertible under the original hypotheses:
Proof. We’ll prove this by converting it into a claim about linear transformations: By Corollary 2 on page
102 of the text, A and B are invertible matrices if and only if the left-multiplication transformations LA
and LB are invertible linear transformations. Furthermore, by the same Corollary, our assumption that
AB is invertible tells us that the composition LA LB = LAB is an invertible linear transformation.
Note first that, because A and B are n × n matrices, each of LA and LB maps Rn to itself. So to show
that each of LA and LB is invertible, we may show either that it is one-to-one or that it is onto—because,
by the Dimension Theorem, the conditions of being one-to-one and onto are equivalent here.
We first show that LB is one-to-one. Note that if LB (x) = 0 , then LA LB (x) = LA (0 ) = 0 . Thus,
N(LB ) ⊆ N(LA LB ) = {0 },
where the last equality follows from the assumption that LA LB is invertible and hence one-to-one. So
N(LB ) = {0 }, and therefore LB is one-to-one.
Next, we show that LA is onto. Indeed, for any y ∈ R(LA LB ), we can write y = LA LB (x) for some
x ∈ Rn . Thus,
Rn = R(LA LB ) ⊆ R(LA ),
where the first equality follows from the assumption that LA LB is invertible and hence onto. So we have
R(LA ) = Rn , and therefore LA is onto.
Thus, LA and LB are invertible linear transformations; as noted above, this is equivalent to saying that
A and B are invertible matrices.
To wrap up this exercise, we construct an example to show that for arbitrary (i.e., not necessarily
square) matrices A and B with AB invertible, A and B need not be invertible. The point here is that, by
definition, an invertible matrix must be a square matrix. So if AB is invertible, it must be square, say of
dimensions n × n; then we must have A ∈ Mn×m (R) and B ∈ Mm×n (R) for some m ∈ N. So we should
look for such examples with m 6= n, since by definition such A and B cannot be invertible as they are not
square matrices. Possibly the simplest example is the following: Take
1
A = 1 0 ∈ M1×2 (R) and B =
∈ M2×1 (R).
0
Then A and B are not square matrices, so they can’t be invertible; on the other hand, AB = 1 = I1 ,
the 1 × 1 identity matrix, which is invertible.
Remark: If you want a principled way to come up with an example instead of just “playing around”
until you find one, try using ideas similar to the proof for the first part of the exercise to prove the following:
Suppose A is an n × m matrix such that LA is onto (i.e., rank(A) = n) and B is an m × n matrix such
that LB is one-to-one. Suppose further that R(LB ) ∩ N(LA ) = {0 }. Then AB ∈ Mn×n (R) is invertible.
(I don’t know what exact language was used in Math 54, but note that R(LB ) is the “column space” of
B and N(LA ) is the “null space” of A. You can easily arrange for the three conditions we just listed by
writing down matrices in “reduced row echelon form.”)
2
Exercise 2.4.15: Let V and W be n-dimensional vector spaces, and let T : V → W be a linear transformation. Suppose that β is a basis for V . Prove that T is an isomorphism if and only if T(β) is a basis for
W.
Proof. We first prove the “only if” implication. So assume that T : V → W is an isomorphism; we
first claim that then T(β) must be a linearly independent set of n vectors in W . To that end, write
β = {v1 , . . . , vn }. Because T is one-to-one, the vectors T(vk ) are distinct for 1 ≤ k ≤ n, and thus
T(β) = {T(v1 ), . . . , T(vn )} is a set of n vectors in W . Now suppose that some linear combination of these
vectors is equal to the zero vector of W ; i.e., suppose
a1 T(v1 ) + . . . + an T(vn ) = 0
for some scalars a1 , . . . , an ∈ R. It suffices to prove that ak = 0 for all k. Now, again because T is a
one-to-one linear transformation, we have N(T) = {0 }; thus,
a1 T(v1 ) + . . . + an T(vn ) = T(a1 v1 + . . . + an vn ) = 0 =⇒ a1 v1 + . . . + an vn = 0 .
Since β is assumed to be a basis, in particular the vk are mutually linearly independent; thus, we must have
ak = 0 for all k as desired. So T(β) is a linearly independent set of n vectors in W ; because dim(W ) = n,
T(β) is a basis for W .
Now let us prove the “if” implication. To that end, assume that T(β) = {T(v1 ), . . . , T(vn )} is a basis
for W . Let y ∈ W be arbitrary. Using the assumed linearity of T, we can find scalars b1 , . . . , bn ∈ R such
that
y = b1 T(v1 ) + . . . + bn T(vn ) = T(b1 v1 + . . . + bn vn ) = T(x),
where x = b1 v1 + . . . + bn vn ∈ V . So we’ve shown that for all y ∈ W , y ∈ R(T). In other words, T is
onto. By the Dimension Theorem, because dim(V ) = dim(W ), T must also be one-to-one; thus, T is an
isomorphism.
Remarks: For the “only if” direction, we chose to show that T(β) must be linearly independent; of
course one could also have chosen to show that T(β) spans W . Similarly, for the “if” direction, it would
have been totally acceptable to show directly that T must be one-to-one and then to deduce that T must
also be onto from the Dimension Theorem. Note also that the “if” implication of this exercise is not true
in general if we don’t assume dim(V ) = dim(W ); try to think of a counterexample.
Exercise 2.4.16: Let B be an n × n invertible matrix. Define Φ : Mn×n (R) → Mn×n (R) by Φ(A) =
B −1 AB. Prove that Φ is an isomorphism.
Proof. We first show that Φ is a linear transformation. Let A1 , A2 ∈ Mn×n (R) and c ∈ R be arbitrary.
Then
Φ(cA1 + A2 ) = B −1 (cA1 + A2 )B = (cB −1 A1 + B −1 A2 )B = cB −1 A1 B + B −1 A2 B = cΦ(A1 ) + Φ(A2 ),
so Φ is linear. Next, let C ∈ Mn×n (R) be any n × n matrix. Then
Φ(BCB −1 ) = B −1 (BCB −1 )B = (B −1 B) C (B −1 B) = In C In = C,
where In denotes the n × n identity matrix. In particular, C ∈ R(Φ); as C was arbitrary, Φ is onto. Since
Φ is a linear transformation from the finite-dimensional vector space Mn×n (R) to itself, by the Dimension
Theorem Φ must also be one-to-one. Thus, Φ is an isomorphism.
3
Exercise 2.4.17: Let V and W be finite-dimensional vector spaces and T : V → W be an isomorphism.
Let V0 be a subspace of V .
(a) Prove that T(V0 ) is a subspace of W .
Proof. We need only use the fact that T is a linear transformation here; i.e., we do not need to use
the full fact that T is an isomorphism. Because V0 is a subspace of V , the zero vector 0V of V is
an element of V0 . So 0W = T(0V ) ∈ T(V0 ). To see that T(V0 ) is closed under addition and scalar
multiplication in W , let w1 , w2 ∈ T(V0 ) and c ∈ R be arbitrary. Then there exist v1 , v2 ∈ V0 such
that T(v1 ) = w1 and T(v2 ) = w2 . Furthermore, because V0 is a subspace of V , we have cv1 + v2 ∈ V0 .
Thus,
cw1 + w2 = cT(v1 ) + T(v2 ) = T(cv1 + v2 ) ∈ T(V0 ).
As c, w1 , and w2 were arbitrary, T(V0 ) is closed under addition and scalar multiplication in W .
Thus, T(V0 ) is a subspace of W .
(b) Prove that dim(V0 ) = dim T(V0 ) .
Proof. Here we need to use the full fact that T is an isomorphism. Consider the restriction TV0 of
T to the subspace V0 . The restriction of a linear transformation to a subspace of its domain was
defined in a previous homework; in this case we may view it as the linear transformation
TV0 : V0 −→ T(V0 )
defined by TV0 (x) := T(x) for all x ∈ V0 . By definition, R TV0 = T(V0 ), so TV0 is onto. Moreover,
if x ∈ N(TV0 ), then
0W = TV0 (x) = T(x) =⇒ x = 0V ,
because T is an isomorphism, hence one-to-one. Thus, TV0 is one-to-one; since we also noted that
it’s onto, TV0 : V0 → T(V0 ) is an isomorphism. In
other words, V0 and T(V0 ) are isomorphic vector
spaces; by Theorem 2.19, dim(V0 ) = dim T(V0 ) .
Exercise 2.4.21: Let V and W be finite-dimensional vector spaces with ordered bases β = {v1 , v2 , . . . , vn }
and γ = {w1 , w2 , . . . , wm }, respectively. By Theorem 2.6 (p. 72), there exist linear transformations Tij :
V → W such that
wi if k = j
Tij (vk ) =
0 if k 6= j.
First prove that {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} is a basis for L(V, W ).
Remark: One could make
this proof a bit shorter by using Theorem 2.20 and its Corollary; that would
tell us that dim L(V, W ) = mn. Once we knew the dimension, we could get away with showing either
linear independence or spanning and then immediately deduce the other. But since the spirit of this exercise
seems to be to reprove those results from a slightly different perspective, we’ll just do everything from
scratch. Note also that the arguments below can be motivated by considering the matrix representations
of linear transformations with respect to the bases β and γ.
4
Proof. We first show that the set {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} is linearly independent. Let T0 : V → W
denote the zero transformation, which is the zero vector of the vector space L(V, W ). Assume there exist
scalars aij ∈ R such that
m X
n
X
aij Tij = T0 ;
i=1 j=1
we aim to show that we must have aij = 0 for all 1 ≤ i ≤ m and 1 ≤ j ≤ n. Our assumption is equivalent
to the condition
m X
n
X
aij Tij (v) = 0
(1)
i=1 j=1
for all v ∈ V . Fix k ∈ {1, . . . , n}. Then plugging v = vk into equation (1) gives
m X
n
X
aij Tij (vk ) =
i=1 j=1
m
X
aik wi = a1k w1 + . . . + amk wm = 0 .
i=1
Since γ = {w1 , . . . , wm } is a basis for W , the wi are linearly independent, and thus we must have aik = 0
for all 1 ≤ i ≤ m. Since k was arbitrary, we have shown aij = 0 for all 1 ≤ i ≤ m and 1 ≤ j ≤ n, as
desired. So the given set is linearly independent in L(V, W ).
Next, we show that {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} spans L(V, W ). Let S ∈ L(V, W ) be an arbitrary
linear transformation. Again using the fact that γ = {w1 , . . . , wm } is a basis for W , for each 1 ≤ k ≤ n
we can find scalars b1k , b2k , . . . , bmk ∈ R such that
S(vk ) = b1k w1 + . . . + bmk wm =
m
X
bik wi .
i=1
Consider the linear transformation
m X
n
X
bij Tij ∈ span {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ m} .
i=1 j=1
Note that, for each 1 ≤ k ≤ n,
m X
n
X
bij Tij (vk ) =
i=1 j=1
m
X
bik wi = S(vk ).
i=1
Since β = {v1 , . . . , vn } is a basis for v, by the Corollary to Theorem 2.6 on page 73 of the text we have
S=
m X
n
X
bij Tij ∈ span {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} .
i=1 j=1
Since S ∈ L(V, W ) was arbitrary, the given set spans L(V, W ). Since we also showed it is linearly independent, it is a basis for L(V, W ).
5
Exercise 2.4.21 (continued): Then let M ij be the m × n matrix with 1 in the ith row and jth column
and 0 elsewhere, and prove that [Tij ]γβ = M ij .
Proof. We just compute to show that the corresponding entries of the two matrices are equal; i.e., we show
that for all 1 ≤ k ≤ m and all 1 ≤ ` ≤ n we have
[Tij ]γβ k` = M ij k` .
To do that, we just use the definition of the matrix representation of a linear transformation with respect
to a pair of bases β and γ. Fix an arbitrary such k and `. Note that the `th column of [Tij ]γβ is the
coordinate vector
[wi ]γ if ` = j
ei ∈ Rm if ` = j
[Tij (v` )]γ =
=
[0 ]γ if ` 6= j
0 ∈ Rm if ` 6= j.
(Here ei denotes the ith standard basis vector in Rm .) So the kth entry of this vector is
1 if ` = j and k = i
γ
[Tij ]β k` =
= M ij k` ,
0 otherwise
which is what we wanted to show.
Exercise 2.4.21 (continued): Again by Theorem 2.6, there exists a linear transformation Φ : L(V, W ) →
Mm×n (R) such that Φ(Tij ) = M ij . Prove that Φ is an isomorphism.
Proof. Note that from the first part of this exercise, in which we showed that {Tij | 1 ≤ i ≤ m, 1 ≤ j ≤ n}
is a basis for L(V, W ) consisting of mn distinct linear transformations, we can deduce
dim L(V, W ) = mn.
Thus, because we also have dim Mm×n (R) = mn, we may apply Exercise 15 above to the linear transformation Φ : L(V, W ) → Mm×n (R). Once we note that the set {M ij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} is just the
standard basis for Mm×n (R), the “if” implication of Exercise 15 tells us that Φ is an isomorphism.
Remark: If you think about it, Φ here is in fact the exact same isomorphism Φ appearing in Theorem
2.20 of the text. We’ve basically given a somewhat lengthy elaboration of the proof of that theorem;
comparing the proof we did here with that appearing in the book may be instructive.
6
Exercise 2.4.23: Let V denote the vector space defined in Example 5 of Section 1.2, and let W = P(R).
Define
n
X
T : V → W by T(σ) =
σ(i)xi ,
i=0
where n is the largest integer such that σ(n) 6= 0. Prove that T is an isomorphism.
Remark 1: To be utterly pedantic, the statement of the exercise does not define T(0V ). (The zero vector
0V of V is the sequence σ0 defined by σ0 (m) = 0 for all m, so it does not make sense to say “n is the
largest integer such that σ0 (n) 6= 0.”) So if we want to be completely precise, we should further define
T(0V ) = 0W , where 0W is the zero polynomial in P(R). This exercise is also confusing as stated because
Example 5 in Section 1.2 defines V to be the space of functions from the positive integers to R; thus,
“σ(0)” does not make sense for σ ∈ V under that definition. The natural thing to do is to modify the
definition of V slightly, taking V to be the vector space of functions from the nonnegative integers to R.
(N.B.: The two definitions yield isomorphic vector spaces. To check your understanding, find a reasonably
obvious isomorphism from “the new V ” to “the old V .”) We’ll use that modified definition of V for the
following proof.
Remark 2: Note that V and W here are infinite-dimensional vector spaces; accordingly, we cannot hope
to make any use of the Dimension Theorem or other finite-dimensional techniques.
Proof. We first show that T is a linear transformation. Let c ∈ R and σ, η ∈ V be arbitrary. Let N be the
largest integer such that either σ(N ) 6= 0 or η(N ) 6= 0; if σ = η = 0V , then set N = 0. Then
T(cσ + η) =
N
X
i
(cσ + η)(i)x = c
i=0
N
X
i
σ(i)x +
i=0
N
X
η(i)xi = cT(σ) + T(η),
i=0
so T is linear.
Next, we show T is one-to-one. Indeed, suppose T(σ) = 0W , the zero polynomial. Then by definition
of the zero polynomial and the map T, we know that:
ˆ σ(n) = 0 for all integers n > 0, and
ˆ σ(0) = 0.
In other words, σ(n) = 0 for all nonnegative integers n, and thus σ = 0V . So N(T) = {0V }, and T is
one-to-one.
Finally, let f ∈ W = P(R) be an arbitrary polynomial, and write
f (x) =
n
X
ai xi ,
i=0
so that deg(f ) = n. Define a sequence σ ∈ V by
ak for k ≤ n
σ(k) =
0 for k > n.
Then T(σ) = f ; since f ∈ W was arbitrary, T is onto.
Thus, we’ve shown that T is a linear transformation that is both one-to-one and onto; in other words,
T is an isomorphism.
7
Section 2.5
Exercise 2.5.2: For each of the following pairs of ordered bases β and β 0 for R2 , find the change of
coordinate matrix that changes β 0 -coordinates into β-coordinates.
(a) β = {e1 , e2 } and β 0 = {(a1 , a2 ) , (b1 , b2 )}.
Solution: For j = 1, 2, the j-th column of the change-of-coordinates matrix Q is given by the
β-coordinate vector of the j-th vector in the basis β 0 . We compute these coordinate vectors as
follows:
a1
(a1 , a2 ) = a1 e1 + a2 e2 ⇐⇒ [(a1 , a2 )]β =
;
a2
b1
(b1 , b2 ) = b1 e1 + b2 e2 ⇐⇒ [(b1 , b2 )]β =
.
b2
So the change-of-coordinates matrix is
Q=
[IR2 ]ββ 0
=
a1 b1
a2 b2
,
where IR2 denotes the identity transformation on the vector space R2 .
(d) β = {(−4, 3) , (2, −1)} and β 0 = {(2, 1) , (−4, 1)}.
Solution: With the same approach as in part (a), we compute the relevant coordinate vectors:
2
a
−4a + 2c = 2
;
= [(2, 1)]β =
⇐⇒
(2, 1) = a(−4, 3) + c(2, −1) ⇐⇒
5
c
3a − c = 1
(−4, 1) = b(−4, 3) + d(2, −1) ⇐⇒
−4b + 2d = −4
3b − d = 1
⇐⇒
So the change-of-coordinates matrix is
Q=
a b
c d
8
=
2 −1
5 −4
.
b
d
= [(−4, 1)]β =
−1
−4
.
Exercise 2.5.3: For each of the following pairs of ordered bases β and β 0 for P2 (R), find the change of
coordinate matrix that changes β 0 -coordinates into β-coordinates.
(b) β = {1, x, x2 } and β 0 = {a2 x2 + a1 x + a0 , b2 x2 + b1 x + b0 , c2 x2 + c1 x + c0 }.
Solution: We follow the same basic approach as in Exercise 2. The β-coordinate vectors of the basis
vectors in β 0 can easily be computed by inspection in this case:






a0
b0
c0
[a2 x2 + a1 x + a0 ]β =  a1  , [b2 x2 + b1 x + b0 ]β =  b1  , [c2 x2 + c1 x + c0 ]β =  c1  .
a2
b2
c2
So the change-of-coordinates matrix is

Q = [IP2 (R) ]ββ 0

a0 b0 c0
=  a1 b1 c1  .
a2 b2 c2
(d) β = {x2 − x + 1 , x + 1 , x2 + 1} and β 0 = {x2 + x + 4 , 4x2 − 3x + 2 , 2x2 + 3}.
Solution: For convenience, write β 0 = {f1 , f2 , f3 } and β = {g1 , g2 , g3 } (as ordered bases). Then we
compute the coordinate vectors:


 

a1 + a3 = 1
a1
2

−a1 + a2 = 1 ⇐⇒ [f1 ]β =  a2  =  3  ;
f1 = a1 g1 + a2 g2 + a3 g3 ⇐⇒

a1 + a2 + a3 = 4
a3
−1


b1 + b3 = 4
−b1 + b2 = −3
f2 = b1 g1 + b2 g2 + b3 g3 ⇐⇒

b1 + b2 + b3 = 2

c1 + c3 = 2

−c1 + c2 = 0
f3 = c1 g1 + c2 g2 + c3 g3 ⇐⇒

c1 + c2 + c3 = 3

 
b1
⇐⇒ [f2 ]β =  b2  = 
b3

 
c1
⇐⇒ [f2 ]β =  c2  = 
c3
So the change-of-coordinates matrix is

Q = [IP2 (R) ]ββ 0
 
a1 b1 c1
2
1
=  a2 b2 c2  =  3 −2
a3 b3 c3
−1
3
9

1
1 .
1

1
−2  ;
3

1
1 .
1
(f) β = {2x2 − x + 1 , x2 + 3x − 2 , −x2 + 2x + 1} and β 0 = {9x − 9 , x2 + 21x − 2 , 3x2 + 5x + 2}.
Solution: Again, write β 0 = {f1 , f2 , f3 } and β = {g1 , g2 , g3 }. Compute the coordinate vectors (in
the interest of saving space I will not show my work in solving the systems of equations, but you
should):


 

2a1 + a2 − a3 = 0
a1
−2

−a1 + 3a2 + 2a3 = 9
⇐⇒ [f1 ]β =  a2  =  3  ;
f1 = a1 g1 + a2 g2 + a3 g3 ⇐⇒

a1 − 2a2 + a3 = −9
a3
−1


  
2b1 + b2 − b3 = 1
b1
1

−b1 + 3b2 + 2b3 = 21 ⇐⇒ [f2 ]β =  b2  =  4  ;
f2 = b1 g1 + b2 g2 + b3 g3 ⇐⇒

b1 − 2b2 + b3 = −2
b3
5

  

2
c1
2c1 + c2 − c3 = 3

−c1 + 3c2 + 2c3 = 5 ⇐⇒ [f3 ]β =  c2  =  1  .
f3 = c1 g1 + c2 g2 + c3 g3 ⇐⇒

2
c3
c1 − 2c2 + c3 = 2
So the change-of-coordinates matrix is

Q = [IP2 (R) ]ββ 0
 
a1 b1 c1
−2



3
= a2 b2 c2 =
a3 b3 c3
−1
1
4
5

2
1 .
2
Exercise 2.5.6: For each matrix A and ordered basis β, find [LA ]β . Also, find an invertible matrix Q
such that [LA ]β = Q−1 AQ.
1
1
1 2
.
,
and β =
(b) A =
−1
1
2 1
Solution: For convenience, write β = {v1 , v2 } (as an ordered basis). Then the j-th column of [LA ]β
is the β-coordinate vector [LA (vj )]β . So we compute
3
3
;
= 3v1 + 0v2 =⇒ [LA (v1 )]β =
LA (v1 ) =
0
3
0
−1
LA (v2 ) =
= 0v1 + (−1)v2 =⇒ [LA (v2 )]β =
.
−1
1
So
[LA ]β =
3
0
0 −1
.
By the Corollary on page 115 of the text, the change-of-coordinates matrix Q that changes βcoordinates into standard-basis-coordinates, given by
1
1
Q=
,
1 −1
satisfies [LA ]β = Q−1 AQ.
10

 
  


1
1
1 
13 1 4








1 , −1 , 1  .
(d) A = 1 13 4 and β =


−2
0
1
4 4 10
Solution: Again, for convenience write β = {v1 , v2 , v3 }. Compute the relevant β-coordinate vectors:


 
6
6
6  = 6v1 + 0v2 + 0v3 =⇒ [LA (v1 )]β = 0 ;
LA (v1 ) = 
−12
0


 
12
0



LA (v2 ) = −12 = 0v1 + 12v2 + 0v3 =⇒ [LA (v2 )]β = 12 ;
0
0
 
 
18
0



LA (v3 ) = 18 = 0v1 + 0v2 + 18v3 =⇒ [LA (v3 )]β = 0  .
18
18
So we have


6 0 0
[LA ]β = 0 12 0  .
0 0 18
Again using the Corollary on page 115 of the text, the change-of-coordinates matrix


1
1
1
1 
Q =  1 −1
−2
0
1
satisfies [LA ]β = Q−1 AQ.
Remark: In both parts of this exercise, it was easy enough to solve (by inspection) the systems of
linear equations required to obtain the relevant coordinate vectors to construct [LA ]β . On the other hand,
if the vectors of the basis β are explicitly presented to you in standard coordinates, the matrix Q is always
easy to construct: its columns are simply the standard coordinate vectors of the basis vectors from β,
arranged in their proper order. So in general, you have a choice of which computation you’d rather do:
(1) solve the linear systems needed to compute the relevant coordinate vectors, or
(2) invert the matrix Q, by whatever technique you desire (probably either by row operations or by
Cramer’s rule).
Which option is the less tedious varies on a case-by-case basis, of course.
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Exercise 2.5.7(b): In R2 , let L be the line y = mx, where m 6= 0. Find an expression for T(x, y), where
T is the projection on L along the line perpendicular to L. (See the definition of projection in the exercises
of Section 2.1.)
Solution: We follows the geometrically intuitive approach of Example 3 on pages 113–114 of the text.
1
Let L⊥ denote the line in R2 perpendicular to L; it has equation y = − m
x. Since L and L⊥ are lines
2
2
through the origin in R , they are one-dimensional subspaces of R . Indeed, β = {(1, m)} is a basis for L,
and β ⊥ = {(−m, 1)} is a basis for L⊥ . We first verify that R2 = L ⊕ L⊥ . Geometrically, it is clear that
the lines L and L⊥ intersect (only) at the origin; that is, L ∩ L⊥ = {(0, 0)}. Moreover, since neither of the
basis vectors (1, m) and (−m, 1) is a scalar multiple of the other, the set γ = {(1, m) , (−m, 1)} is linearly
independent in R2 ; since dim(R2 ) = 2 the set must also span R2 , which implies R2 = L + L⊥ . Note that
γ is in fact a basis of R2 .
By definition of the projection on L along L⊥ , we see that T(1, m) = (1, m) and T(−m, 1) = (0, 0). So
we have
1 0
[T]γ =
.
0 0
Moreover, we can form the change-of-coordinates matrix Q that converts γ-coordinates into standardbasis-coordinates, and we can take its inverse using the familiar formula for the inverse of a 2 × 2 matrix:
1
1 −m
1 m
−1
, Q = 2
.
Q=
m
1
m + 1 −m 1
Then, as in Example 3, T = LA , where A is the matrix
A = Q [T]γ Q
−1
1
= 2
m +1
1 m
m m2
.
So we just compute
1
x + my mx + m2 y
x
x + my
= 2
A
=⇒ T(x, y) =
,
.
y
m + 1 mx + m2 y
m2 + 1
m2 + 1
Remark: If you have learned elsewhere how to take orthogonal projections using “dot products,”
you can verify that T(x, y) is indeed the same orthogonal projection of (x, y) to L as computed using the
dot-product method.
Exercise 2.5.10: Prove that if A and B are similar n × n matrices, then tr(A) = tr(B). Hint: Use
Exercise 13 of Section 2.3.
Proof. Exercise 13 of Section 2.3 gives the following “commutativity” property of the trace: For any
C, D ∈ Mn×n (R), we have tr(CD) = tr(DC). Since A and B are similar matrices, there exists some
invertible n × n matrix Q such B = Q−1 AQ, by definition. So we just compute, using the “commutativity”
property:
tr(B) = tr Q−1 (AQ) = tr (AQ)Q−1 = tr A(QQ−1 ) = tr(AI) = tr(A),
where I denotes the n × n identity matrix, as desired.
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