Friedberg, Insel, and Spence Linear algebra, 4th ed. SOLUTIONS REFERENCE
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MATH 115a
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MATH 115a
Friedberg, Insel, and Spence
Linear algebra, 4th ed.
SOLUTIONS REFERENCE
Preface
The aim of this document is to serve as a reference of problems and solutions from the fourth edition
of “Linear Algebra” by Friedberg, Insel and Spence. Originally, I had intended the document to
be used o...
4th ed solutions reference preface the aim of this document is to serve as a ref
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Friedberg, Insel, and Spence
Linear algebra, 4th ed.
SOLUTIONS REFERENCE
Michael L. Baker
<mbaker@lambertw.com>
UNIVERSITY OF WATERLOO
January 23, 2011
Preface
The aim of this document is to serve as a reference of problems and solutions from the fourth edition
of “Linear Algebra” by Friedberg, Insel and Spence. Originally, I had intended the document to
be used only by a student who was well-acquainted with linear algebra. However, as the document
evolved, I found myself including an increasing number of problems. Therefore, I believe the
document should be quite comprehensive once it is complete.
I do these problems because I am interested in mathematics and consider this kind of thing to be
fun. I give no guarantee that any of my solutions are the “best” way to approach the corresponding
problems. If you find any errors (regardless of subtlety) in the document, or you have different
or more elegant ways to approach something, then I urge you to contact me at the e-mail address
supplied above.
This document was started on July 4, 2010. By the end of August, I expect to have covered up to
the end of Chapter 5, which corresponds to the end of MATH 146, “Linear Algebra 1 (Advanced
Level)” at the University of Waterloo. This document is currently a work in progress.
2 Linear Transformations and Matrices 30
2.1 Linear Transformations, Null spaces, and Ranges . . . . . . . . . . . . . . . . . . . . 30
2.2 The Matrix Representation of a Linear Transformation . . . . . . . . . . . . . . . . . 34
2.3 Composition of Linear Transformations and Matrix Multiplication . . . . . . . . . . 36
2
,1 Vector Spaces
1.1 Introduction
Section 1.1 consists of an introductory, geometrically intuitive treatment of vectors (more specifi-
cally, Euclidean vectors). The solutions to the exercises from this section are very basic and as such
have not been included in this document.
1.2 Vector Spaces
Section 1.2 introduces an algebraic structure known as a vector space over a field, which is then used
to provide a more abstract notion of a vector (namely, as an element of such an algebraic structure).
Matrices and n-tuples are also introduced. Some elementary theorems are stated and proved, such
as the Cancellation Law for Vector Addition (Theorem 1.1), in addition to a few uniqueness results
concerning additive identities and additive inverses.
8. In any vector space V, show that (a + b)(x + y) = ax + ay + bx + by for any x, y ∈ V and any
a, b ∈ F .
Solution. Noting that (a + b) ∈ F , we have
(a + b)(x + y) = (a + b)x + (a + b)y (by VS 7)
= ax + bx + ay + by (by VS 8)
= ax + ay + bx + by (by VS 1)
as required.
9. Prove Corollaries 1 and 2 [uniqueness of additive identities and additive inverses] of Theorem
1.1 and Theorem 1.2(c) [a0 = 0 for each a ∈ F ].
Solution. First, let 0, 00 ∈ V be additive identities, and let x ∈ V. We have, by VS 3,
x = x + 0 = x + 00
Whereupon VS 1 yields
0 + x = 00 + x
By Theorem 1.1 [Cancellation Law for Vector Addition], we obtain 0 = 00 , proving that the
additive identity is unique.
Second, let x ∈ V and y, y 0 ∈ V be such that x + y = x + y 0 = 0. Then we obtain, by VS 1,
y + x = y0 + x
By Theorem 1.1 [Cancellation Law for Vector Addition], we obtain y = y 0 , proving that each
additive inverse is unique.
Third, let a ∈ F . We have, by VS 3 and VS 7,
a0 = a(0 + 0) = a0 + a0
By Theorem 1.1 [Cancellation Law for Vector Addition], we obtain a0 = 0, as required.
3
, 10. Let V denote the set of all differentiable real-valued functions defined on the real line. Prove
that V is a vector space with the operations of addition and scalar multiplication defined in
Example 3.
Solution. Let f, g, h ∈ V and c, d, s ∈ R. Due to rules taught in elementary calculus, we
have f + g ∈ V as well as cf ∈ V for all f, g ∈ V and c ∈ R. Now, note that by the
commutativity of addition in R, we have f + g = f (s) + g(s) = g(s) + f (s) = g + f for
all s. Therefore, V satisfies VS 1. Next, by the associativity of addition in R, we have
(f + g) + h = (f (s) + g(s)) + h(s) = f (s) + (g(s) + h(s)) = f + (g + h) for all s. Therefore,
V satisfies VS 2. Next, let 0V (s) = 0F for all s, where 0F is the additive identity of R.
Then f + 0V = f (s) + 0F = f (s) = f for all s. Therefore V satisfies VS 3. Now, for each
s we have f (s) ∈ R and so we are guaranteed an additive inverse, say −f (s), such that
f (s) + (−f (s)) = 0F . Now, 0V (s) = 0F for all s also, so we note f + (−f ) = 0V for all f .
Therefore V satisfies VS 4. Next, let 1F represent R’s multiplicative identity. Then we have
1F ·f = 1F ·f (s) = f (s) = f for all s. Therefore, 1F ·f = f for all f . So VS 5 is satisfied. Note
also that (cd)f = (cd) · f (s) = c · (df (s)) = c(df ) for all s. This is because multiplication in R
is associative. So VS 6 is satisfied. We also have c(f +g) = c·(f (s)+g(s)) = c·f (s)+c·g(s) =
cf +cg. This is because multiplication in R is distributive (over addition). So VS 7 is satisfied.
Finally, note (c + d)f = (c + d) · f (s) = c · f (s) + d · f (s) = cf + df . This is again because
multiplication in R is distributive (over addition). So VS 8 is satisfied. This completes the
proof that V, together with addition and scalar multiplication as defined in Example 3, is a
vector space.
22. How many matrices are there in the vector space Mm×n (Z2 )?
Solution. There are 2mn vectors in this vector space.
1.3 Subspaces
Section 1.3 examines subsets of vector spaces, and defines a special type of subset, known as a
subspace, which is a subset of a vector space that can be considered a vector space in its own right
(only certain subsets satisfy this property). A certain theorem that is referred to as the subspace test
is stated and proven, which provides a fast way of checking whether a given subset is a subspace.
More concepts relating to matrices are introduced as well. The section closes with a proof that the
intersection of two subspaces is itself a subspace.
3. Prove that (aA + bB)t = aAt + bB t for any A, B ∈ Mm×n (F ) and any a, b ∈ F .
Solution. We have (aA)ij = a · Aij , (bB)ij = b · Bij , so (aA + bB)ij = a · Aij + b · Bij . So
[(aA + bB)t ]ij = a · Aji + b · Bji
Now, (At )ij = Aji , (B t )ij = Bji , (aAt )ij = a · Aji , (bB t )ij = b · Bji . So
[aAt + bB t ]ij = a · Aji + b · Bji
Therefore (aA + bB)t = aAt + bB t as required.
4. Prove that (At )t = A for each A ∈ Mm×n (F ).
4
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