Comps Study Guide for Linear Algebra
Department of Mathematics and Statistics
Amherst College
September, 2017
This study guide was written to help you prepare for the linear algebra portion of the Comprehensive
and Honors Qualifying Examination in Mathematics. It is based on the Syllabus for the Comprehensive
Examination in Linear Algebra (Math 271/272) available on the Department website.
Each topic from the syllabus is accompanied by a brief discussion and examples from old exams. When
reading this guide, you should focus on three things:
• Understand the ideas. If you study problems and solutions without understanding the underlying ideas,
you will not be prepared for the exam.
• Understand the strategy of each proof. Most proofs in this guide are short—the hardest part is often
knowing how to start. Focus on the setup step rather than falling into the trap of memorizing proofs.
• Understand the value of scratchwork. Sometimes scratchwork is needed to explore possible approaches
to a computation or proof. That is, sometimes it is helpful to work out some details on the side by
trial and error before writing up a clear, presentable solution.
The final section of the guide has some further suggestions for how to prepare for the exam.
1 Vector Spaces and Subspaces
Some basic things to be aware of, although they do not arise directly on the comps:
• The definition of a vector space: A set with operations called addition and scalar multiplication (by
elements of R) satisfying a certain long list of axioms.
• One of the equivalent definitions of a subspace: as a subset W of a vector space V such that W is a
vector space in its own right, with the same addition and scalar multiplication operations.
The elements of a vector space V are called vectors, even if those elements are functions, matrices, or other
objects. The additive identity of V is called the zero vector, and it is usually denoted 0 or 0V .
Simple vector space examples. Here are some vector spaces you should know, each with standard
addition and scalar multiplication operations. You don’t need to memorize the notation (other than for Rn ,
which you surely already know), because any such notation will be defined on the exam if it appears.
• Rn is the vector space of ordered n-tuples of real numbers. Sometimes denoted Rn . Sometimes its
elements are written as row vectors (x1 , . . . , xn ) and sometimes as columns. Note: dim(Rn ) = n.
• Pn (R) is the vector space of polynomials of degree less than or equal to n. Sometimes denoted Pn or
P n or Pn or some other such variation. Note: dim(Pn (R)) = n + 1.
• Mm×n (R) is the vector space of m × n matrices with real entries. Sometimes denoted Mm×n or some
other such variation. Note: dim(Mm×n (R)) = mn.
1
,There are also some infinite-dimensional vector spaces which arise occasionally. For example, P (R) is the
vector space of all polynomials, F (R) is the vector space of all functions f : R → R, and C(R) is the vector
space of all continuous functions f : R → R. Again, don’t bother memorizing the notation, which will be
fully defined if it appears; just be ready to work with such spaces if they show up on the exam.
Subspace Theorem. The following Theorem is usually used to check whether a given subset is a subspace;
in fact, some books use it as the definition of a subspace.
Theorem. Let V be a vector space. A subset W ⊆ V is a subspace if it satisfies the following properties:
1. W 6= ∅.
2. For all x, y ∈ W and all c ∈ R, we have cx + y ∈ W .
Property 1 is usually verified by proving that 0V ∈ W .
Property 2 can be replaced by two separate statements:
Closed under addition: For all x, y ∈ W , we have x + y ∈ W .
Closed under scalar multiplication: For all x ∈ W and c ∈ R, we have cx ∈ W .
That is, a subset W of a vector space V is a subspace if and only if W is nonempty, closed under addition,
and closed under scalar multiplication.
Note:
• The empty set ∅ is not a vector space. Instead, the smallest vector space is the trivial space, {0}.
• Every vector space V has two obvious subspaces: the trivial subspace {0} ⊆ V , and the improper
subspace V ⊆ V . (Obviously, the two coincide if and only if V = {0} is trivial.)
1 (March 2006) Let U and V be subspaces of a vector space W .
(a) Prove that U ∩ V is a subspace of W .
(b) Prove that U + V = {u + v : u ∈ U, v ∈ V } is a subspace of W .
(c) Give an example to show that U ∪ V need not be a subspace of W .
Proof. (a): (Nonempty) We have 0W ∈ U and 0W ∈ V , since both U and V are subspaces. Hence
0W ∈ U ∩ V .
(Closure) Given x, y ∈ U ∩V and c ∈ R, we have cx +y ∈ U and cx +y ∈ V since both are subspaces.
Hence cx + y ∈ U ∩ V . QED (a)
(b): (Nonempty) We have 0W ∈ U and 0W ∈ V , since both are subspaces. Hence 0W + 0W ∈ U + V .
(Closure) Given x, y ∈ U + V and c ∈ R, there exist u1 , u2 ∈ U and v1 , v2 ∈ V such that x = u1 + v1
and y = u2 + v2 . Thus,
cx + y = c(u1 + v1 ) + (u2 + v2 ) = (cu1 + u2 ) + (cv1 + v2 ) ∈ U + V QED (b)
(c): Let W = R2 , let U = Span({(1, 0)}), and let V = Span({(0, 1)}). [That is, U is the x-axis, and
V is the y-axis.] Then U and V are subspaces because they are each the span of a set. However,
(1, 0) ∈ U ⊆ U ∪ V and (0, 1) ∈ V ⊆ U ∪ V , but (1, 0) + (0, 1) = (1, 1) 6∈ U ∪ V . Thus, U ∪ V is not
closed under addition and hence is not a subspace. QED (c)
Comment 1. In the “Nonempty” step of (b), there was no need to observe that 0W + 0W = 0W .
Yes, that’s true, and it wouldn’t hurt to say it, but it would be unnecessary. We’re simply trying to
show that U + V 6= ∅, so all we need to do is produce an element; we aren’t required to simplify that
element. On the other hand, in part (c), it was important to simplify the sum (1, 0) + (0, 1), to verify
that the result is not an element of U ∪ V .
Comment 2. In part (c), we made use of Span (to be discussed below) because it made the proof
shorter, rather than verifying by hand that both U and V are subspaces. Also in part (c), there are
many ways to do this. (Any choice of W with any subspaces U and V will work, as long as U 6⊆ V
and V 6⊆ U .) But it’s generally best to pick as simple an example as you can.
2
, Linear combinations, etc. Let V be a vector space, and let S = {v1 . . . , vn } ⊆ V be a finite set of
vectors in V .
• A linear combination of elements of S is an expression a1 v1 +· · ·+an vn for some scalars a1 , . . . , an ∈ R.
• The span of S, denoted Span(S), is the set of all linear combinations of elements of S. That is,
Span(S) = {a1 v1 + · · · + an vn | a1 , . . . , an ∈ R}.
– Fact: Span(∅) = {0}.
– Fact: Span(S) is a subspace of V .
– Fact: If W is a subspace and S ⊆ W , then Span(S) ⊆ W .
– If Span(S) = W , we say that S spans W .
• The set S is linearly dependent if there exist scalars a1 , . . . , an ∈ R that are not all zero such that
a1 v1 + · · · + an vn = 0.
S is linearly independent if it is not linearly dependent. Equivalently, for any scalars a1 , . . . , an ∈ R
such that a1 v1 + · · · + an vn = 0, we must have a1 = · · · = an = 0.
• The set S ⊆ V is a basis for V if S is linearly independent and Span(S) = V .
• The dimension of V is the number dim(V ) of elements in a basis for V . (It is a Theorem that any two
bases for V have the same number of elements.) If V has no finite basis, we say dim(V ) = ∞.
Side notes: Linear independence really should include an extra specification that we first ensure v1 , . . . , vn
are all distinct; but this subtlety does not arise on the comps. Also, there are notions of linear combinations,
span, and linear (in)dependence for an infinite set S ⊆ V , but again, these do not arise on the comps.
2 (March 2007) Suppose that V is a vector space and {v1 , . . . , vn } ⊆ V is linearly independent.
Also assume that v ∈ V is not contained in the span of {v1 , . . . , vn }. Prove that {v, v1 , . . . , vn } is
linearly independent.
Proof. Given scalars a0 , a1 , . . . , an ∈ R such that a0 v + a1 v1 + · · · + an vn = 0, we need to show that
a0 = a1 = · · · = an = 0.
Case 1 : Suppose a0 = 0. Then a1 v1 + · · · + an vn = 0, and hence a1 = · · · = an = 0, because
{v1 , . . . , vn } is linearly independent. Thus, a0 = a1 = · · · = an = 0, as desired.
Case 2 : Suppose a0 6= 0. Then a0 v = −a1 v1 − · · · − an vn , and multiplying both sides by 1/a0 (which
a1 an
is a real number, since a0 6= 0), we have v = − v1 −· · ·− vn ∈ Span({v1 , . . . , vn }), contradicting
a0 a0
the hypotheses. Thus, Case 2 cannot happen, and we are done by Case 1. QED
Comment 1. Don’t start a proof by regurgitating what the hypotheses mean. Instead, focus on the
statement you’re asked to deduce: for this problem, that the set {v, v1 , . . . , vn } is linearly indepen-
dent. So our first line should be to suppose we are given scalars ai for which a0 v+a1 v1 +· · ·+an vn = 0,
and our last line should be to conclude that a0 = a1 = · · · = an = 0. Once you know the first and
last line, then you can go back and try to figure out how the hypotheses could help you get there.
Comment 2. In this problem, two cases are required. We discovered that by doing some scratchwork,
and realizing that the span hypothesis could only be used in one situation, and the linear independence
hypothesis can be used in the other. Scratchwork is important!
Row reduction. Many problems in linear algebra end up requiring row reduction. You will need to know:
• How to set up a system of equations when necessary.
• The three kinds of row reduction steps. (Switching two rows, multiplying a row by a nonzero scalar,
and adding a multiple of one row to another.)
• The overall strategy, leading to a matrix in echelon form.
• How to interpret the echelon form to solve your problem.
3
Department of Mathematics and Statistics
Amherst College
September, 2017
This study guide was written to help you prepare for the linear algebra portion of the Comprehensive
and Honors Qualifying Examination in Mathematics. It is based on the Syllabus for the Comprehensive
Examination in Linear Algebra (Math 271/272) available on the Department website.
Each topic from the syllabus is accompanied by a brief discussion and examples from old exams. When
reading this guide, you should focus on three things:
• Understand the ideas. If you study problems and solutions without understanding the underlying ideas,
you will not be prepared for the exam.
• Understand the strategy of each proof. Most proofs in this guide are short—the hardest part is often
knowing how to start. Focus on the setup step rather than falling into the trap of memorizing proofs.
• Understand the value of scratchwork. Sometimes scratchwork is needed to explore possible approaches
to a computation or proof. That is, sometimes it is helpful to work out some details on the side by
trial and error before writing up a clear, presentable solution.
The final section of the guide has some further suggestions for how to prepare for the exam.
1 Vector Spaces and Subspaces
Some basic things to be aware of, although they do not arise directly on the comps:
• The definition of a vector space: A set with operations called addition and scalar multiplication (by
elements of R) satisfying a certain long list of axioms.
• One of the equivalent definitions of a subspace: as a subset W of a vector space V such that W is a
vector space in its own right, with the same addition and scalar multiplication operations.
The elements of a vector space V are called vectors, even if those elements are functions, matrices, or other
objects. The additive identity of V is called the zero vector, and it is usually denoted 0 or 0V .
Simple vector space examples. Here are some vector spaces you should know, each with standard
addition and scalar multiplication operations. You don’t need to memorize the notation (other than for Rn ,
which you surely already know), because any such notation will be defined on the exam if it appears.
• Rn is the vector space of ordered n-tuples of real numbers. Sometimes denoted Rn . Sometimes its
elements are written as row vectors (x1 , . . . , xn ) and sometimes as columns. Note: dim(Rn ) = n.
• Pn (R) is the vector space of polynomials of degree less than or equal to n. Sometimes denoted Pn or
P n or Pn or some other such variation. Note: dim(Pn (R)) = n + 1.
• Mm×n (R) is the vector space of m × n matrices with real entries. Sometimes denoted Mm×n or some
other such variation. Note: dim(Mm×n (R)) = mn.
1
,There are also some infinite-dimensional vector spaces which arise occasionally. For example, P (R) is the
vector space of all polynomials, F (R) is the vector space of all functions f : R → R, and C(R) is the vector
space of all continuous functions f : R → R. Again, don’t bother memorizing the notation, which will be
fully defined if it appears; just be ready to work with such spaces if they show up on the exam.
Subspace Theorem. The following Theorem is usually used to check whether a given subset is a subspace;
in fact, some books use it as the definition of a subspace.
Theorem. Let V be a vector space. A subset W ⊆ V is a subspace if it satisfies the following properties:
1. W 6= ∅.
2. For all x, y ∈ W and all c ∈ R, we have cx + y ∈ W .
Property 1 is usually verified by proving that 0V ∈ W .
Property 2 can be replaced by two separate statements:
Closed under addition: For all x, y ∈ W , we have x + y ∈ W .
Closed under scalar multiplication: For all x ∈ W and c ∈ R, we have cx ∈ W .
That is, a subset W of a vector space V is a subspace if and only if W is nonempty, closed under addition,
and closed under scalar multiplication.
Note:
• The empty set ∅ is not a vector space. Instead, the smallest vector space is the trivial space, {0}.
• Every vector space V has two obvious subspaces: the trivial subspace {0} ⊆ V , and the improper
subspace V ⊆ V . (Obviously, the two coincide if and only if V = {0} is trivial.)
1 (March 2006) Let U and V be subspaces of a vector space W .
(a) Prove that U ∩ V is a subspace of W .
(b) Prove that U + V = {u + v : u ∈ U, v ∈ V } is a subspace of W .
(c) Give an example to show that U ∪ V need not be a subspace of W .
Proof. (a): (Nonempty) We have 0W ∈ U and 0W ∈ V , since both U and V are subspaces. Hence
0W ∈ U ∩ V .
(Closure) Given x, y ∈ U ∩V and c ∈ R, we have cx +y ∈ U and cx +y ∈ V since both are subspaces.
Hence cx + y ∈ U ∩ V . QED (a)
(b): (Nonempty) We have 0W ∈ U and 0W ∈ V , since both are subspaces. Hence 0W + 0W ∈ U + V .
(Closure) Given x, y ∈ U + V and c ∈ R, there exist u1 , u2 ∈ U and v1 , v2 ∈ V such that x = u1 + v1
and y = u2 + v2 . Thus,
cx + y = c(u1 + v1 ) + (u2 + v2 ) = (cu1 + u2 ) + (cv1 + v2 ) ∈ U + V QED (b)
(c): Let W = R2 , let U = Span({(1, 0)}), and let V = Span({(0, 1)}). [That is, U is the x-axis, and
V is the y-axis.] Then U and V are subspaces because they are each the span of a set. However,
(1, 0) ∈ U ⊆ U ∪ V and (0, 1) ∈ V ⊆ U ∪ V , but (1, 0) + (0, 1) = (1, 1) 6∈ U ∪ V . Thus, U ∪ V is not
closed under addition and hence is not a subspace. QED (c)
Comment 1. In the “Nonempty” step of (b), there was no need to observe that 0W + 0W = 0W .
Yes, that’s true, and it wouldn’t hurt to say it, but it would be unnecessary. We’re simply trying to
show that U + V 6= ∅, so all we need to do is produce an element; we aren’t required to simplify that
element. On the other hand, in part (c), it was important to simplify the sum (1, 0) + (0, 1), to verify
that the result is not an element of U ∪ V .
Comment 2. In part (c), we made use of Span (to be discussed below) because it made the proof
shorter, rather than verifying by hand that both U and V are subspaces. Also in part (c), there are
many ways to do this. (Any choice of W with any subspaces U and V will work, as long as U 6⊆ V
and V 6⊆ U .) But it’s generally best to pick as simple an example as you can.
2
, Linear combinations, etc. Let V be a vector space, and let S = {v1 . . . , vn } ⊆ V be a finite set of
vectors in V .
• A linear combination of elements of S is an expression a1 v1 +· · ·+an vn for some scalars a1 , . . . , an ∈ R.
• The span of S, denoted Span(S), is the set of all linear combinations of elements of S. That is,
Span(S) = {a1 v1 + · · · + an vn | a1 , . . . , an ∈ R}.
– Fact: Span(∅) = {0}.
– Fact: Span(S) is a subspace of V .
– Fact: If W is a subspace and S ⊆ W , then Span(S) ⊆ W .
– If Span(S) = W , we say that S spans W .
• The set S is linearly dependent if there exist scalars a1 , . . . , an ∈ R that are not all zero such that
a1 v1 + · · · + an vn = 0.
S is linearly independent if it is not linearly dependent. Equivalently, for any scalars a1 , . . . , an ∈ R
such that a1 v1 + · · · + an vn = 0, we must have a1 = · · · = an = 0.
• The set S ⊆ V is a basis for V if S is linearly independent and Span(S) = V .
• The dimension of V is the number dim(V ) of elements in a basis for V . (It is a Theorem that any two
bases for V have the same number of elements.) If V has no finite basis, we say dim(V ) = ∞.
Side notes: Linear independence really should include an extra specification that we first ensure v1 , . . . , vn
are all distinct; but this subtlety does not arise on the comps. Also, there are notions of linear combinations,
span, and linear (in)dependence for an infinite set S ⊆ V , but again, these do not arise on the comps.
2 (March 2007) Suppose that V is a vector space and {v1 , . . . , vn } ⊆ V is linearly independent.
Also assume that v ∈ V is not contained in the span of {v1 , . . . , vn }. Prove that {v, v1 , . . . , vn } is
linearly independent.
Proof. Given scalars a0 , a1 , . . . , an ∈ R such that a0 v + a1 v1 + · · · + an vn = 0, we need to show that
a0 = a1 = · · · = an = 0.
Case 1 : Suppose a0 = 0. Then a1 v1 + · · · + an vn = 0, and hence a1 = · · · = an = 0, because
{v1 , . . . , vn } is linearly independent. Thus, a0 = a1 = · · · = an = 0, as desired.
Case 2 : Suppose a0 6= 0. Then a0 v = −a1 v1 − · · · − an vn , and multiplying both sides by 1/a0 (which
a1 an
is a real number, since a0 6= 0), we have v = − v1 −· · ·− vn ∈ Span({v1 , . . . , vn }), contradicting
a0 a0
the hypotheses. Thus, Case 2 cannot happen, and we are done by Case 1. QED
Comment 1. Don’t start a proof by regurgitating what the hypotheses mean. Instead, focus on the
statement you’re asked to deduce: for this problem, that the set {v, v1 , . . . , vn } is linearly indepen-
dent. So our first line should be to suppose we are given scalars ai for which a0 v+a1 v1 +· · ·+an vn = 0,
and our last line should be to conclude that a0 = a1 = · · · = an = 0. Once you know the first and
last line, then you can go back and try to figure out how the hypotheses could help you get there.
Comment 2. In this problem, two cases are required. We discovered that by doing some scratchwork,
and realizing that the span hypothesis could only be used in one situation, and the linear independence
hypothesis can be used in the other. Scratchwork is important!
Row reduction. Many problems in linear algebra end up requiring row reduction. You will need to know:
• How to set up a system of equations when necessary.
• The three kinds of row reduction steps. (Switching two rows, multiplying a row by a nonzero scalar,
and adding a multiple of one row to another.)
• The overall strategy, leading to a matrix in echelon form.
• How to interpret the echelon form to solve your problem.
3
