† MAT1341 I NTRODUCTION TO L INEAR A LGEBRA E LIZABETH M ALTAIS

1. Vectors in Rn

We begin MAT1341 with a friendly review of vectors, starting with concrete geometric examples
in 2- and 3-space. Once comfortable in those spaces, we will be ready to explore the abstract realm
of “vector spaces”. By reframing vectors in terms of their abstract definitions and properties, we
will be able to capture the true essence of what vectors really are, and we will discover that there
are many surprising and interesting examples, beyond Rn .

V ECTORS ( AS PRESENTED IN HIGH SCHOOL )

We start with the high school definition:

Definition 1.1. A V ECTOR is an object that has:

We picture vectors as arrows, where the length of the arrow corresponds to its magnitude and the
direction the arrow points is the direction of the vector.

We can represent a vector by its picture, or give it a name, such as v.

Note: Usually, the name of a vector is typed using boldface: ex: v x a
But, if we are writing a vector’s name by hand, we tend to add an arrow on top (because it’s
difficult to convey boldface by hand).
Two vectors are E QUAL if they have the same magnitude and direction.

We have two operations: V ECTOR A DDITION and S CALAR M ULTIPLICATION .

• We add vectors tail-to-tip.
C LOSED U NDER A DDITION : The sum of two vectors is again a vector.

†
These notes are intended for students registered in MAT1341. ©EJM
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 • We multiply a vector v by a S CALAR k (which is a real number) by creating a parallel
vector with the correspondingly scaled length. If the scalar is positive, the new vector has
the same direction as v. If the scalar is negative, the new vector has the opposite direction
as v.
C LOSED U NDER S CALAR M ULTIPLICATION :
A vector multiplied by a scalar is again a vector.

Vectors have a “zero concept” called THE Z ERO V ECTOR , denoted 0. The Z ERO V ECTOR has no
magnitude (i.e. zero magnitude) and it does not have a well-defined direction.
What if we add a vector v to the zero vector 0 ?

Each vector v has a N EGATIVE , denoted v. The N EGATIVE of v has the same magnitude as v
but the opposite direction.

What if we add a vector v to its negative v?

Recall: Vector addition and scalar multiplication also satisfy the following properties.
Let u, v, w be vectors and let c, d 2 R be scalars.
V ECTOR A DDITION IS C OMMUTATIVE

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V ECTOR A DDITION IS A SSOCIATIVE

S CALAR M ULTIPLICATION D ISTRIBUTES OVER V ECTOR A DDITION

A DDITION OF S CALARS D ISTRIBUTES OVER S CALAR M ULTIPLICATION

M ULTIPLICATION OF S CALARS IS C OMPATIBLE WITH S CALAR M ULTIPLICATION

U NITY L AW

Note: What we’re really saying is that the arrows we draw, equipped with our geometric rules for
vector addition and scalar multiplication, satisfy the A XIOMS OF A V ECTOR S PACE ( OVER R) !
For now, we will focus on how these properties work in R, R2 , R3 ,. . . , Rn , . . . , both algebraically
and geometrically. Later, we will reconsider these properties in terms of abstract vector spaces.
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We also use coordinate systems to help us represent vectors.

• We can refer to a vector using its I NITIAL P OINT (the TAIL of the arrow) and its
T ERMINAL P OINT (the T IP or H EAD of the arrow). The coordinates of these two points
indicate the vector’s position relative to the axes.
!
Ex. Points: A(1, 1) and B(3, 4) Vector AB
!
Ex. Points: O(0, 0) and D(2, 3) Vector OD
! !
Note: AB = OD since they have the same magnitude and direction.
• We can refer to a vector using its C OORDINATES , which indicate the position of its tip
relative to its tail.
Ex. v = (2, 3)
• We often write vectors as a C OLUMN V ECTOR which is a matrix with one column:

2
Ex. v =
3
Using a column vector is nicer and this notation is compatible with matrix multiplication
(we’ll see more about this later in the course).
• But, it saves space on paper to write vectors as comma-separated rows, or by using the
matrix-transpose notation (the T stands for T RANSPOSE , which changes the rows of a
matrix into columns and vice versa):

2 ⇥ ⇤T
v= = (2, 3) = 2 3
3

2
We can see by the Pythagorean theorem that the L ENGTH of the vector v = is
3


x1
For x = , we have:
x2

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 Rn

While the arrows we draw work best on T HE P LANE (i.e. R2 ), or in 3- SPACE (i.e. R3 ), the
coordinate system representation of vectors generalizes quite naturally to higher (or lower)
dimensions!
n=1 R T HE R EAL L INE

n=2 R2 T HE P LANE

n=3 R3 3- SPACE

n=4 R4 4- SPACE

..
.

n2Z Rn n- SPACE
n>0

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 V ECTOR A DDITION AND S CALAR M ULTIPLICATION

2 3 2 3
u1 v1
Let u = 4 .. 5, and v = 4 ... 5 be vectors in Rn . Let k 2 R be a scalar.
.
un vn
Note: The E NTRIES or C OMPONENTS of these vectors are scalars, i.e. u1 , . . . , un , v1 , . . . , vn 2 R.
Again, we have two operations: V ECTOR A DDITION and S CALAR M ULTIPLICATION .
• We add vectors in Rn C OMPONENT- WISE .

C LOSED U NDER A DDITION : The sum of two vectors in Rn is again a vector in Rn .

• We scalar-multiply a vector in Rn C OMPONENT- WISE .

C LOSED U NDER S CALAR M ULTIPLICATION :

A vector in Rn multiplied by a scalar in R is again a vector in Rn .
What is the Z ERO V ECTOR in Rn ? It should satisfy 0 + v = v for all v 2 Rn .

For a vector v 2 Rn , what is its N EGATIVE v? The negative of v should satisfy v + ( v) = 0.

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Proposition 1.2. Let u, v, w 2 Rn and let c, d 2 R. Then the following properties hold:
• u+v =v+u
• u + (v + w) = (u + v) + w
• c(u + v) = cu + cv
• (c + d)u = cu + du
• c(du) = (cd)u
• 1u = u

Proof that scalar multiplication distributes over vector addition:

E XERCISE : Prove that the other properties hold.

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 L INEAR C OMBINATIONS

With vector addition and scalar multiplication, what can we accomplish?

More precisely, if we are given m vectors in Rn , say u1 , u2 , . . . , um 2 Rn , what other vectors can we
create by scaling and/or adding these vectors together? That is, what vectors v can be expressed
as a L INEAR C OMBINATION of the vectors u1 , . . . , um ?

Definition 1.3. Let u1 , . . . , um be vectors in Rn and let k1 , . . . , km be scalars in R.

A L INEAR C OMBINATION of u1 , . . . , um is a vector v of the form

  
3 1 2
Example 1.4. In R , is v =
2
a linear combination of the vectors u1 = , u2 = ?
12.6 0 3

h i h i h i
1 0 0
E XERCISE ! Show that every vector in R is a linear combination of 0 , 1 , 0 .
3
0 0 1
8
 2 3 2 3 2 3
1 2 0
Example 1.5. Is w = 1 a linear combination of v1 = 0 , v2 = 15 ?
4 5 4 5 4
1 1 2

D OT P RODUCT, N ORM , AND D ISTANCE

Recall: for two vectors x = (x1 , x2 , x3 ) and y = (y1 , y2 , y3 ) in R3 , their D OT P RODUCT is defined as:

x · y = x1 y1 + x2 y2 + x3 y3
We can then view the L ENGTH or N ORM of x as follows:

p p
kxk = x·x= x1 2 + x2 2 + x3 2
We then generalize this to Rn .

Definition 1.6. Let x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) be vectors in Rn . Their D OT P RODUCT is

defined as:

The N ORM of x is defined as:

The dot product and norm, as defined above, have some nice properties:

Proposition 1.7. Let u, v, w 2 Rn and let k 2 R. Then

• u·v =v·u
• u·0=0 (a dot product with the zero vector yields 0 the number zero!)
• (ku) · v = k(u · v) = u · (kv)
• (u + v) · w = u · w + v · w
• kvk 0
• kvk = 0 () v = 0
• kkvk = |k| kvk E XERCISE ! Verify each of the statements in Prop. 1.7.
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  
2 2
Example 1.8. Let u = ,v= .
4 1

Find u · v.

Find kvk.

Find a U NIT V ECTOR parallel to v.

In general, for v 2 Rn such that v 6= 0, we can create a U NIT V ECTOR v̂ in the same direction as v
as follows:

Although vectors are “portable”, using the norm, we define the D ISTANCE between two vectors
as follows:

Definition 1.9. Let x, y 2 Rn . The D ISTANCE between x and y is defined

In two dimensions, we see that kx yk represents the distance between the tips of the vectors,
when they are placed tail-to-tail:

2 3 2 3
1 1
E XERCISE ! Find the distance between 4 2 5 and 4 0 5.
3 1
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 O RTHOGONALITY AND A NGLES BETWEEN V ECTORS
Not only does the dot product give us a way to compute norms of vectors and distances between
vectors, but it also allows us to find A NGLES between vectors.
We’ll start with one of the nicest angles to consider: ⇡/2 (90 )

Definition 1.10. Let u, v be two vectors in Rn .

We say that u and v are O RTHOGONAL (or P ERPENDICULAR ) if

Note: if u = 0 or if v = 0, then u · v = 0.
Thus, the zero vector 0 is considered to be orthogonal to all other vectors (even though in the
picture, we cannot convey a right angle between a vector and the zero vector).

What about other angles? We can imagine placing two nonzero vectors u and v tail-to-tail and
measuring the angle formed between them as follows:

We use these ideas to generalize what we mean by “angle” between two vectors in spaces beyond
R2 and R3 .

Definition 1.11. Let u, v be two nonzero vectors in Rn . The A NGLE B ETWEEN u AND v is defined
to be the number ✓ such that:

In order for the above definition to make sense, we need kuk = 6 0 and 1  kukkvk
6 0, kvk = u·v
 1.
The norms will both be nonzero since u 6= 0 and v 6= 0. The bounds on kukkvk follow from the
u·v

theorem below.

Theorem 1.12. (C AUCHY–S CHWARZ I NEQUALITY ) If u, v 2 Rn , then |u · v|  kuk · kvk

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Corollary 1.13. (T RIANGLE I NEQUALITY ) If If u, v 2 Rn , then ku + vk  kuk + kvk

Proof:

2 3 2 3
1 0
Example 1.14. Find the angle between u = 4 1 5 and v = 35.
4
2 1

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 P ROJECTION

The way we defined the angle between two vectors generalizes what we see in triangles in R2 :

In the above illustration, we decomposed the “v-edge of the triangle” into the vector p and v p
exactly so that p and u p were orthogonal (in the picture, we dropped a perpendicular to show
the height of the triangle, making a 90 angle with the base).
Now, with our definition for angles between vectors in Rn , we want to generalize this idea, as
follows:
Question:

Answer:

Definition 1.15. Let u and v be nonzero vectors in Rn . Then the P ROJECTION OF u ONTO v ,
denoted projv (u), is the unique vector which satisfies:
• projv (u) is parallel to v (i.e. projv (u) is a scalar multiple of v), and
• u projv (u) is orthogonal to v (i.e. v · (u projv (u)) = 0)

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Proposition 1.16. Let u and v be nonzero vectors in Rn . Then
u·v
projv (u) = k · v where k=
kvk2
u·v
That is projv (u) = v
kvk2

Proof:

Example 1.17. Let u = (1, 3) and v = (2, 1). Find projv (u) and find proju (v).

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