NOTES WEEK 03 DAY 2
SCOT ADAMS
DEFINITION 0.1. Let n P N. We define Rn :“ Rt1,...,nu . For all
a P Rn , we have a : t1, . . . , nu Ñ R and, for all j P t1, . . . , nu, we use
aj to denote apjq. For all a P Rn , we use pa1 , . . . , an q to denote a.
So, for example, if a “ p4, 5, 6q, then a P R3 “ Rt1,2,3u . Then we have
a : t1, 2, 3u Ñ R and a1 “ ap1q “ 4, a2 “ ap2q “ 5, a3 “ ap3q “ 6.
DEFINITION 0.2. For all n P N, we define 0n :“ p0, . . . , 0q P Rn .
DEFINITION 0.3. Let m, n P N. Define Rmˆn :“ Rt1,...,muˆt1,...,nu .
For all A P Rn , we have A : t1, . . . , mu ˆ t1, . . . , nu Ñ R and, for
all j P t1, . . . , mu, k P t1, .»
. . , nu, we use Ajk or
fi Aj,k to denote Apj, kq.
A11 ¨ ¨ ¨ A1n
— ..
.. ffi to denote A.
mˆn
For all A P R
, we use – .
. fl
Am1 ¨ ¨ ¨ Amn
»
fi
2 3
So, for example, if A “ – 4 5 fl, then A P R3ˆ2 . Then we have
6 7
A : t1, 2, 3u ˆ t1, 2u Ñ R and, for example, A31 “ A3,1 “ Ap3, 1q “ 6.
»
fi
0 ¨¨¨ 0
—
.. ffi P Rmˆn .
DEFINITION 0.4. For all m, n P N, 0mˆn :“ – ...
. fl
0 ¨¨¨ 0
DEFINITION 0.5. For all m, n P N, for all A P Rmˆn , we define
At P Rn ˆ m by pAt qjk “ Akj .
»
fit
„

2 3
2
4
6
So, for example – 4 5 fl “
.
3 5 7
6 7
DEFINITION 0.6. Let n P N, A P Rnˆn . Then A is symmetric
means At “ A, and A is anti-symmetric means At “ ´A.
Date: February 2, 2017
Printout date: February 20, 2017.
1
2
SCOT ADAMS
nˆn
DEFINITION 0.7. Let n P N. Then Rsym
:“ tA P Rnˆn | At “ Au,
and Rnˆn
:“ tA P Rnˆn | At “ ´Au.
as
»
fi
u1
—
ffi
DEFINITION 0.8. Let n P N. For all u P Rn , we define uV :“ – ... fl
H
“
‰
nˆ1
un
Ñ
Ý
, u :“ pu11 , . . . , un1 q. For
and u :“ u1 ¨ ¨ ¨ un . For all u P R
Ý
all u P R1ˆn , Ñ
u :“ pu11 , . . . , u1n q.
» fi
1
“
‰
So, for example p1, 2, 3qV “ – 2 fl and p1, 2, 3qH “ 1 2 3 . Also,
3
ÝÝÝÝÑ
»
fi
4
“ÝÝÝÝÝÝÝÝщ
–
we have
5 fl “ p4, 5, 6q and 2 7 8 “ p2, 7, 8q. Such a long ar6
row is“ sometimes
hard on the eyes, and we may choose, instead, to
‰Ñ
Ý
write 2 7 8
“ p2, 7, 8q.
DEFINITION 0.9. Let m,
, . . . , un P Rmˆ1 , we
» n P N. For all u1fi
pu1 q1 ¨ ¨ ¨ pun q1
‰
“
— ..
.. ffi P Rmˆn .
:“ – .
define u1 ¨ ¨ ¨ un
. fl
ˆ
»
pu1 qm ¨ ¨ ¨
fi
pun qm
»
fi
1
4
For example, let u1 :“ – 2 fl and let u2 :“ – 5 fl. Then
3
6
»
fi
1 4
“
‰
u1 u2 ˆ “ – 2 5 fl .
3 6
nˆ1
DEFINITION
»
fi 0.10.
» Let m, n P N. Forfi all u1 , . . . , um P R , we
u1
pu1 q1 ¨ ¨ ¨ pu1 qn
— .. ffi
— ..
.. ffi P Rnˆm .
define – . fl :“ – .
. fl
um ˆ
pum q1 ¨ ¨ ¨ pum qn
“
‰
“
‰
For example, let u1 :“ 1 2 3 and let u2 :“ 4 5 6 . Then
„

„

u1
1 2 3
“
.
u2 ˆ
4 5 6
NOTES WEEK 03 DAY 2
3
DEFINITION 0.11. Let m, n P N and let L P Rmˆn . Then, for
all j P t1, . . . , mu, we define Lj‚ P Rn by pLj‚ qk “ Ljk . Also, for
all k P t1, . . . , nu, we define L‚k P Rn by pL‚k qj “ Ljk .
DEFINITION 0.12. A vector space is a set V together with
(α) a zero element 0V P V
(β) a negation v ÞÑ ´v : V Ñ V
(γ) a vector addition pv, wq ÞÑ v ` w : V ˆ V Ñ V
and
(δ) a scalar multiplication pc, vq ÞÑ cv : R ˆ V Ñ V
such that
(1) @v, w P V , v ` w “ w ` v
(2) @v, w, x P V , pv ` wq ` x “ v ` pw ` xq
(3) @v P V , 0V ` v “ v
(4) @v P V , v ` p´vq “ 0V
(5) @v P V , 1v “ v
(6) @a, b P R, @v P V , pabqv “ apbvq
(7) @a, b P R, @v P V , pa ` bqv “ av ` bv
and
(8) @a P R, @v, w P V , apv ` wq “ av ` aw.
What we call a vector space here is often called a real vector
space or R-vector space or vector space over the reals. In this
exposition, we will not be using other scalar fields besides R.
For any vector space V , the vector addition (γ) and scalar multiplication (δ) are called the linear operations in V , and we will refer to
the zero element (α) and the negation (β) as the extras in V . It’s the
linear operations in a vector space that are most important; the extras
are used, but, as we’ll see, aren’t nearly as important.
Requirements (1)-(4) are expressed by saying that pV, 0V , ´, `q is an
additive group. Requirements (5)-(6) are expressed by saying that the
monoid pR, 1q acts on V by scalar multiplication. Requirement (7) is
expressed by saying that scalar multiplication distributes over scalar
addition (from the right). Requiment (8) is expressed by saying that
scalar multiplication distributes over vector addition (from the left).
So we have four group laws, two action laws and two distributive laws.
When specifying a vector space, we often just give V and expect
the reader to intuit what (α), (β), (γ) and (δ) are. Occasionally, in
case of confusion, we specify the linear operations (γ) and (δ), but
almost always leave it to the reader to find the unique zero element
(α) and negation (β). So, for example, R is a vector space, where
4
SCOT ADAMS
vector addition is just the usual addition of real numbers and scalar
multiplication is just the usual multiplication of real numbers. Below,
we will explain how, for every n P N, Rn is a vector space. Also, we
will explain how, for every set S, RS is a vector space.
Let V be a vector space. For all v, w P V , we define v´w :“ v`p´wq.
For all c P R, for all v P V , we define c ¨ v :“ cv. For all v P V , for
all h P Rzt0u, we define v{h :“ r1{hs ¨ v
The easiest example of a vector space is the zero vector space t0u Ď
R. The second easiest is R itself, using the regular addition, multiplication, 0 and negation.
Let S be a set and V a vector space. We noted that the set V S of
functions V Ñ W is a vector space, under the usual addition of maps
(pf ` gqpsq “ rf psqs ` rgpsqs) and the usual scalar multiplcation on
maps pcf qpsq “ c ¨ rf psqs).
Keep in mind, as a special case: For any set S, RS is a vector space.
Let n P N. Then Rn is a vector space, because Rn “ Rt1,...,nu , so
we can use S “ t1, . . . , nu in the preceding paragraph. The vector
addition defined by pv ` wqj “ vj ` wj and the scalar multiplication by
pcvqj “ c ¨ vj . Negation is also defined component by component, and
0Rn “ 0n .
Let m, n P N. Then Rmˆn is a vector space, because Rmˆn “
t1,...,muˆt1,...,nu
R
, so we can use S “ t1, . . . , mu ˆ t1, . . . , nu two paragraphs ago. The vector addition defined by pA ` Bqjk “ Ajk ` Bjk and
the scalar multiplcation by pcAqjk “ c ¨ Ajk . Negation is also defined
entry by entry, and 0Rmˆn “ 0mˆn .
DEFINITION 0.13. For all n P N, for all u, v P Rn , we define
u ¨ v “ u1 v1 ` ¨ ¨ ¨ ` un vn .
Here’s a game: I pick a secret u P R3 . You tell me k P N and
v1 , . . . , vk P R3 . I tell you u ¨ v1 , . . . , u ¨ vk , which are real numbers. I
tell you a w P R3 . You guess u ¨ w, which is a real number. If you guess
right, you win. Otherwise, I win.
After a brief discussion, we decided on a strategy. Let k “ 3 and let
v1 , v2 , v3 be the standard ordered basis of R3 . That is, v1 “ p1, 0, 0q,
v2 “ p0, 1, 0q and v3 “ p0, 0, 1q. This forces me to tell you u ¨ v1 “ u1 ,
u¨v2 “ u2 and u¨v3 “ u3 , from which you can figure out u “ pu1 , u2 , u3 q.
Then you don’t have to guess; you can calculate u ¨ w “ u1 w1 ` u2 w2 `
NOTES WEEK 03 DAY 2
5
u3 w3 . (When reading this, keep in mind that u1 , u2 , u3 , w1 , w2 , w3 P R
are scalars, but u, v1 , v2 , v3 , w P R3 are vectors.)
DEFINITION 0.14. Let V and W be vector spaces. Let L : V Ñ W .
Then L is linear means:
(i) @u, v P V , Lpu ` vq “ rLpuqs ` rLpvqs,
(ii) @c P R, @v P V , Lpcvq “ c ¨ rLpvqs.
and
Sometimes a linear map is called a linear transformation.
Keep in mind that there are two ways that the word “linear” gets
used in mathematics: Define f : R Ñ R by f pxq “ 2x ` 3. In high
school and freshman calculus, the function f is typically said to be
“linear”, even though it’s not linear in our sense here. (Note that
f p0 ` 0q ‰ rf p0qs ` rf p0qs.) It would be better to call f an “affine”
map, which is the usual name for a polynomial of degree ď 1. When all
of the terms of a polynomial have the same degree, then the polynomial is said to be “homogeneous”, and f is a typical example of an
inhomogeneous polynomial. The function g : R Ñ R defined by
gpxq “ 4x would be a homogeneous polynomial. Next, define maps
φ, ψ : R2 Ñ R by φpx, yq “ 5x ` 6y ` 7 and ψpx, yq “ 8x ` 9y. Then
φ is an inhomogeneous polynomial of degree ď 1, meaning that every
term in 5x ` 6y ` 7 has degree ď 1. Note that the term 7 has degree
0. On the other hand, ψ is a homogeneous polynomial of degree “ 1,
meaning every term in 8x ` 9y has degree “ 1. Traditionally, in high
school and freshman calculus (and sometimes even in second year calculus), “linear” means “of degree ď 1, but not of degree 0”. In higher
level mathematics, the term of art for this is “affine”, and the word
“linear” is reserved only for homogeneous polynomials of degree “ 1.
Eventually, students move from the world of Euclidean spaces R, R2 ,
. . . , to the world of vector spaces, and another transition has to occur;
“linear” then means “respects the linear operations”. For Euclidean
spaces, this is equivalent to “homogeneous of degree “ 1”.
DEFINITION 0.15. Let V and W be vector spaces. Then LpV, W q
denotes the set of all linear maps V Ñ W .
DEFINITION 0.16. Let V be a vector space. Then LpV q :“ LpV, Rq.
DEFINITION 0.17. Let U be a vector space and let S Ď U . Assume
S ‰ H. By S is a subspace of U , we mean:
6
SCOT ADAMS
(a) S ` S Ď S,
(b) RS Ď S,
i.e.,
i.e.,
( @r, s P S, r ` s P S ).
( @c P R, @s P S, cs P S ).
Any subspace of a vector space is again a vector space, by restriction
vector addition and scalar multiplication. (Unassigned exercise: Show,
for any subspace S of a vector space U , that 0U P S and ´S Ď S. We
convey this by saying, “the extras come along for free”.)
Let V be a set and W a vector space. We noted that the set W V of
functions V Ñ W is a vector space, under the usual addition of maps
(pf `gqpvq “ rf pvqs`rgpvqs) and the usual scalar multiplcation on maps
pcf qpvq “ c ¨ rf pvqs). When V is a vector space, then the linearity two
linearity conditions (i) and (ii) in Definition 0.14 determine a subset
LpV, W q of W V . This subset is a subspace, and LpV, W q is a vector
space by restricting vector addition and scalar multiplication.
Here’s a game: I pick a secret L P LpR2 , R3 q. You tell me k P N and
v1 , . . . , vk P R2 . I tell you Lpv1 q, . . . , Lpvk q, which are vectors in R3 .
I tell you a w P R3 . You guess Lpwq, which is a vector in R3 . If you
guess right, you win. Otherwise, I win.
After a brief discussion, we decided on a strategy. Let k “ 2 and
let v1 , v2 be the standard ordered basis of R2 . That is, v1 “ p1, 0q and
v2 “ p0, 1q. This forces me to tell you Lpv1 q and Lpv2 q. Then you don’t
have to guess; because w “ w1 v1 ` w2 v2 , you can calculate Lpwq “ w1 ¨
rLpv1 qs`w2 ¨rLpv2 qs. (When reading this, keep in mind that w1 , w2 P R
are scalars, but v1 , v2 , v3 , w P R2 and Lpv1 q, Lpv2 q, Lpv3 q, Lpwq P R3 are
vectors.
We actually played the game to show that the strategy works.
DEFINITION 0.18. Let m P N, L P LpRm , Rn q. Let e1 , . . . , em be
the standard ordered basis of Rm . Then the matrix of L is
“
‰
rLs :“
pLpe1 qqV ¨ ¨ ¨ pLpem qqV ˆ P Rnˆm .
WARNING: In Definition 0.18, you see L P LpRm , Rn q, with m on the
left of n. However, you also see rLs P Rnˆm , with n on the left of m.
So the order of m and n reverses when you go from L to rLs.
Let m, n P N. Recall that, for any two vector spaces V and W , we discussed how LpV, W q is again a vector space. In particular, LpRm , Rn q
is a vector space. Recall, also, that Rmˆn is a vector space. Unassigned
HW: Show that L ÞÑ rLs : LpRm , Rn q Ñ Rmˆn is linear.
NOTES WEEK 03 DAY 2
7
DEFINITION 0.19. Let m, n P N, A P Rnˆm . For all j P t1, . . . , nu,
let vj :“ Aj‚ P Rm denote the jth row of A. We define LA P LpRm , Rn q
by LA pxq “ pv1 ¨ x, . . . , vn ¨ xq.
Assigned HW#13 and HW#14.
HW#14 asserts that the two maps
and
L Ñ
Þ
rLs
A Ñ
Þ
LA
:
:
LpRm , Rn q Ñ Rnˆm
Rnˆm Ñ LpRm , Rn q
are inverses. So, since they are both linear, they are both vector space
isomorphisms. In other words, from a linear algebra point of view,
there’s no difference between LpRm , Rn q and Rnˆm .
DEFINITION 0.20. Let V , W , X be vector spaces. Then BpV, W, Xq
denotes the set of maps B : V ˆ W Ñ X such that both
(i) @v P V , the map Bpv, ‚q : W Ñ X is linear
(ii) @w P W , the map Bp‚, wq : V Ñ X is linear.
and
A map B P BpV, W, Xq is said to be an “X-valued bilinear map
on V ˆ W ”. Sometimes a bilinear map is called a bilinear form.
In Definition 0.20, condition (i) is called “linearity in the second
variable”, or, more explicitly, “linearity in the second variable, with
the first variable fixed”. In Definition 0.20, condition (ii) is called
“linearity in the first variable”, or, more explicitly “linearity in the
first variable, with the second variable fixed”.
A very important case is when we simplify by using X “ R:
DEFINITION 0.21. Let V and W be vector spaces. Then we define
BpV, W q :“ BpV, W, Rq.
Let m : R ˆ R Ñ R be multiplication, defined by mpx, yq “ xy.
Then m P BpR, Rq, and m is the easiest example of a bilinear map.
The properties of bilinearity are reminiscent of multiplication, and so,
when studying specific vector spaces V , W and X and a particular
bilinear map B P BpV, W, Xq, it’s common to represent B by ˚ and to
declare, for all v P V and w P W , that v ˚ w “ Bpv, wq.
We »
worked fian example: Let V :“ R1ˆ2 and W :“ R3ˆ1 . Let
2 5
–
M :“ 3 6 fl. Define B P BpV, W q by Bpv, wq “ pvM wq11 . (Note:
4 7
8
SCOT ADAMS
v P R1ˆ2 , M P R2ˆ3 and w P R3ˆ1 , so vM w P R1ˆ1 , so pvM wq11 P R
is the unique entry in the 1 ˆ 1 matrix vM w.) We calculated: for
all p, q, x, y, z P R, we have
¨
» fi ˛
x
“
‰
B ˝ p q , – y fl ‚ “ 2px ` 5qx ` 3py ` 6qy ` 4pz ` 7qz.
z
Here’s a game: I pick a secret B P BpR2 , R3 q. For all v P R2 , for
all w P R3 , we agree to define v ˚ w :“ Bpv, wq. So this multiplication, denoted ˚, takes in two real numbers, then three more, and then
computes, from all five, a single real number, in a bilinar way. You
tell me k P N and v1 , . . . , vk P R2 and w1 , . . . , wk P R3 . I tell you
v1 ˚ w1 , . . . , vk ˚ wk , which real numbers. I tell you an x P R2 and a
y P R3 . You guess x ˚ y, which is a real number. If you guess right, you
win. Otherwise, I win.
After a brief discussion, we decided on a strategy. Let e1 , e2 be the
standard ordered basis of R2 . Let f1 , f2 , f3 be the standard ordered
basis of R3 . Let k “ 6. Let
v1 :“ e1 , v2 :“ e1 , v3 :“ e1 , v4 :“ e2 , v5 :“ e2 , v6 :“ e2 ,
w1 :“ f1 , w2 :“ f2 , w3 :“ f3 , w4 :“ f1 , w5 :“ f2 , w6 :“ f3 .
This forces me to tell you e1 ˚ f1 , e1 ˚ f2 , e1 ˚ f3 , e2 ˚ f1 , e2 ˚ f2 , e2 ˚ f3 .
Then you don’t have to guess; because x “ x1 e1 ` x2 e2 and because
y “ y1 f1 ` y2 f2 ` y3 f3 , you can calculate
x ˚ y “ px1 e1 ` x2 e2 q ˚ py1 f1 ` y2 f2 ` y3 f3 q
“ px1 y1 qpe1 ˚ f1 q ` px1 y2 qpe1 ˚ f2 q ` px1 y3 qpe1 ˚ f3 q `
px2 y1 qpe2 ˚ f1 q ` px2 y2 qpe2 ˚ f2 q ` px2 y3 qpe2 ˚ f3 q.
(When reading this, keep in mind that x1 , x2 , y1 , y2 , y3 P R are scalars,
but v1 , . . . , v6 , x P R2 and w1 , . . . , w6 , y P R3 are vectors.)