2. Linear Algebra for Machine Learning: Basis and Dimension

Seminar Series on
Linear Algebra for Machine Learning
Part 2: Basis and Dimension
Dr. Ceni Babaoglu
Ryerson University
cenibabaoglu.com
Dr. Ceni Babaoglu cenibabaoglu.com
Linear Algebra for Machine Learning: Basis and Dimension

Overview
1 The span of a set of vectors
2 Linear dependence and independence
3 Basis and Dimension
4 Change of Basis
5 Changing Coordinates
6 Orthogonal and Orthonormal Bases
7 References

Vector Space
A real vector space is a set V of elements on which we have two
operations: ⊕ and deﬁned with the following properties.
If x and y are any elements in V , then x ⊕ y ∈ V . (V is
closed under ⊕)
A1. x ⊕ y = y ⊕ x for x, y ∈ V .
A2. (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) for x, y, z ∈ V .
A3. There exists an element 0 in V such that x ⊕ 0 = x for x ∈ V .
A4. For x ∈ V , there exists an element −x such that x ⊕ (−x) = 0.
If x is any element in V and α is any real number, then
α ⊕ x ∈ V . (V is closed under )
A5. α (x ⊕ y) = α x ⊕ α y for α ∈ R and x, y ∈ V .
A6. (α + β) x = α x ⊕ β x for α, β ∈ R and x, y ∈ V .
A7. (α β) x = α (β x) for α, β ∈ R and x, y ∈ V .
A8. 1 x = x for x ∈ V .

The span of a set of vectors
The set of all linear combinations of a set of vectors
{v1, v2, · · · , vn} is called the span.
Let S = {v1, v2, · · · , vn} be a subset of a vector space V . We say
S spans V , if every vector in V can be written as a linear
combination of vectors in S:
v = c1v1 + c2v2 + · · · + cnvn.
Note: A linear combination is in the form
c1v1 + c2v2 + · · · + cnvn
where ci ’s are real numbers.

Linear dependence and independence
The vectors
{v1, v2, · · · , vn}.
are linearly dependent if there exist scalars,
ci , i = 1, 2, · · · , n,
not all zero, such that
c1v1 + c2v2 + · · · + cnvn = 0.
The vectors
{v1, v2, · · · , vn}
are linearly independent if and only if
ci = 0, i = 1, 2, · · · , n,
is the only solution to
c1v1 + c2v2 + · · · + cnvn = 0.

Basis
The vectors
v1, v2, · · · , vn
form a basis for a vector space V if and only if
(i) v1, v2, · · · , vn are linearly independent.
(ii) v1, v2, · · · , vn span V.
If {v1, v2, · · · , vn} is a spanning set for a vector space V , then
any collection of m vectors in V, where m > n, is linearly
dependent.
If {v1, v2, · · · , vn} and {u1, u2, · · · , um} are both bases for a
vector space V , then n = m.

Dimension
If a vector space V has a basis consisting of n vectors, we say that
V has dimension n.
If V is a vector space of dimension n, then
(I) Any set of n linearly independent vectors spans V .
(II) Any n vectors that span V are linearly independent.
(III) No set of less than n vectors can span V .
(IV) Any subset of less than n linearly independent vectors can be
extended to form a basis for V .
(V) Any spanning set containing more than n vectors can be pared
down to form a basis for V .

Example
Rn is n-dimensional. Let x = [x1 x2 x3 · · · xn]T ∈ Rn. Then





x1
x2
...
xn





= x1







1
0
0
...
0







+ x2







0
1
0
...
0







+ · · · + xn







0
0
0
...
1







≡ x1i1 + x2i2 + · · · + xnin.
Therefore span{i1, i2, · · · , in} = Rn.
i1, i2, · · · , in are linearly independent.
They form a basis and this implies the dimension of R is n.

Note
The basis in the previous example is the standard basis for Rn.
In R3 the following notation is used for the standard basis:
i =


1
0
0

 , j =


0
1
0

 , k =


0
0
1

 .
In an n-dimensional vector space V , set of n elements that span V
must be independent and any set of n independent elements must
span V .

Change of Basis
The standard basis for R2 is {e1, e2}. Any vector x ∈ R2 can be
expressed as a linear combination
x = x1e1 + x2e2.
The scalars x1 and x2 can be thought of as the coordinates of x
with respect to the standard basis.
For any basis {y, z} for R2, a given vector x can be represented
uniquely as a linear combination,
x = αy + βz.
[y, z]: ordered basis
(α, β)T : the coordinate vector of x with respect to [y, z]
If we reverse the order of the basis vectors and take [z, y], then
we must also reorder the coordinate vector and take (β, α)T .

Example
y = (2, 1)T and z = (1, 4)T are linearly independent and form a
basis for R2.
x = (7, 7)T can be written as a linear combination:
x = 3y + z
The coordinate vector of x with respect to [y, z] is (3, 1)T .
Geometrically, the coordinate vector speciﬁes how to get from the
origin O(0, 0) to the point P(7, 7), moving ﬁrst in the direction of
y and then in the direction of z .

Changing Coordinates
Suppose, for example, instead of using [e1, e2] for R2, we wish to
use a different basis, say
u1 =
3
2
, u2 =
1
1
.
I. Given a vector c1u1 + c2u2, let’s find its coordinates with
respect to e1 and e2.
II. Given a vector x = (x1, x2)T , let’s find its coordinates with
respect to u1 and u2.

I. Given a vector c1u1 + c2u2, let’s ﬁnd its coordinates with
respect to e1 and e2.
u1 = 3e1 + 2e2, u2 = e1 + e2
c1u1 + c2u2 = (3c1e1 + 2c1e2) + (c2e1 + c2e2)
= (3c1 + c2)e1 + (2c1 + c2)e2
x =
3c1 + c2
2c1 + c2
=
3 1
2 1
c1
c2
U = (u1, u2) =
3 1
2 1
U: the transition matrix from [u1, u2] to [e1, e2]
Given any coordinate c with respect to [u1, u2], the corresponding
coordinate vector x with respect to [e1, e2] by
x = Uc

II. Given a vector x = (x1, x2)T , let’s find its coordinates with
respect to u1 and u2.
We have to find the transition matrix from [e1, e2] to [u1, u2].
The matrix U is nonsingular, since its column vectors u1 and u2
are linearly independent.
c = U−1
x
Given vector x,
x = (x1, x2)T
= x1e1 + x2e2
We need to multiply by U−1 to find its coordinate vector with
respect to [u1, u2].
U−1: the transition matrix from [e1, e2] to [u1, u2]

Example
Let E = (1, 2)T , (0, 1)T and F = (1, 3)T , (−1, 2)T be ordered
basis for R2.
(i) Find the transition matrix S from the basis E to the basis F.
(ii) If the vector x ∈ R2 has the coordinate vector [x]E = (5, 0)T
with respect to the ordered basis E, determine the coordinate
vector [x]F with respect to the ordered basis F.

Example
(i) We should ﬁnd the matrix S =
a c
b d
such that
[x]F = S [x]E for x ∈ R2. We solve the following system:
1
2
= a
1
3
+ b
−1
2
0
1
= c
1
3
+ d
−1
2
⇒ 1 = a − b 2 = 3a + 2b
0 = c − d 1 = 3c + 2d
⇒ a = 4/5 b = −1/5 c = d = 1/5
⇒ S =
4/5 1/5
−1/5 1/5

Example
E -V (1, 0)T , (0, 1)T
?
U−1
Z
Z
Z
Z
Z
Z
Z
ZZ~
U−1V
F
where V =
1 0
2 1
and U =
1 −1
3 2
. Thus we obtain
U−1 =
2/5 1/5
−3/5 1/5
and then S = U−1 V =
4/5 1/5
−1/5 1/5
.

Example
(ii) Since [x]E = (5, 0)T ,
[x]F = S [x]E =
4/5 1/5
−1/5 1/5
5
0
=
4
−1

Orthogonal and Orthonormal Bases
n linearly independent real vectors span Rn and they form a basis
for the space.
An orthogonal basis, a1, · · · , an satisﬁes
ai · aj = 0, if i = j
An orthonormal basis, a1, · · · , an satisﬁes
ai · aj = 0, if i = j
ai · aj = 1, if i = j

References
Linear Algebra With Applications, 7th Edition
by Steven J. Leon.
Elementary Linear Algebra with Applications, 9th Edition
by Bernard Kolman and David Hill.

2. Linear Algebra for Machine Learning: Basis and Dimension

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