3. Linear Algebra for Machine Learning: Factorization and Linear Transformations

Seminar Series on
Linear Algebra for Machine Learning
Part 3: Factorization and Linear Transformations
Dr. Ceni Babaoglu
Data Science Laboratory
Ryerson University
cenibabaoglu.com
Dr. Ceni Babaoglu cenibabaoglu.com
Linear Algebra for Machine Learning: Factorization and Linear Transformations

Overview
1 Row and Column Spaces
2 Rank of a Matrix
3 Rank and Singularity
4 Inner Product Spaces
5 Gram-Schmidt Process
6 Factorization
7 Linear Transformation
8 Linear Transformation and Singularity
9 Similar Matrices
10 References

Row and Column Spaces
Let
A =





a11 a12 a13 . . . a1n
a21 a22 a23 . . . a2n
...
...
...
...
...
am1 am2 am3 . . . amn





be an m × n matrix.
The rows of A, considered as vectors in Rn, span a subspace
of Rn called the row space of A.
Similarly, the columns of A, considered as vectors in Rm, span
a subspace of Rm called the column space of A.
If A and B are two m × n row (column) equivalent matrices,
then the row (column) spaces of A and B are equal.

Rank of a Matrix
The dimension of the row (column) space of A is called the
row (column) rank of A.
The row rank and column rank of the m × n matrix A = [aij ]
are equal.

Rank and Singularity
A is nonsingular.
Ax = 0 has only the trivial solution.
A is row (column) equivalent to In.
For every vector b in Rn, the system Ax = b has a unique
solution.
det(A) = 0.
The rank of A is n.
The rows of A form a linearly independent set of vectors in Rn.
The columns of A form a linearly independent set of vectors in
Rn.

Example
Let A =


1 −1 2 0 −3
0 1 0 4 0
2 −1 4 4 −6

. Find the following:
(i) A basis for the column space of A and its dimension.
(ii) A basis for the row space of A and its dimension.

Example


1 −1 2 0 −3
0 1 0 4 0
2 −1 4 4 −6

 S3−2S1→S3
−−−−−−−→


1 −1 2 0 −3
0 1 0 4 0
0 1 0 4 0


S1+S2→S1
S3−S2→S3
−−−−−−→


1 0 2 4 −3
0 1 0 4 0
0 0 0 0 0


(i) The column space of A is spanned by the vectors (1, 0, 2)T
and
(−1, 1, −1)T
. These vectors are linearly independent.
{(1, 0, 2)T
, (−1, 1, −1)T
} is a basis for this space and its dimension is 2.
(ii) The row space of A is spanned by the vectors (1, 0, 2, 4, −3) and
(0, 1, 0, 4, 0). These vectors are linearly independent.
{(1, 0, 2, 4, −3), (0, 1, 0, 4, 0)} is a basis for this space and its
dimension is 2.

Inner Product Spaces
Let V be a real vector space. An inner product on V is a function
that assigns to each ordered pair of vectors u, v in V real number
(u, v) satisfying the following properties:
(u, u) ≥ 0, (u, u) = 0 if and only if u = 0v
(v, u) = (u, v) for any u, v in V
(u + v, w) = (u, w) + (v, w) for any u, v, w in V
(cu, v) = c(u, v) for u, v in V and c, a real scalar
A real vector space that has an inner product defined on it is
called an inner product space. If the space is finite
dimensional it is called a Euclidean space.
In an inner product space we define the length of
a vector u by u = (u, u).

Gram-Schmidt Process
Let V be an inner product space and W = {0} an m-dimensional
subspace of V . Then there exists an orthonormal basis
T = {w1, w2, · · · , wm} for W .
Let S = {u1, u2, · · · , um} be any basis for W . Construct an
orthogonal basis T∗ = {v1, v2, · · · , vm} for W. Select any one of
the vectors in S, say u1 and call it v1. Look for a vector v2 in the
subspace W1 of W spanned by {u1, u2} that is orthogonal to v1.
v2 = u2 −
(u2, v1)
(v1, v1)
v1

Gram-Schmidt Process
Next, we look for a vector v3 in the subspace W2 of W spanned by
{u1, u2, u3} that is orthogonal to both v1 and v2.
v2 = u2 −
(u2, v1)
(v1, v1)
v1
v3 = u3 −
(u3, v1)
(v1, v1)
v1 −
(u3, v2)
(v2, v2)
v2

Example
Let’s use the Gram-Schmidt process to ﬁnd an orthonormal basis
for the subspace of R4 with basis u1 = (1, 1, 1, 0)T ,
u2 = (−1, 0, −1, 1)T and u3 = (−1, 0, 0, −1)T . First let v1 = u1,
v2 = u2 −
(u2, v1)
(v1, v1)
v1 =




−1
0
−1
1



 − (−
2
3
)




1
1
1
1



 =




−1/3
2/3
−1/3
1




Multiplying v2 by 3 to clear fractions, we get




−1
2
−1
3





Example
v3 = u3 −
(u3, v1)
(v1, v1)
v1 −
(u3, v2)
(v2, v2)
v2 =




−4/5
3/5
1/5
−3/5




Multiplying v3 by 5 to clear fractions, we get (−4, 3, 1, −3)T . An
orthogonal basis:
{v1, v2, v3} =




1
1
1
0



 ,




−1
2
−1
3



 ,




−4
3
1
−3




An orthonormal basis:
{w1, w2, w3} =




1/
√
3
1/
√
3
1/
√
3
0



 ,




−1/
√
15
2/
√
15
−1/
√
15
3/
√
15



 ,




−4/
√
35
3/
√
35
1/
√
35
−3/
√
35





Factorization
If A is an m × n matrix with linearly independent columns, then A
can be factored as
A = QR,
Q: an m × n matrix whose columns form an orthonormal basis for
the column space of A,
R: an n × n nonsingular upper triangular matrix.

Example
Let’s ﬁnd the factorization of




1 −1 −1
1 0 0
1 −1 0
0 1 −1



 .
Let’s deﬁne the columns of A as the vectors u1, u2, u3.
The orthonormal basis for the column space of A is
w1 =




1/
√
3
1/
√
3
1/
√
3
0



 , w2 =




−1/
√
15
2/
√
15
−1/
√
15
3/
√
15



 , w3 =




−4/
√
35
3/
√
35
1/
√
35
−3/
√
35



 .

Example
Q =




1/
√
3 −1/
√
15 −4/
√
35
1/
√
3 2/
√
15 3/
√
35
1/
√
3 −1/
√
15 1/
√
35
0 3/
√
15 −3/
√
35




R =


r11 r12 r13
0 r22 r23
0 0 r33


where rji = (ui , wj ).
R =


3/
√
3 −2/
√
3 −1/
√
3
0 5/
√
15 −2/
√
15
0 0 7/
√
35


A = QR

Linear Transformation
A mapping L : V −→ W is said to be a linear transformation or a
linear operator if
L(αv1 + βv2) = αL(v1) + βL(v2)
OR
L(v1 + v2) = L(v1) + L(v2), (α = β = 1)
L(αv) = αL(v) (v = v1, β = 0)

Example
L(x) = 3x, x ∈ R2
.
L(x + y) = 3(x + y) = L(x + y)
L(αx) = 3(αx) = αL(x)
L is a linear transformation.
α : positive scalar
F(x) = αx can be thought of as a stretching or shrinking by a
factor of α.

Example
L(x) = x1e1, x ∈ R2
.
If x = (x1, x2)T
, then L(x) = (x, 0)T
If y = (y1, y2)T
, then αx + βy =
αx1 + βy1
αx2 + βy2
L(αx + βy) = (αx1 + βy1)e1 = α(x1e1) + β(y1e1) = αL(x) + βL(y)
L is a linear transformation, a projection onto the x1 axis.

Example
L(x) = (−x2, x1)T , x = (x1, x2)T ∈ R2.
L(αx + βy) =
−(αx2 + βy2)
αx1 + βy1
= α
−x2
x1
+ β
−y2
y1
= αL(x) + βL(y)
L is a linear transformation. It has the eﬀect of rotating each vector in R2
by
90◦
in the counterclockwise direction.

Example
M(x) = (x2
1 + x2
2 )1/2, M : R2 −→ R.
M(αx) = (α2
x2
1 + α2
x2
2 )1/2
=| α | M(x),
αM(x) = M(αx), α < 0, x = 0.
M is not a linear transformation.

Linear One-to-one Transformations
A linear transformation L : V → W is called one-to-one if it is
a one-to-one function; that is, if v1 = v2 implies that
L(v1) = L(v2).
An equivalent statement is that L is one-to-one if
L(v1) = L(v2) implies that v1 = v2.

Linear Onto Transformations
If L : V → W is a linear transformation of a vector space V
into a vector space W , then the range of L or image of V
under L, denoted by range L , consists of all those vectors in
W that are images under L of vectors in V .
Thus w is in range L if there exists some vector v in V such
that L(v) = w. The linear transformation L is called onto if
range L = W .

Linear Transformation and Singularity
A is nonsingular.
Ax = 0 has only the trivial solution.
A is row (column) equivalent to In.
For every vector b in Rn, the system Ax = b has a unique
solution.
det(A) = 0.
The rank of A is n.
The rows of A form a linearly independent set of vectors in Rn.
The columns of A form a linearly independent set of vectors in
Rn.
The linear transformation L : Rn −→ Rn deﬁned by
L(x) = A(x), for x in Rn, is one-to-one and onto.

Similar Matrices
If A and B are n × n matrices, we say that B is similar to A if
there is a nonsingular matrix P such that B = P−1AP.
Let V be any n−dimensional vector space and let A and B be
any n × n matrices. Then A and B are similar if and only if A
and B represent the same linear transformation L : V → V
with respect to two ordered bases for V .
If A and B are similar n × n matrices, then rank A = rank B.

Example
Let L : R3 → R3 be deﬁned by
L([ u1 u2 u3 ]) = [ 2u1 − u3 u1 + u2 − u3 u3 ]
and S = {[1 0 0], [0 1 0], [0 0 1]} be the natural basis
for R3. The representation of L with respect of S is
A =


2 0 −1
1 1 −1
0 0 1

 .
Considering S = {[1 0 1], [0 1 0], [1 1 0]} as ordered
bases for R3, the transition matrix P from S to S is
P =


1 0 1
0 1 1
1 0 0

 P−1
=


0 0 1
−1 1 1
1 0 −1

 .

Example
Then the representation of L with respect to S is
B = P−1
AP =


1 0 0
0 1 0
0 0 2


The matrices
A =


2 0 −1
1 1 −1
0 0 1

 and B =


1 0 0
0 1 0
0 0 2


are similar.

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

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