The singular value decomposition (SVD) of a matrix allows us to decompose any (not necessarily square) matrix into a product of three terms:
a unitary matrix;
a matrix having positive entries on its main diagonal and zero entries elsewhere;
another unitary matrix.
Easily recognizable subsets of the columns of the two unitary matrices involved in the decomposition have the remarkable property of being orthonormal bases for the column space and the null space of the original matrix.
The singular value decomposition is derived by using results about the diagonalization of a positive definite matrix. We therefore recommend to revise the lecture on positive definite matrices before reading this one.
Here is the decomposition.
Proposition
Let
be a
matrix. Then, there exist a
unitary matrix
and an
unitary matrix
such
that
where
denotes the conjugate
transpose of
,
is a
matrix such that
if
and
if
.
We start by observing that the
matrixis
Hermitian (i.e., equal to its conjugate transpose), hence
normal and unitarily
diagonalizable (i.e.,
unitarily similar to a diagonal
matrix). Moreover, for any
vector
,
we have
As
a consequence
is positive semi-definite, which implies that its
eigenvalues are
real non-negative numbers. Thus, we can diagonalize
as
where
is a unitary matrix and
is a diagonal matrix such that its diagonal entries are real non-negative
numbers. When we diagonalize a matrix, we can arbitrarily choose the order in
which the eigenvalues appear on the diagonal of
.
Therefore, we can perform the diagonalization in such a manner
that
where
the block
is a diagonal matrix having all the strictly positive eigenvalues of
on its main diagonal. Suppose that there are
strictly positive eigenvalues, that is,
is
.
Partition
into two
blocks:
where
is
.
By using the rules for
the multiplication of block matrices, we
obtain
Note
that
and
the last equation
implies
As
a consequence, the squared norms of the columns of
(which are on the main diagonal of the matrix on the left-hand side of the
equation) are all zero. Hence,
by
the positive definiteness of the
norm. Define the
matrix
where
is a diagonal matrix whose diagonal entries are equal to the square roots of
the diagonal entries of
.
Note that
so
that the columns of
are orthonormal. Find any
matrix
having orthonormal columns and such
that
In
other words, the columns of
are chosen so as to "complete the orthonormal basis", that is, so as to form
an orthonormal basis together with the columns of
.
Stated differently, the columns of
span the
orthogonal complement of
the span of the columns of
.
Define
Thus,
is a unitary matrix.
Then,
By
pre-multiplying both sides by
and post-multiplying both sides by
,
we
obtain
The diagonal entries of
(i.e., the entries
for
)
are called singular values of
.
Example
Ifthen
the singular values are
,
and
.
Example
LetThen,
the singular values are
,
and
.
Example
Ifthen
the singular values are
,
and
.
As shown in the proof above, the singular value decomposition of
is obtained from the diagonalization of
.
But the diagonalization is not unique (as discussed in the lecture on
diagonalization).
Therefore, also the SVD is not unique.
A more compact form of the SVD, called compact SVD, can be obtained as follows.
Proposition
Let
be a
matrix. Let
be the rank of
.
Then, there exist a
matrix
and an
matrix
,
both having orthonormal columns, such
that
where
denotes the conjugate transpose of
and
is an
diagonal matrix with strictly positive entries on its main diagonal.
In this proof we use all the matrices
defined in the above proof of the non-compact SVD. Note that
(i.e., the number of non-zero singular values) is the rank of
and, as a consequence, also the rank of
(because multiplication by
the full-rank matrices
and
does not change the rank of
).
We have
where
we have defined
.
If the matrix
is real (i.e., all its entries are real numbers), then the SVD
is
where
and
are real orthogonal matrices (we just need to replace conjugate transposition
with simple transposition in the proof of the SVD).
Let
be a
matrix and let
be the space of all
column vectors.
Remember that the column
space
is the span of the columns of
:
Proposition
Let
be a
matrix. Let
be the rank of
.
Let
be
an SVD of
such that the
non-negative singular values of
are the first
entries on the main diagonal of
.
Let
be the block formed by the first
columns of
.
Then,
where
is the span of the columns of
and
is the span of the columns of
.
The proof of this proposition is in Solved Exercise 1 below.
Note that
is the matrix appearing in the compact SVD above.
Since the columns of
are orthonormal, they are also linearly independent. Thus, the matrix
in the compact SVD exposes an
orthonormal basis for the
column space of
.
Remember that the null
space
is the kernel of the linear
transformation defined by
:
Proposition
Let
be a
matrix. Let
be the rank of
.
Let
be
an SVD of
such that the
non-negative singular values of
are the first
entries on the main diagonal of
.
Let
be the block formed by the last
columns of
.
Then,
where
is the null space of
and
is the span of the columns of
.
The proof of this proposition is in Solved Exercise 2 below.
Below you can find some exercises with explained solutions.
Prove the above proposition about the column space of
.
Suppose that
.
Then, there exists an
vector
such
that
where
we have used the compact SVD of
.
Thus,
and
where
is the span of the columns of
(i.e., the column space of
).
Now, suppose that
.
Then, there exists an
vector
such
that
where
again we have used the compact SVD of
.
Hence,
and
But
(1) and (2)
imply
Prove the above proposition about the null space of
.
Suppose that
.
The columns of
and those of
together form a basis for the space of
vectors. As a consequence, there exist an
vector
and an
vector
such
that
We
can write the equation
as
By
pre-multiplying both sides of the equation by
,
we
obtain
because
and
,
as previously demonstrated. But
Thus,
we
have
Since
and
are both full-rank, the last equation
implies
which
in turn
implies
Thus,
.
We have just demonstrated
that
Now,
suppose that
.
Hence, there exists a
vector
such that
By
pre-multiplying both sides of the equation by
,
we
get
which
implies that
.
Thus,
But
(1) and (2)
imply
Please cite as:
Taboga, Marco (2021). "Singular value decomposition", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/singular-value-decomposition.
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