For any given matrix, the Schur decomposition (or factorization) allows us to find another matrix that is similar to the given one and is upper triangular. Moreover, the change-of-basis matrix used in the similarity transformation is unitary.
In other words, the Schur decomposition shows that each matrix is unitarily similar to an upper triangular matrix.
Let us revise some notions that are essential to understand the Schur decomposition.
We say that two square matrices
and
are similar if and only if there
exists an invertible
matrix
such
that
The matrix
involved in the similarity transformation is called a
change-of-basis matrix.
Two similar matrices have the same rank, trace, determinant and eigenvalues. Moreover, the algebraic and geometric multiplicities of their eigenvalues coincide.
We say that a square matrix
is unitary if and only if its
columns have unit norm and are
orthogonal to each other, in
which case
has the property
that
where
denotes the conjugate
transpose of
.
When
is unitary, the similarity transformation above
becomes
and
we say that
and
are unitarily similar.
In the proof that any square matrix
is unitarily similar to an upper triangular matrix
we are going to repeatedly use so-called
Householder matrices.
Householder matrices are among the most powerful tools in modern linear algebra. We therefore recommend to study them in detail. However, here is a summary of what we need to know in order to understand the proof of the Schur decomposition:
if
is a
column vector and
is the
identity matrix, then the
matrix
is
called a Householder matrix
(
is the norm of
and
its conjugate transpose);
a Householder matrix
is unitary and
;
for any
vector
,
we have that
if and only if
;
for any non-zero
vector
,
there is a Householder matrix
such
that
where
is a non-zero scalar and
is the first vector of the canonical basis (i.e., a
vector that has the first entry equal to 1 and all the other entries equal to
0).
We are now ready to present the Schur decomposition.
Proposition
Let
be a square
matrix. Then,
is unitarily similar to an upper
triangular
matrix
,
that
is,
where
is a
unitary matrix.
Let
be any eigenvalue of
and
one of its associated
eigenvectors (hence,
).
We can find a Householder matrix
such
that
where
.
Moreover,
Partition
so as to form a
block-matrix
where
is the first column of
and
contains all the other columns. Since
,
we
have
By
using the rules for
the multiplication of block matrices, we
obtain
where
in step
we have used the fact that
is an eigenvector of
.
We can
write
where
is a
vector of zeros,
is a
vector of possibly non-zero entries (equal to the first row of
),
and
is a
matrix (containing all the rows of
except the first one). This was the first step of the triangularization
process. Let us now start the second step. Let
be any eigenvalue of
and
one of its associated eigenvectors. We construct a Householder matrix
such
that
where
.
Note that
is now shorter than in the previous step: its first entry is 1 and all its
other entries are 0, but the number of 0s decreases by one unit. To be
rigorous, we should change notation, but we keep the same notation for
clarity's sake. As before, the symmetric
equality
holds
and, similarly to what we have demonstrated above, we
have
We
create a new
block-matrix
The
matrix
,
being a Householder matrix, is unitary. The first column of
has unit norm and it is orthogonal to all the other columns. Thus,
is unitary. We can now compute the
product
where
we have denoted various new vectors of possibly non-zero entries by
.
This ends the second step of the triangularization process. The successive
steps are straightforward modifications of the second one. We keep doing
Householder transformations until we
obtain
Since
the product of unitary matrices is unitary, the
matrix
is
unitary.
Moreover,
because
all the matrices involved in the product are derived from Householder matrices
(which are Hermitian), possibly by adding blocks of zeros and ones which are
unaffected by conjugate transposition.
Therefore,
where
is the upper triangular matrix above (with entries
on the main diagonal).
The Schur decomposition is not unique. This can be seen easily from the
algorithm used in the constructive proof above: at each step we choose an
eigenvalue arbitrarily; as a consequence, there are different possible
orderings of the eigenvalues of
on the main diagonal of
.
More in general, if
is
a Schur decomposition of
,
we can take any unitary matrix
such
that
is
upper triangular, and use it to construct another Schur
decomposition
A triangular matrix has the property that its diagonal entries are equal to
its eigenvalues. Moreover, two similar matrices have the same eigenvalues.
Therefore, the Schur decomposition
allows us
to read the eigenvalues of
on the main diagonal of
,
which is upper triangular and similar to
.
We should compare the results derived here with those presented in the lecture
on matrix
diagonalization, where we have shown that if
has no defective eigenvalues, then it is similar to a diagonal matrix having
the eigenvalues of
on the main diagonal.
Here are the similarity and differences:
all matrices have a Schur decomposition, but only those that have no defective eigenvalues are diagonalizable;
in the Schur decomposition we can read the eigenvalues from an upper triangular matrix, while in a matrix diagonalization we read them from a (simpler) diagonal matrix;
in the Schur decomposition the change-of-basis matrix is unitary, while in a diagonalization it need only be invertible.
Each of the two decompositions has its own strengths. As a consequence, they are useful in different situations.
A simple example follows.
Example
DefineThen,
the Schur decomposition of
is
where
and
From
we can read the eigenvalues of
,
which are
and
.
Please cite as:
Taboga, Marco (2021). "Schur decomposition", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/Schur-decomposition.
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