Schur's Decomposition

The eigenvector of the matrix determines in the space a one-dimensional subspace that is invariant with respect to the multiplication by the matrix A on the left.

Definition 2.6.1. The subspace is called invariant with respect to the multiplication by the matrix A on the left if

Proposition 2.6.1. If and AX=XB, then the space of the matrix X is invariant with respect to the multiplication by the matrix A on the left and the space is invariant with respect to the multiplication by the matrix B on the right. In addition, the following connections

and

holds.

Proof. If then

and



and

Therefore, the space of the column-vectors of the matrix X is invariant with respect to the multiplication by the matrix A on the left. Analogously, one can prove that the space of the row-vectors is invariant with respect to the multiplication by the matrix B on the right. If then

i.e., is an eigenvalue of the matrix A if is the eigenvalue of the matrix B. Naturally,

therefore, if the column-vectors of the matrix X are linearly independent, then If and X is a regular square matrix , then it follows from the equality AX=XB that A=XBX^-1 and

i.e., every eigenvalue of the matrix A is an eigenvalue of the matrix B, , and thus

Definition 2.6.2. Matrices are said to be similar if there exists a regular matrix such that A=XBX^-1.

Due to proposition 2.6.1 (the last assertion), the spectres of two similar matrices are equal. We can get this result also directly:

Problem 2.6.1.^* Are the matrices A and B similar if

Proposition 2.6.2. If and

then

Proof. If i.e., and then

If then

If then

Thus,

Since the potencies of the sets and are equal, then the proposition holds.

Example 2.6.1. Using proposition 2.6.2, let us find the spectre of the matrix

First, we find the eigenvalues of the matrices and :



Thus, the spectre of the matrix is

Problem 2.6.2.^* Find by the use of proposition 2.6.2 the spectre of the matrix A if

Definition 2.6.3. A matrix is called a unitary matrix if Q^HQ=QQ^H=I.

Problem 2.6.3.^* Is the matrix Q a unitary matrix if

Proposition 2.6.3 (the QR factorisation theorem) . If , then the matrix A can be expressed in form A=QR, where matrix is unitary matrix and matrix is an upper triangular matrix.

Proposition 2.6.4. If
 
and rank(X)=p, then there exists a unitary matrix such that

where

Proof. Let us consider for the matrix its QR factorization where and Substracting the factorization of the matrix X into equality (20), we get

The spectres of the matrices Q^HAQ and A coincide, i.e. Representing the matrix Ain form

we find

Therefore, the proposition holds.

Remark 2.6.1. Proposition 2.6.4 makes it possible, if we know an invariant subspace of the given matrix, to transform it by unitary similarity transformations into a triangular block form.

Proposition 2.6.5 (Schur's decomposition). If then there exists a unitary matrix such that
 
where and is a strictly upper triangular matrix, i.e., an upper triangular matrix with zeros on the main diagonal. The matrix Q can be formed so that the eigenvalues of the matrix A are in the given order on the main diagonal of the matrix D.

To prove this assertion we will use the method of complete induction. As the assertion holds for n=1, the base for the induction exists. We are going now to show the admissibility of the induction steps. We suppose that the assertion holds for the matrices which order is less or equal to k-1. Let us show that the assertion will be valid for k, too. If and then, by lemma 2.6.4, choosing , there exists a unitary matrix U such that

Since the assertion is valid for this matrix, i.e., there exists a unitary matrix such that is an upper triangular matrix. If then





and so the matrix Q^HAQ is an upper triangular matrix.

Example 2.6.2. Let

Let us show that Q is unitary matrix. For this we, first, find the product Q^HAQ. The checking of the matrix Q for unitarity gives:

Let us find the product



Consequently, we have obtained the Schur decomposition of the matrix A.

Now (21) can be represnted in the form AQ=QT. Replacing where the vectors are called Schur vectors, into the last equality , we get

or



or

From this equality it turns out that all subspaces ( k = 1: n) are invariant with respect to multiplication by the matrix A on the left, and Schur vector is an eigenvector of the matrix A if and only if in the -th column of the matrix there are only zeros.

Definition 2.6.4. If and A^HA=AA^H, then A is called a normal matrix.

Exercise 2.6.4.* Check the normality of A if

Proposition 2.6.6. A matrix is normal matrix if there exists a unitary matrix satisfying the condition

Proof. If the matrix A is unitarily similar to the diagonal matrix D, then



and since diagonal matrices are commutative, then A^HA=AA^H and the matrix A is normal . Vice versa, if the matrix A is normal and the Schur decomposition of this matrix is Q^HAQ=T, then T is also normal because

and

Since a triangular matrix is normal only if it is a diagonal matrix, then it has been proved that a unitary matrix is similar to a diagonal matrix.

Proposition 2.6.7 (block-diagonal-decomposition). Let

be the Schur decomposition of the matrix , where the blocks are square matrices. If then there exists a regular matrix such that

Corollary 2.6.1. If then there exists a regular matrix X such that

where , and each is a strictly upper triangular matrix.

Proposition 2.6.8 (Jordan decomposition). If then there exists a regular matrix such that where ,

is an Jordan block,and the matrix J is called the Jordan normal form of the matrix A .

Proof. See Lankaster (1982, p. 143).

Example 2.6.3. Using ''Maple'' , we find the Jordan decompositions A=XJX^-1 of two matrices:

and