A matrix is normal if and only if either pre-multiplying or post-multiplying it by its conjugate transpose gives the same result. It turns out that a matrix is normal if and only if it is unitarily similar to a diagonal matrix.
In other words, not only normal matrices are diagonalizable, but the change-of-basis matrix used to perform the diagonalization is unitary. Said in layman terms, normal matrices are those that behave nicely as far as diagonalization is concerned.
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Let us start with a formal definition of normality.
Definition A matrix is said to be normal if and only ifwhere denotes the conjugate transpose of .
A simple example follows.
Example DefineThe conjugate transpose of isThe product of and isThe product of and isTherefore, and is normal.
Several important kinds of matrices are normal.
Proposition Let be a matrix. If is unitary, then it is normal.
By the definition of unitary matrix, we havewhere is the identity matrix.
Remember that a matrix is Hermitian if and only if it is equal to its conjugate transpose. Since complex conjugation leaves real numbers unaffected, a real matrix is Hermitian when it is symmetric (equal to its transpose).
Proposition Let be a matrix. If is Hermitian, then it is normal.
By the definition of Hermitian matrix, we have
A matrix is said to be skew-Hermitian if and only if
Proposition Let be a matrix. If is skew-Hermitian, then it is normal.
By the definition of skew-Hermitian matrix, we have
Another useful fact follows.
Proposition Let be a matrix. If is diagonal, then it is normal.
If is diagonal, the products and are diagonal. The main diagonals of these products contain entries of the formTherefore, .
Remember that two matrices and are said to be unitarily similar if and only ifwhere is a unitary matrix.
Proposition Let be a matrix. Let be unitarily similar to . If is normal, then is normal.
The proof is as follows:where: in steps and we have used the fact that since is unitary; in step we have used the fact that is normal.
The following proposition will be used below to prove the main result about the diagonalization of normal matrices.
Proposition A triangular matrix is normal if and only if it is diagonal.
We will prove the proposition for upper triangular matrices. The proof is by induction on the dimension of the matrix. A matrix is diagonal by definition and normal because the product of scalars is commutative. Now, suppose that upper triangular matrices are normal if and only if they are diagonal. We need to prove that the claim is true for matrices. In order to prove the "only if" part, we choose a triangular and normal matrix . We partition so as to form the following block-matrix:where is a scalar, is a vector of zeros, is a vector and is a matrix. Then,andSince is normal, , which implies that each block of must be equal to the corresponding block of . The equalityimplieswhich, in turn, implies by the positive definiteness of the norm. Moreover, the equalityimplieswhich means that is normal. Furthermore, is the diagonal block of an upper triangular matrix. As a consequence, it is upper triangular and, by the induction hypothesis, diagonal. Thus, we havewhich is a diagonal matrix because is a scalar and is diagonal. This proves the "only if" part. The "if" one is trivial because we have already proved a proposition that states that diagonal matrices are normal. The proof for lower triangular matrices is analogous.
We are now ready to prove the most important result.
Remember that a matrix is said to be diagonalizable if and only if there exists an invertible matrix such thatand is diagonal. In other words, is similar to a diagonal matrix .
It turns out that the diagonal entries of are the eigenvalues of and the columns of are the eigenvectors of .
When is unitary, the diagonalization becomesand we say that is unitarily diagonalizable.
Proposition A matrix is unitarily diagonalizable if and only if it is normal.
We first prove the "only if" part, starting from the hypothesis that is unitarily diagonalizable, that is, unitarily similar to a diagonal matrix . We have proved above that 1) unitary similarity preserves normality and 2) diagonal matrices are normal. As a consequence, must be normal. We can now prove the "if" part, starting from the hypothesis that is normal. The Schur decomposition theorem says that any square matrix is unitarily similar to an upper triangular matrix :Since is normal and unitary similarity preserves normality, must be normal. But an upper triangular matrix can be normal only if it is diagonal. Therefore, must be diagonal.
Recall that a diagonalizable matrix is not defective, that is, it possesses linearly independent eigenvectors. In the case of a normal matrix , the matrix of eigenvectors is unitary, which means that the columns of are orthonormal. In other words, a normal matrix possesses a set of orthonormal eigenvectors. Said differently, there is a basis of orthonormal vectors for the eigenspace of a normal matrix.
When the matrix being diagonalized is real and symmetric, then both the matrix of eigenvalues and the change-of-basis matrix are real.
Proposition Let be a real and symmetric matrix. Then it can be diagonalized aswhere both and are real, is diagonal and is orthogonal.
In the lecture on the properties of eigenvalues and eigenvectors, we have shown that all the eigenvalues of a symmetric real matrix are real. This implies that for any eigenvalue , we can find a real eigenvector by searching for a real solution to the equationThe solution is guaranteed to exist because is rank-deficient by the definition of eigenvalue. The real eigenvectors thus found can be used in the algorithm that leads to the Schur decomposition. In particular, if we look into the proof of the Schur decomposition, we can see that the Householder matrices used there are real if we choose real eigenvectors for . As a consequence, the Schur decompositioncan be performed in such a way that the matrices and are real. The matrix is unitary (i.e., ), but since it is also real, we have andthat is, is orthogonal. Moreover, since is real and symmetric, it is Hermitian and therefore normal. From the proof of the previous proposition, we know that the matrix in the Schur decomposition is diagonal when is normal.
Below you can find some exercises with explained solutions.
Is the matrixunitarily diagonalizable?
The matrix is upper triangular. If it was unitarily diagonalizable, it would be normal. But an upper triangular matrix is normal only if it is diagonal. The matrix is not diagonal, hence it is not normal and it is not unitarily diagonalizable.
Is the matrixunitarily diagonalizable?
The matrix is symmetric (it is equal to its transpose). Since it is real, it is also Hermitian. We know that Hermitian matrices are normal. Therefore, is normal and unitarily diagonalizable.
Check whether the matrix is normal.
The conjugate transpose of isWe need to compute the productsandThus, and is not normal.
Please cite as:
Taboga, Marco (2017). "Normal matrix", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/normal-matrix.
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