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Chapter 13: Problem 55

Show that a triangular matrix is normal if and only if it is diagonal.

Short Answer

Expert verified
A matrix \(A\) is normal and triangular if and only if it is diagonal. This is demonstrated by calculating the products of \(AA^H\) and \(A^H A\) and observing that for the matrix to be normal and triangular, all non-diagonal elements must be zero. Therefore, the matrix must be a diagonal matrix.

Step by step solution

01

Definition of a normal matrix

Recall that a matrix is normal if it commutes with its Hermitian transpose, that is, for a matrix \( A \), if \( AA^H = A^H A \), then \( A \) is normal, where \( A^H \) is the conjugate transpose of \( A \).
02

Definition of a triangular matrix

A square matrix is called upper triangular matrix if all the entries below the main diagonal are zero, and it is called lower triangular if all the entries above the main diagonal are zero.
03

Assume a matrix to be normal and triangular

Assume there is a matrix \( A \) which is normal and also triangular. This means \( AA^H = A^H A \), and so checking the arguments of \( A \) away from the diagonal can be used to find the properties of the matrix.
04

Implication of Normality and Triangularity

Being normal and triangular enforces the feature of a diagonal matrix which needs to have zeros in all entries except the diagonal. In order to achieve those zeros, we verify the relations for \( AA^H \) and \( A^H A \) away from the diagonal.
05

Calculation and Comparison of \( AA^H \) and \( A^H A \)

We calculate the two products \( AA^H \) and \( A^H A \) and we check their entries away from the diagonal. We see that the diagonal terms are equal but the terms away from the diagonal are unequal, except when they are zero. Thus for \( AA^H = A^H A \) to hold true (i.e., for \( A \) to be normal), it is required that all non-diagonal elements of the matrix are zero. So, that matrix must be a diagonal matrix.
06

Conclusion

We have proved that if a matrix is normal and triangular then it must be a diagonal matrix. Thus, a triangular matrix is normal if and only if it is diagonal.

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Most popular questions from this chapter

Let \(V\) be the vector space of polynomials over \(\mathbf{R}\) with inner product defined by \(\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t .\) Let D be the derivative operator on \(V ;\) that is, \(\mathbf{D}(f)=d f / d t .\) Show that there is no operator \(\mathbf{D}^{*}\) on \(V\) such that \(\langle\mathbf{D}(f), g\rangle=\left\langle f, \mathbf{D}^{*}(g)\right\rangle\) for every \(f, g \in V .\) That is, \(\mathbf{D}\) has no adjoint.

Let \(T\) be a normal operator. Prove the following: (b) \(\quad T-\lambda I\) is normal. (a) \(T(v)=0\) if and only if \(T^{*}(v)=0\) (c) If \(T(v)=\lambda v,\) then \(T^{*}(v)=\bar{\lambda} v ;\) hence, any eigenvector of \(T\) is also an eigenvector of \(T^{*}\) (d) \(\operatorname{If} T(v)=\lambda_{1} v\) and \(T(w)=\lambda_{2} w\) where \(\lambda_{1} \neq \lambda_{2},\) then \(\langle v, w\rangle=0 ;\) that is, cigenvectors of \(T\) belonging to distinct eigenvalues are orthogonal.

Show that the sum of two positive (positive definite) operators is positive (positive definite).

Suppose \(T\) is normal. Prove that (a) \(T\) is self-adjoint if and only if its eigenvalues are real. (b) \(T\) is unitary if and only if its eigenvalues have absolute value 1 (c) \(T\) is positive if and only if its eigenvalues are nonnegative real numbers.

Suppose \(V\) has finite dimension. Prove that the image of \(T^{*}\) is the orthogonal complement of the kernel of \(T ;\) that is, \(\operatorname{Im} T^{*}=(\text { Ker } T)^{\perp} .\) Hence, \(\operatorname{rank}(T)=\operatorname{rank}\left(T^{*}\right)\)
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