Question

If \(A_{1}, A_{2}, \ldots . A_{n}\) are operators on a dense domain such that $$ \sum_{i=1}^{n} A_{1}^{*} A_{1}=0 $$ show that \(A_{1}=A_{2}=\cdots=A_{n}=0 .\)

Short answer

Applying the properties of operators and their adjoints, along with comparisons to the properties of inner products, demonstrates that each operator \( A_{i}\) must be zero.

Step-by-step solution

Understand the given condition

The problem begins with the condition \( \sum_{i=1}^{n} A_{1}^{*} A_{1} = 0 \). This can be understood as the sum of the product of each operator with its own adjoint being zero.

Apply the properties of inner product

The product of an operator and its adjoint closely resembles the concept of inner product in linear algebra. Recall that in this context, the inner product of a vector with itself equals zero if and only if the vector is zero. This implies that for every \(i\), we must have \( A_{i}^{*} A_{i} = 0\).

Conclude that each operator is zero

Recall that the only way for an operator's product with its own adjoint (or for an inner product of a vector with itself) to be zero is for the operator (or vector) to be zero. Therefore, each \(A_{i}\) must be zero.

Key concepts

Adjoint Operators

Adjoint operators are key concepts in operator theory. These operators are the "transpose and conjugate" version of a given linear operator. In simple terms, for an operator \( A \), the adjoint \( A^* \) represents a transformation that reverses \( A \) while accounting for complex conjugates.
Understanding the adjoint is crucial because it provides insights into the properties of operators.
For instance, an operator is called self-adjoint if \( A = A^* \). Adjoint operators feature prominently in quantum mechanics and functional analysis, as they help explore properties like symmetry and unitarity.
In exercises involving adjoint operators, like the one presented, you often use properties such as \( (A^*)^* = A \) and \( (AB)^* = B^* A^* \).
These properties assist in simplifying expressions and drawing conclusions.

Inner Product

The inner product is a generalization of the dot product in vector spaces. It's a way to multiply vectors to obtain a scalar, offering measurements of angles and lengths.
In Euclidean spaces, the inner product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is usually \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \cdots + u_nv_n \).
However, in the context of linear operators, inner products connect with adjoint operators through expressions like \( \langle A\mathbf{x}, \mathbf{y} \rangle = \langle \mathbf{x}, A^*\mathbf{y} \rangle \).
In the problem given, comparing the product of an operator with its adjoint to the inner product's nature is critical. The condition \( A_i^* A_i = 0 \) implies that \( A_i \) acts like a vector with a zero inner product, which means \( A_i = 0 \).
Thus, understanding inner products helps decode relationships between operators and offers tools for showing operator properties like being zero.

Dense Domain

When discussing operators in functional analysis, the domain where these operators act is crucial. A dense domain is one where every element of the space can be approximated as closely as desired by elements within this domain.
In simple terms, if you imagine the whole space as points on a map, a dense domain ensures that you can "get close" to any point on the map by staying within a particular subset.
The significance of the dense domain in operator theory, as illustrated in the problem exercise, is that operators defined on a dense domain often ensure the operators adjoint or closure properties.
This is critical in proving results like the one in the exercise, where the operators are confirmed to be zero. Having a dense domain means that if an operator behaves in a certain way on this subset, it implies similar behavior across the entire space.