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If \(T\) is a bounded and self-adjoint operator on a Hilbert space and \(T^{2}=T\), show that \(T\) is the orthogonal projection onto its range.

Short Answer

Expert verified
\(T\) is an orthogonal projection onto its range.

Step by step solution

01

Understanding the Problem

We are given a bounded and self-adjoint operator \(T\) on a Hilbert space, with the property that \(T^2 = T\). We need to show that \(T\) is an orthogonal projection onto its range.
02

Checking Idempotency and Self-Adjointness

Since \(T^2 = T\), \(T\) is idempotent. We are also given that \(T\) is self-adjoint, meaning \(T = T^*\). We need to use these properties to prove orthogonal projection.
03

Showing \(T\) Projects Onto Its Range

Consider any vector \(x\) in the Hilbert space. Let \(y = T(x)\). Then \(T(y) = T(T(x)) = T^2(x) = T(x) = y\). Hence, every vector in the range of \(T\) is a fixed point of \(T\).
04

Proving Orthogonality

To show orthogonal projection, we need to prove that for any vector \(x\), the vector \(x - T(x)\) is orthogonal to the range of \(T\). For any \(z\) in the range of \(T\), there exists some \(w\) such that \(z = T(w)\). We have: \[\langle x - T(x), T(w) \rangle = \langle x, T(w) \rangle - \langle T(x), T(w) \rangle = \langle T(x), w \rangle - \langle T(x), T(w) \rangle = \langle T(x), w - T(w) \rangle = 0\]The last equality comes from the fact that \(T(x) = T^*(x)\) and \(w - T(w)\) is zero in the range of \(T\).
05

Conclusion: T is an Orthogonal Projection

We have shown that \(T\) satisfies two critical properties: idempotency (\(T^2 = T\)), and it preserves orthogonality, as shown by the orthogonality theorem. Thus, \(T\) is the orthogonal projection onto its range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hilbert Space
A Hilbert space is a special kind of vector space that extends the concept of Euclidean space. It allows for infinite dimensions while retaining properties such as length and angle, which one might associate with Euclidean spaces. What makes Hilbert spaces unique is that they come equipped with an inner product— a way to multiply vectors together to get a scalar that provides information on length and angle. This allows us to say when two vectors are orthogonal or perpendicular, based on their inner product being zero.
In mathematical terms, a Hilbert space is a complete inner product space. Completeness means that any Cauchy sequence of vectors (sequences where vectors get "closer" and closer to each other) converges within the space. It's these properties that make Hilbert spaces so powerful and useful in functional analysis, particularly in the study of operator theory and quantum mechanics.
Hilbert spaces are important in the context of linear operators, like the self-adjoint operator mentioned in the problem, because they provide a structured framework where these operators can act and be analyzed. The behavior and properties of functions or elements in these spaces can be deeply explored with tools such as projections and adjoint operators, which are cornerstones of many proofs and concepts within functional analysis.
Orthogonal Projection
Orthogonal projection is a fundamental concept in the realm of Hilbert spaces. Given a linear operator, an orthogonal projection is one that maps every vector in a space to its nearest point in a given subspace, such that the difference (or error) between the vector and its projection is perpendicular to the subspace. This serves as an extension of the geometric idea of dropping a perpendicular from a point to a plane.
The distinctive feature of orthogonal projections is that they are both idempotent and self-adjoint. An idempotent operator satisfies the condition: \(P^2 = P\). This means if you apply the projection twice, there is no change beyond the first application. Self-adjointness (or that the operator is equal to its own adjoint, \(P = P^*\)) ensures that the projection respects the angle properties of the space.
In the context of the exercise, we have a bounded and self-adjoint operator \(T\), which is proving to be an orthogonal projection, as the vectors are fixed upon application of \(T\). Any vector \(x\) not in the range of \(T\) is mapped in such a way that \(x - T(x)\) is orthogonal to the range. This proves the orthogonal characteristics based on the properties of both idempotency and preservation of orthogonality, confirming the operator as an orthogonal projection.
Self-Adjoint Operator
Self-adjoint operators are significant in the study of Hilbert spaces due to their symmetric properties. An operator \(A\) on a Hilbert space is self-adjoint if it equals its own adjoint, that is, \(A = A^*\). This property implies that the operator offers a mirrored effect around its range within the space, preserving the orthogonality and lengths when interacting with vectors.
Self-adjoint operators are important for various reasons, especially since their eigenvalues are real. This is due to the inner product which dictates that \(\langle Ax, x \rangle = \langle x, Ax \rangle\) is always a real number. In quantum mechanics, self-adjoint operators correspond to observable physical quantities, reinforcing their utility beyond pure mathematics.
In our exercise, the self-adjointness of \(T\) combined with its idempotent property proves crucial because it ensures that \(T\) is not just any linear operator but one that guarantees orthogonal projection onto its range. The symmetry and self-administering characteristics of \(T\) secure its role in maintaining precise orthogonality within the Hilbert space, thus connecting vectors logically and geometrically within its framework.

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Most popular questions from this chapter

Show that a diagonal operator on a Hilbert space is an orthogonal projection if and only if its diagonal consists of 0 's and 1 's.

Suppose that \(T\) is an operator in \(\mathscr{B}(\mathscr{H})\) for some Hilbert space \(\mathscr{H}\) and suppose that \(T=T^{-1}\) and \(T=T^{*}\). Show that the sets $$ \mathscr{H}_{1} \equiv\\{h+T h: h \in \mathscr{H}\\} $$ and $$ \mathscr{H}_{2} \equiv\\{h-T h: h \in \mathscr{H}\\} $$ are closed subspaces of \(\mathscr{H}\) with \(\mathscr{H}=\mathscr{H}_{1} \oplus \mathscr{H}_{2}\), and that the restriction of \(T\) to \(\mathscr{H}_{1}\) is the identity \(I\) while the restriction of \(T\) to \(\mathscr{H}_{2}\) is \(-I\). Conversely, show that if \(\mathscr{H}\) is the direct sum of two subspaces \(\mathscr{H}_{1}\) and \(\mathscr{H}_{2}\) with \(T(h)=h\) for \(h \in \mathscr{H}_{1}\) and \(T(h)=-h\) for \(h \in \mathscr{H}_{2}\), then \(T=T^{-1}\) and \(T^{*}=T\).

Suppose \(T\) is a bounded linear operator on a Hilbert space \(\mathscr{H}\) and suppose further that the range of \(T\) is one-dimensional. Show that there are vectors \(x\) and \(y\) in \(\mathscr{H}\) so that $$ T z=\langle z, x\rangle y $$ for all \(z \in \mathscr{H}\). This operator is sometimes written as \(y \otimes x\). Identify \(T^{*}\) in this case.

Show that a linear surjective isometry from one Hilbert space \(\mathscr{H}\) to another Hilbert space \(\mathscr{X}\) is unitary.

Thus far, the only topology we have considered on \(\mathscr{B}(\mathscr{H})\) for \(\mathscr{H}\) a Hilbert space is the topology that comes from the operator norm; this is called the norm topology or the uniform operator topology. However, there are other useful topologies on \(\mathscr{B}(\mathscr{H})\), and in this problem we introduce two of them by discussing sequential convergence in two new senses. Definition. Given \(\left\\{T_{n}\right\\}\) in \(\mathscr{B}(\mathscr{H})\), we say \(T_{n} \rightarrow T \in \mathscr{B}(\mathscr{H})\) in the strong operator topology if \(T_{n} h \rightarrow T h\) \(T_{n} \rightarrow T\) (SOT). for each \(h \in \mathscr{H}\). This is abbreviated \(T_{n} \rightarrow T\) (SOT). We say \(T_{n}\) converges to \(T\) in the weak operator topology, denoted \(T_{n} \rightarrow T\) (WOT), if $$ \left\langle T_{n} h, g\right\rangle \rightarrow\langle T h, g\rangle $$ for each fixed \(h, g\) in \(\mathscr{H}\). (a) Show that if \(S\) is the forward shift operator on \(\ell^{2}\), then \(S^{n} \rightarrow 0\) (WOT), but \(S^{n}\) does not converge to 0 in either the norm or strong operator topology. (b) If \(P_{n}: \ell^{2} \rightarrow \ell^{2}\) by $$ P_{n}\left(x_{1}, x_{2}, \ldots\right)=\left(0,0, \ldots, x_{n+1}, x_{n+2}, \ldots\right) $$ show that \(P_{n} \rightarrow 0\) (SOT), but not in the norm topology. (c) Show that if \(T_{n} \rightarrow T\) (SOT) for \(T_{n}, T \in \mathscr{B}(\mathscr{H})\), then \(T_{n} \rightarrow T\) (WOT). (d) Is the mapping from \(\mathscr{B}(\mathscr{H}) \rightarrow \mathbb{C}\) which sends \(T\) to \(\|T\|\) continuous if we use the strong operator topology (respectively, the weak operator topology) on \(\mathscr{B}(\mathscr{H})\) ?

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