Adjoint operator and orthogonal projection

• MHB
• mathzero
In summary: However, if we take $T(y) = y + x$ for all $y\in H$, then $T^*(y) = y - x$ for all $y\in H$, which satisfies the equation $\operatorname{Pr}_C(x + C) = \operatorname{Pr}_C(y)$.
mathzero
Hello,

I want to show $T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$

where $T \in B(H)$ and $TT^{*}=I$ , $H$ is Hilbert space and $C$ is a closed convex non empty set.

but i don't know how to start, or what tricks needed to solve this type of problems.

also i want know how to construct $T$ to satisfying

$Pr_{x + C}(y) = Pr_{C}(x + y)$ and $Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

Thanks!

mathzero said:
Hello,

I want to show $T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$

where $T \in B(H)$ and $TT^{*}=I$ , $H$ is Hilbert space and $C$ is a closed convex non empty set.
Since $TT^* = I$, then $\|T^*x\| = \|x\|$ for all $x\in H$. Indeed, given $x\in H$, $\|T^*x\|^2 = \langle T^*x, T^*x\rangle = \langle TT^*x, x\rangle = \langle x,x\rangle = \|x\|^2$. Show that
$$\inf_{c\in C} \|T^*c - T^*y\| = \inf_{c\in C} \|c - y\|$$
and deduce $T^*(\operatorname{Pr}_C(y))$ is a point of $T^*(C)$ closest to $T^*y$. Then the result follows.
mathzero said:
also i want know how to construct $T$ to satisfying

$Pr_{x + C}(y) = Pr_{C}(x + y)$ and $Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$
How does $T$ relate to those equations?

Thank you!

Euge said:
How does $T$ relate to those equations?

i want apply this result $T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$, to

$Pr_{x + C}(y) = Pr_{C}(x + y)$ and $Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

by creating $T$ first and later deducing $T^{*}$ (i don't know if this is possible!),

i imagine maybe, just by letting $Ty = y - x$ then $T^{*}y= y + x$? this can be possible?

Last edited:
mathzero said:
Thank you!
i want apply this result $T^{*}(Pr_{C}(y)) = Pr_{T^{*}(C)}(T^{*}y)$, to

$Pr_{x + C}(y) = Pr_{C}(x + y)$ and $Pr_{\lambda C}(\lambda y) = \lambda Pr_{C}(y)$

by creating $T$ first and later deducing $T^{*}$ (i don't know if this is possible!),

i imagine maybe, just by letting $Ty = y - x$ then $T^{*}y= y + x$? this can be possible?

In the case $\operatorname{Pr}_C(\lambda C) = \lambda \operatorname{Pr}_C(y)$, you can take $T(y) = \lambda y$, for all $y\in H$, which defines an element of $B(H)$ with $\|T\| = \lvert \lambda \rvert$. For the other equation, you suggest letting $Ty = y - x$, but this does not define a linear operator unless $x = 0$.

1. What is the adjoint operator?

The adjoint operator is a linear transformation that maps a vector space onto itself in a way that preserves inner products. It is also known as the Hermitian adjoint or the conjugate transpose.

2. How is the adjoint operator related to the orthogonal projection?

The adjoint operator is closely related to the orthogonal projection because it is used to find the orthogonal projection of a vector onto a subspace. The adjoint operator of an orthogonal projection is equal to the projection itself.

3. What is the significance of the adjoint operator in linear algebra?

The adjoint operator plays a key role in linear algebra, particularly in the study of inner product spaces. It is used to define important concepts such as self-adjoint and unitary operators, and is a fundamental tool in solving systems of equations and finding eigenvalues and eigenvectors.

4. How is the adjoint operator related to the conjugate of a complex number?

The adjoint operator is closely related to the conjugate of a complex number. In fact, the adjoint operator of a linear transformation in a complex vector space is equivalent to taking the conjugate of the matrix representation of the transformation.

5. Can the adjoint operator be applied to non-square matrices?

No, the adjoint operator is only defined for square matrices. It is a map from a vector space onto itself, so the input and output dimensions must be the same. However, the concept of the adjoint can be extended to rectangular matrices in some cases, such as when using the Moore-Penrose pseudoinverse.

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