which finite-dimensional inner product space? And what is the codomain of T?julydecember said:Homework Statement
If T is a linear transformation on the finite-dimensional inner product space over complex numbers
what does normal mean?julydecember said:and is normal,
Do you mean "prove that the image of T is convex"?julydecember said:then prove that the numerical range of T is convex.
The numerical range of a normal operator T is always convex. This means that for any two points in the numerical range, the line segment connecting them is also contained within the numerical range. In other words, the numerical range is a convex set.
A set is convex if for any two points in the set, the line segment connecting them is also contained within the set.
To prove convexity of the numerical range for a normal operator T, we can use the spectral theorem to show that the numerical range is equal to the convex hull of the eigenvalues of T. This means that any point in the numerical range can be expressed as a convex combination of the eigenvalues, and therefore the line segment connecting any two points is also contained within the numerical range.
No, convexity of the numerical range is not unique to normal operators. In fact, the numerical range of any operator is always convex, regardless of whether the operator is normal or not.
The convexity of the numerical range is closely related to the spectrum of a normal operator T. The spectrum of T is the set of all possible values that T can take on, and the numerical range is a subset of the spectrum. Therefore, the convexity of the numerical range is a property of the spectrum of T.