Suppose that 𝒥 is a closed two-sided ideal in ℬ(ℋ), for ℋ a Hilbert space. The goal of this prob

Suppose that 𝒥 is a closed two-sided ideal in ℬ(ℋ), for ℋ a...

Suppose that $\mathscr{J}$ is a closed two-sided ideal in $\mathscr{B}(\mathscr{H})$, for $\mathscr{H}$ a Hilbert space. The goal of this problem is to show that either $\mathscr{J}=\{0\}$, or $\mathscr{J}$ contains $\mathscr{K}(\mathscr{H})$, the ideal of compact operators on $\mathscr{H}$. (Compare this with the statement in Exercise 5.22.) (a) Suppose $T$ is a nonzero operator in $\mathscr{J}$. Find vectors $f_{0}, f_{1}$ with $T f_{0}=f_{1}$ and $f_{1} \neq 0$. Show that if $g_{0}, g_{1}$ are any pair of nonzero vectors in $\mathscr{H}$, then the rank one operator $S$ defined by $$ S f=\frac{\left\langle f, g_{0}\right\rangle g_{1}}{\left\|g_{0}\right\|^{2}} $$ is in $\mathscr{J}$. By Exercise $2.6$ in Chapter 2 , this will show that $\mathscr{J}$ contains all rank 1 operators. Hint: Let $$ A f=\frac{\left\langle f, g_{0}\right\rangle f_{0}}{\left\|g_{0}\right\|^{2}} $$ and $$ B f=\frac{\left\langle f, f_{1}\right\rangle g_{1}}{\left\|f_{1}\right\|^{2}} $$ and compute $B T A$. (b) Apply Exercise $4.9$ of Chapter 4 to show that $\mathscr{J}$ contains all finite rank operators.

Suppose that $\mathscr{J}$ is a closed two-sided ideal in $\mathscr{B}(\mathscr{H})$, for $\mathscr{H}$ a Hilbert space. The goal of this problem is to show that either $\mathscr{J}=\{0\}$, or $\mathscr{J}$ contains $\mathscr{K}(\mathscr{H})$, the ideal of compact operators on $\mathscr{H}$. (Compare this with the statement in Exercise 5.22.)
(a) Suppose $T$ is a nonzero operator in $\mathscr{J}$. Find vectors $f_{0}, f_{1}$ with $T f_{0}=f_{1}$ and $f_{1} \neq 0$. Show that if $g_{0}, g_{1}$ are any pair of nonzero vectors in $\mathscr{H}$, then the rank one operator $S$ defined by
$$
S f=\frac{\left\langle f, g_{0}\right\rangle g_{1}}{\left\|g_{0}\right\|^{2}}
$$
is in $\mathscr{J}$. By Exercise $2.6$ in Chapter 2 , this will show that $\mathscr{J}$ contains all rank 1 operators. Hint: Let
$$
A f=\frac{\left\langle f, g_{0}\right\rangle f_{0}}{\left\|g_{0}\right\|^{2}}
$$
and
$$
B f=\frac{\left\langle f, f_{1}\right\rangle g_{1}}{\left\|f_{1}\right\|^{2}}
$$
and compute $B T A$.
(b) Apply Exercise $4.9$ of Chapter 4 to show that $\mathscr{J}$ contains all finite rank operators.

Key Concepts

Rank-One Operators

Rank-one operators are a special case of finite rank operators, mapping any vector into a scalar multiple of a fixed vector. Despite their simplicity, they can be used to approximate more complicated operators and play a critical role in the structure theory of operator algebras. They are often instrumental in constructing examples and counterexamples, as well as in proofs that involve element-wise manipulation in ideals.

Compact Operators

Compact operators on a Hilbert space form an important ideal within B(H). They can be approximated in operator norm by finite rank operators and share many properties with matrices, such as having a discrete spectrum. The structure of compact operators and their role as building blocks is crucial in various areas including spectral theory and the study of Fredholm operators.

Finite Rank Operators

Finite rank operators are operators whose range is a finite-dimensional subspace of the Hilbert space. They are particularly significant because they form a dense subset in the ideal of compact operators and are often used as simple approximants in more complex operator theoretical investigations. Their structure helps in understanding how more general operators can be approximated or constructed.

Bounded Linear Operators

Bounded linear operators are maps between Hilbert spaces that are linear and continuous. The set of all bounded operators on a Hilbert space, denoted B(H), forms a Banach algebra and, in the context of operator algebras, is a canonical example of a C*-algebra. These operators are central to understanding how different substructures, such as ideals, function within the algebra of all operators.

Hilbert Space

A Hilbert space is a complete inner product space that provides the setting in which many problems in functional analysis and quantum mechanics are formulated. It allows the use of concepts like orthogonality and projection and serves as the domain and codomain for linear operators, making it fundamental for studying operator algebras.

Closed Two-Sided Ideals

Closed two-sided ideals in an algebra are subspaces that are invariant under multiplication from elements from both the left and right by any element in the algebra, and they are closed in the topology induced by the operator norm. In the context of B(H), determining the structure and nature of these ideals reveals key aspects of the algebra's internal makeup, such as simplicity or the existence of certain canonical subalgebras like the compact operators.

Transcript

00:01 We're given a set of vectors s with vectors u1, u2, up to u .n, which is a basis of a vector space v, and we're told that functions f and g are linear operators on v, such that the matrix of f relative to the basis s has zeros on and below diagonal.

00:43 Well that's a product of his size, right? he's like four foot a.

00:47 But it's upper triangular, but also zero is diagonal, and the matrix of g relative to the basis s has has an entry a, non -zero on the super diagonal and zeros elsewhere.

01:22 So in other words, if you write them of matrices, the matrix of f relative to s is the matrix 0, 0, a2, 1.

01:34 A 3 -1 all the way up to a n1, 0 -0 -0 -2 all the way up a n2, all the way up, a n2, and so on, down to 0 -0 ,000, all the way over to a n -minus 1, and the final row is just all 0 -0.

02:11 And likewise, the matrix of g relative to s is given by, well, a diagonal zero.

02:22 So we have zero, then we have the constant a, then the rest of zeros.

02:27 We have zero, zero, then a again on super diagonal, but the rest are zeros.

02:38 And we continue this pattern all the way down.

02:40 So we have in the second of the last row, zeros, all the way until we reach the last entry in the second to last row, which will be a.

02:51 And finally, the last row is just going to be all zeros.

02:58 Okay.

03:02 So there's a few parts to this exercise.

03:06 Part a, we're asking to show that the m power of f is zero.

03:15 So i go to like, dead stop.

03:19 Well, we have plugging in u1, just from our basis with respect to s, you see that f of u1 is going to be the zero vector.

03:33 And we have for r greater than one, we see from the matrix of f with respect s, the f of ur, this is the r column.

03:49 And therefore is a linear combination of the vectors preceding you are.

04:04 In boston, i almost got, checked out there when it came out to boston, in our set s.

04:11 This is because we have zeros in the diagonals and zero is below the diagonal...

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