Diagonals of Compact Operators

John Jasper

South Dakota State University

The Pythagorean Theorem

Theorem. If $$\Delta$$ is a right triangle with side lengths $$c\geq b\geq a$$, then $a^{2}+b^{2}=c^{2}.$

$$a$$

$$b$$

$$c$$

The Pythagorean Theorem

Theorem. If $$v$$ and $$w$$ are orthogonal vectors, then $\|v\|^2 + \|w\|^2 = \|v+w\|^2.$

$$v$$

$$w$$

$$v+w$$

(The Standard Generalization)

Theorem. If $$v_{1},v_{2},\ldots,v_{k}$$ are pairwise orthogonal vectors, then

$\|v_{1}\|^2 + \|v_{2}\|^2 + \cdots + \|v_{k}\|^{2} = \|v_{1}+v_{2}+\cdots+v_{k}\|^2.$

Similar Triangles!

$\|(I-P)e_{2}\| = \|Pe_{1}\|$

$\|Pe_{1}\|^{2} + \|Pe_{2}\|^{2} = 1$

The Pythagorean Theorem

Theorem. If $$P$$ is an orthogonal projection onto a $$1$$-dimensional subspace $$V$$, and $$\{e_{1},e_{2},\ldots,e_{n}\}$$ is an orthonormal basis, then

$\sum_{i=1}^{n}\|Pe_{i}\|^{2} = 1.$

Proof.

• Let $$v\in V$$ be a unit vector.
• $$Px = \langle x,v\rangle v,$$
• $$\|Pe_{i}\|^{2} = |\langle e_{i},v\rangle|^{2}$$
• $$\displaystyle{\sum_{i=1}^{n}\|Pe_{i}\|^{2} = \sum_{i=1}^{n}|\langle e_{i},v\rangle|^{2} = \|v\|^{2} = 1}.$$

$$\Box$$

The Pythagorean Theorem and Diagonals

If $$P$$ is an orthogonal projection onto a subspace $$V$$, and $$(e_{i})_{i=1}^{n}$$ is an orthonormal basis, then

$\|Pe_{i}\|^{2} = \langle Pe_{i},Pe_{i}\rangle = \langle P^{\ast}Pe_{i},e_{i}\rangle = \langle P^{2}e_{i},e_{i}\rangle = \langle Pe_{i},e_{i}\rangle$

$= \left[\begin{array}{cccc} \|Pe_{1}\|^{2} & \ast & \cdots & \ast\\ \overline{\ast} & \|Pe_{2}\|^{2} & & \vdots\\ \vdots & & \ddots & \vdots\\ \overline{\ast} & \cdots & \cdots & \|Pe_{n}\|^{2}\end{array}\right]$

$P= \left[\begin{array}{cccc} \langle Pe_{1},e_{1}\rangle & \langle Pe_{2},e_{1}\rangle & \cdots & \langle Pe_{n},e_{1}\rangle \\ \langle Pe_{1},e_{2}\rangle & \langle Pe_{2},e_{2}\rangle & & \vdots\\ \vdots & & \ddots & \vdots\\ \langle Pe_{1},e_{n}\rangle & \cdots & \cdots & \langle Pe_{n},e_{n}\rangle \end{array}\right]$

The Pythagorean Theorem

Theorem. If $$P$$ is an orthogonal projection onto a $$1$$-dimensional subspace $$V$$, and $$\{e_{1},e_{2},\ldots,e_{n}\}$$ is an orthonormal basis, then

$\sum_{i=1}^{n}\|Pe_{i}\|^{2} = 1.$

Proof. $\sum_{i=1}^{n}\|Pe_{i}\|^{2} = \text{tr}(P) = \dim V = 1.$

$$\Box$$

The Pythagorean Theorem

Theorem. If $$P$$ is an orthogonal projection onto a $$k$$-dimensional subspace $$V$$, and $$\{e_{1},e_{2},\ldots,e_{n}\}$$ is an orthonormal basis, then

$\sum_{i=1}^{n}\|Pe_{i}\|^{2} = k.$

Proof. $\sum_{i=1}^{n}\|Pe_{i}\|^{2} = \text{tr}(P) = \dim V = k.$

$$\Box$$

Theorem. If $$P$$ is an orthogonal projection onto a $$k$$-dimensional subspace $$V$$, and $$\{e_{1},e_{2},\ldots,e_{n}\}$$ is an orthonormal basis, then

$\sum_{i=1}^{n}\|Pe_{i}\|^{2} = k.$

Proof. $\sum_{i=1}^{n}\|Pe_{i}\|^{2} = \text{tr}(P) = \dim V = k.$

$$\Box$$

Corollary. If $$P$$ is an orthogonal projection matrix onto a $$k$$-dimensional subspace, and $$(d_{i})_{i=1}^{n}$$ is the sequence on the diagonal of $$P$$, then

$$d_{i}\in[0,1]$$ for each $$i$$, and

$\sum_{i=1}^{n}d_{i} \in\Z$

The Carpenter's Theorem

Theorem. If $$\Delta$$ is a triangle with side lengths $$c\geq b\geq a$$, such that

$$a^{2}+b^{2}=c^{2},$$ then $$\Delta$$ is a right triangle.

$$a$$

$$b$$

$$c$$

Proof. Law of cosines

$c^2=a^2+b^2-2ab\cos(\theta).$

$$\theta$$

Theorem. If $$v$$ and $$w$$ are vectors in a real Hilbert space such that

$$\|v\|^2 + \|w\|^2 = \|v+w\|^2,$$ then $$\langle v,w\rangle = 0.$$

The Carpenter's Theorem

Proof.

$\|v+w\|^2 = \langle v+w,v+w\rangle = \|v\|^2+2\langle v,w\rangle + \|w\|^{2}$

$\|v+w\|^2=\|v\|^2+\|w\|^2 \quad \Rightarrow\quad 2\langle v,w\rangle = 0. \quad \Box$

Theorem. If $$d_{1},d_{2}$$ are two numbers in $$[0,1]$$ such that $$d_{1}+d_{2} = 1,$$ then there is a projection $$P$$ such that $$d_{1} = \|Pe_{1}\|^2$$ and $$d_{2} = \|Pe_{2}\|^2$$, that is,

$P = \begin{bmatrix} d_{1} & \alpha\\ \overline{\alpha} & d_{2}\end{bmatrix}.$

Proof.

• By the intermediate value theorem there is a projection $$P$$ such that $$\|Pe_{1}\|^{2}=d_{1}.$$
• By the Pythagorean theorem                $$\|Pe_{1}\|^2+\|Pe_{2}\|^2=1.$$
• Therefore, $\|Pe_{2}\|^2=d_{2}.$

Theorem. If $$(d_{i})_{i=1}^{n}$$ is a sequence of numbers in $$[0,1]$$ such that $\sum_{i=1}^{n}d_{i}\in\N\cup\{0\},$ then there is an $$n\times n$$ projection $$P$$ such that $\|Pe_{i}\|^{2} = d_{i} \quad\text{for}\quad i=1,\ldots,n.$

$\langle Pe_{i},e_{i}\rangle =$

Note that this means that the sequence on the diagonal of the matrix $$P$$ is $$(d_{i})_{i=1}^{n}$$.

Example. Consider the sequence

$\left(\frac{5}{7},\frac{5}{7},\frac{3}{7},\frac{1}{7}\right).$

$\left[\begin{array}{rrrr}\frac{5}{7} & -\frac{\sqrt{15}}{21} & -\frac{\sqrt{30}}{21} & \frac{\sqrt{5}}{7}\\[1ex] -\frac{\sqrt{15}}{21} & \frac{5}{7} & -\frac{2\sqrt{2}}{7} & -\frac{\sqrt{3}}{21}\\[1ex] -\frac{\sqrt{30}}{21} & -\frac{2\sqrt{2}}{7} & \frac{3}{7} & -\frac{\sqrt{6}}{21}\\[1ex] \frac{\sqrt{5}}{7} & -\frac{\sqrt{3}}{21} & -\frac{\sqrt{6}}{21} & \frac{1}{7}\end{array}\right]$

Challenge: Construct a $$4\times 4$$ projection with this diagonal.

Theorem. Assume $$(d_{i})_{i=1}^{n}$$ is a sequence of numbers in $$[0,1].$$ There is an $$n\times n$$ projection $$P$$ with diagonal $$(d_{i})_{i=1}^{n}$$ if and only if

$\sum_{i=1}^{n}d_{i} \in\N\cup\{0\}.$

Diagonality

Definition. Given an operator $$E$$ on a Hilbert space, a sequence $$(d_{i})_{i\in I}$$ is a diagonal of $$E$$ if there is an orthonormal basis $$(e_{i})_{i\in I}$$ such that

$d_{i} = \langle Ee_{i},e_{i}\rangle \quad\text{for all }i\in I.$

The problem: Given an operator $$E$$, characterize the set of diagonals of $$E$$, that is, the set

$\big\{(\langle Ee_{i},e_{i}\rangle )_{i\in I} : (e_{i})_{i\in I}\text{ is an orthonormal basis}\big\}$

In particular, we want a characterization in terms of linear inequalities between the diagonal sequences and the spectral information of $$E$$.

Note: If we fix an orthonormal basis $$(f_{i})_{i\in I}$$, then the set of diagonals of $$E$$ is also

$\big\{(\langle UEU^{\ast}f_{i},f_{i}\rangle )_{i\in I} : U\text{ is unitary}\big\}$

Projections in infinite dimensions

Examples. Let $$(e_{i})_{i=1}^{\infty}$$ be an orthonormal basis.

Set $$\displaystyle{v = \sum_{i=1}^{\infty}\sqrt{\frac{1}{2^{i}}}e_{i}},$$ then

$I-P = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2^{3/2}} & -\frac{1}{2^{2}} & \cdots \\[1ex] -\frac{1}{2^{3/2}} & \frac{3}{4} & -\frac{1}{2^{5/2}} & \cdots\\[1ex] -\frac{1}{2^{2}} & -\frac{1}{2^{5/2}} & \frac{7}{8} & \cdots\\ \vdots & \vdots & \vdots & \ddots\end{bmatrix}$

$P = \langle \cdot,v\rangle v = \begin{bmatrix} \frac{1}{2} & \frac{1}{2^{3/2}} & \frac{1}{2^{2}} & \cdots \\[1ex] \frac{1}{2^{3/2}} & \frac{1}{4} & \frac{1}{2^{5/2}} & \cdots\\[1ex] \frac{1}{2^{2}} & \frac{1}{2^{5/2}} & \frac{1}{8} & \cdots\\[1ex] \vdots & \vdots & \vdots & \ddots\end{bmatrix}$

Corank 1 projection

Diagonal: $$\displaystyle{\left(\frac{1}{2},\frac{3}{4},\frac{7}{8},\ldots\right)}$$

Rank 1 projection

Diagonal: $$\displaystyle{\left(\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots\right)}$$

Projections in infinite dimensions

Examples.

$\frac{1}{2}J_{2} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2}\\[1ex] \frac{1}{2} & \frac{1}{2}\end{bmatrix}$

$Q = \bigoplus_{i=1}^{\infty}\frac{1}{2}J_{2} = \begin{bmatrix} \frac{1}{2}J_{2} & \mathbf{0} & \mathbf{0} & \cdots\\ \mathbf{0} & \frac{1}{2}J_{2} & \mathbf{0} & \cdots\\ \mathbf{0} & \mathbf{0} & \frac{1}{2}J_{2} & \\ \vdots & \vdots & & \ddots\end{bmatrix}$

$$\infty$$-rank and $$\infty$$-corank

Diagonal: $$\displaystyle{\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\ldots\right)}$$

$$\infty$$-rank and $$\infty$$-corank

Diagonal: $$\displaystyle{\left(\ldots,\frac{1}{8},\frac{1}{4},\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{7}{8},\ldots\right)}$$

$P\oplus (I-P)$

Theorem. Assume $$(d_{i})_{i=1}^{n}$$ is a sequence of numbers in $$[0,1].$$ There is an $$n\times n$$ projection $$P$$ with diagonal $$(d_{i})_{i=1}^{n}$$ if and only if

$\sum_{i=1}^{n}d_{i} \in\N\cup\{0\}.$

Diagonals of Projections Redux

$\sum_{i=1}^{k}d_{i} - \sum_{i=k+1}^{n}(1-d_{i})\in\Z$

$\Updownarrow$

Theorem. Assume $$(d_{i})_{i=1}^{n}$$ is a sequence of numbers in $$[0,1].$$ There is an $$n\times n$$ projection $$P$$ with diagonal $$(d_{i})_{i=1}^{n}$$ if and only if

$\sum_{i=1}^{k}d_{i} - \sum_{i=k+1}^{n}(1-d_{i})\in\Z.$

Diagonals of Projections Redux

Theorem (Kadison '02). Assume $$(d_{i})_{i=1}^{\infty}$$ is a sequence of numbers in $$[0,1],$$ and set

$a=\sum_{d_{i}<\frac{1}{2}}d_{i}\quad\text{and}\quad b=\sum_{d_{i}\geq \frac{1}{2}}(1-d_{i})$

There is a projection $$P$$ with diagonal $$(d_{i})_{i=1}^{\infty}$$ if and only if one of the following holds:

1. $$a=\infty$$
2. $$b=\infty$$
3. $$a,b<\infty$$ and $$a-b\in\Z$$

Examples.

• There is a projection with every number in $$\mathbb{Q}\cap[0,1]$$ on the diagonal.
• There is a projection with diagonal $$(\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{4},\ldots)$$.
• There is no projection with diagonal $\left(\ldots,\frac{1}{25},\frac{1}{16},\frac{1}{9},\frac{1}{4},\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\ldots\right)$

Diagonals of $$2\times 2$$ self-adjoint matrices

Up to unitary equivalence $$E=\begin{bmatrix} \lambda_{1} & 0\\ 0 & \lambda_{2}\end{bmatrix}$$, with $$\lambda_{1}\geq \lambda_{2}$$.

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix} \lambda_{1} & 0\\ 0 & \lambda_{2}\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}$

$= \begin{bmatrix}\alpha\lambda_{1}+(1-\alpha)\lambda_{2} & \ast\\ \overline{\ast} & (1-\alpha)\lambda_{1} + \alpha\lambda_{2} \end{bmatrix},\quad (\alpha = \cos^{2}\theta)$

Hence $$(d_{1},d_{2})$$, with $$d_{1}\geq d_{2}$$, is a diagonal of $$E$$ if and only if

and

$\lambda_{1}+\lambda_{2}=d_{1}+d_{2}$.

$$\lambda_{2}\leq d_{1}\leq\lambda_{1}$$

$d_{1}\in\operatorname{conv}(\{\lambda_{1},\lambda_{2}\})$

$$d_{1}\leq\lambda_{1}$$

Givens rotations

$\begin{bmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} \lambda_{1} & 0 & 0\\ 0 & \lambda_{2} & 0\\ 0 & 0 & \lambda_{3}\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{bmatrix}$

$= \begin{bmatrix}\alpha\lambda_{1}+(1-\alpha)\lambda_{2} & \ast & 0\\ \overline{\ast} & (1-\alpha)\lambda_{1} + \alpha\lambda_{2} & 0\\ 0 & 0 & \lambda_{3}\end{bmatrix},\quad (\alpha = \cos^{2}\theta)$

Suppose $$E = \left[\begin{smallmatrix} \lambda_{1} & 0 & 0\\ 0 & \lambda_{2} & 0\\ 0 & 0 & \lambda_{3}\end{smallmatrix}\right]$$ with $$\lambda_{1}\geq \lambda_{2}\geq\lambda_{3}$$

$\begin{bmatrix}\cos\theta & 0 & -\sin\theta\\ 0 & 1 & 0\\ \sin\theta & 0 & \cos\theta\end{bmatrix}\begin{bmatrix} \mu_{1} & \ast & 0\\ \overline{\ast} & \mu_{2} & 0\\ 0 & 0 & \mu_{3}\end{bmatrix}\begin{bmatrix}\cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta\end{bmatrix}$

$= \begin{bmatrix}\alpha\mu_{1}+(1-\alpha)\lambda_{3} & \ast & \ast\\ \overline{\ast} & \mu_{2} & \ast\\ \overline{\ast} & \overline{\ast} & (1-\alpha)\mu_{1} + \alpha\mu_{3}\end{bmatrix},\quad (\alpha = \cos^{2}\theta)$

Diagonals of $$3\times 3$$ & $$4\times 4$$ self-adjoint matrices

Suppose $$E = \left[\begin{smallmatrix} \lambda_{1} & 0 & 0\\ 0 & \lambda_{2} & 0\\ 0 & 0 & \lambda_{3}\end{smallmatrix}\right]$$ with $$\lambda_{1}\geq \lambda_{2}\geq\lambda_{3}$$

Then $$(d_{1},d_{2},d_{3})$$, with $$d_{1}\geq d_{2}\geq d_{3}$$ is a diagonal of $$E$$ if and only if

$d_{i}\in\operatorname{conv}\{\lambda_{1},\lambda_{2},\lambda_{3}\}\quad\forall\,i$

and

$d_{1}+d_{2}+d_{3}=\lambda_{1}+\lambda_{2}+\lambda_{3}.$

However, $$(7,6,1,1)$$ is not a diagonal of $$E = \left[\begin{matrix} 8 & 0 & 0 & 0\\ 0 & 4 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 1\end{matrix}\right]$$

$$d_{1}\leq \lambda_{1}$$,            $$d_{1}+d_{2}\leq \lambda_{1}+\lambda_{2}$$,

Even though $7,6,1\in\operatorname{conv}(\{8,4,2,1\})=[1,8]$

and $7+6+1+1=15=8+4+2+1$

$$7+6=13\not\leq 12=8+4$$

The Schur-Horn Theorem

Theorem (Schur '23, Horn '54). Let $$(d_{i})_{i=1}^{n}$$ and $$(\lambda_{i})_{i=1}^{n}$$ be nonincreasing sequences. There is a self-adjoint matrix $$E$$ with diagonal $$(d_{i})_{i=1}^{n}$$ and eigenvalues $$(\lambda_{i})_{i=1}^{n}$$ if and only if

$\sum_{i=1}^{k}d_{i}\leq \sum_{i=1}^{k}\lambda_{i}\quad\text{for}\quad k=1,2,\ldots,n-1$

and

$\sum_{i=1}^{n}d_{i} = \sum_{i=1}^{n}\lambda_{i}.$

(1)

(2)

If (1) and (2) hold, then we say that $$(\lambda_{i})_{i=1}^{n}$$ majorizes $$(d_{i})_{i=1}^{n}$$, and we write $$(\lambda_{i})_{i=1}^{n}\succeq (d_{i})_{i=1}^{n}$$

$$(\lambda_{i})_{i=1}^{n}\succeq (d_{i})_{i=1}^{n}$$ is equivalent to saying that $$(d_{i})_{i=1}^{n}$$ is in the convex hull of the permutations of $$(\lambda_{i})_{i=1}^{n}$$.

Theorem (Arveson, Kadison '06, Kaftal, Weiss '10). Let $$(\lambda_{i})_{i=1}^{\infty}$$ be a positive nonincreasing sequence, and let $$(d_{i})_{i=1}^{\infty}$$ be a nonnegative nonincreasing sequence. There exists a positive compact operator with positive eigenvalues $$(\lambda_{i})_{i=1}^{\infty}$$ and diagonal $$(d_{i})_{i=1}^{\infty}$$ if and only if

$\sum_{i=1}^{k}d_{i}\leq \sum_{i=1}^{k}\lambda_{i}\quad\text{for all}\quad k\in\N$

and

$\sum_{i=1}^{\infty}d_{i} = \sum_{i=1}^{\infty}\lambda_{i}.$

Open question: What are the diagonals of positive compact operators with positive eigenvalues

$\left(1,\frac{1}{2},\frac{1}{3},\ldots\right)$ and a $$1$$-dimensional kernel.

Non-positive operators

Example. Consider the diagonal matrix $$E = \operatorname{diag}(-1,1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots)$$

$\begin{bmatrix} -1 & 0 & 0 & 0 & \cdots\\ 0 & 1 & 0 & 0 & \cdots\\ 0 & 0 & \frac{1}{2} & 0 & \cdots\\ 0 & 0 & 0 & \frac{1}{4}\\ \vdots & \vdots & \vdots & & \ddots\end{bmatrix} \simeq \begin{bmatrix} -\frac{1}{2} & \ast & 0 & 0 & \cdots\\ \overline{\ast} & \frac{1}{2} & 0 & 0 & \cdots\\ 0 & 0 & \frac{1}{2} & 0 & \cdots\\ 0 & 0 & 0 & \frac{1}{4}\\ \vdots & \vdots & \vdots & & \ddots\end{bmatrix} \simeq \begin{bmatrix} -\frac{1}{4} & \ast & \ast & 0 & \cdots\\ \overline{\ast} & \frac{1}{2} & \ast & 0 & \cdots\\ \overline{\ast} & \overline{\ast} & \frac{1}{4} & 0 & \cdots\\ 0 & 0 & 0 & \frac{1}{4}\\ \vdots & \vdots & \vdots & & \ddots\end{bmatrix}$

$\simeq\cdots\simeq \begin{bmatrix} 0 & \ast & \ast & \ast & \cdots\\ \overline{\ast} & \frac{1}{2} & \ast & \ast & \cdots\\ \overline{\ast} & \overline{\ast} & \frac{1}{4} & \ast & \cdots\\ \overline{\ast} & \overline{\ast} & \overline{\ast} & \frac{1}{8}\\ \vdots & \vdots & \vdots & & \ddots\end{bmatrix}$

Hence, $$(0,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots)$$ is a diagonal of $$E$$.

Definition. Let $$\boldsymbol\lambda =\{\lambda_i\}_{i\in I}\in c_0$$. Define its positive part $$\boldsymbol\lambda_+ =\{\lambda^{+}_i\}_{i\in I}$$ by $$\lambda^+_i=\max(\lambda_i,0)$$. The negative part is defined as $$\boldsymbol\lambda_-=(-\boldsymbol \lambda)_+.$$
If $$\boldsymbol\lambda \in c_0^+$$, then define its decreasing rearrangement $$\boldsymbol\lambda^{\downarrow} =\{\lambda^{\downarrow}_i\}_{i\in \N}$$ by taking $$\lambda^{\downarrow}_i$$ to be the $$i$$th largest term of $$\boldsymbol \lambda.$$ For the sake of brevity, we will denote the $$i$$th term of $$(\boldsymbol\lambda_{+})^{\downarrow}$$ by $$\lambda_{i}^{+\downarrow}$$, and similarly for $$(\boldsymbol\lambda_{-})^{\downarrow}$$.

$-\lambda^{-\downarrow}_{1}\leq -\lambda^{-\downarrow}_{2}\leq-\lambda^{-\downarrow}_{3}\leq\cdots\leq 0\leq\cdots\leq \lambda^{+\downarrow}_{3}\leq \lambda^{+\downarrow}_{2}\leq \lambda^{+\downarrow}_{1}$

Hence, we have "rearranged" the sequence $$\boldsymbol\lambda$$ as follows:

Note:

If $$\boldsymbol\lambda$$ has infinitely many positive terms, then $$\boldsymbol\lambda_{+}$$ has no zeros.

If $$\boldsymbol\lambda$$ has finitely many positive terms, then $$\boldsymbol\lambda_{+}$$ has infinitely many zeros.

Excess

Given a compact self-adjoint operator $$E$$ with eigenvalue list $$\boldsymbol\lambda$$, it is straightforward to show that if $$\boldsymbol{d}\in c_{0}$$ is a diagonal of $$E$$, then

$\sum_{i=1}^n \lambda_i^{+ \downarrow} \geq \sum_{i=1}^n d_i^{+ \downarrow} \quad\text{and }\quad \sum_{i=1}^n \lambda_i^{- \downarrow} \ge \sum_{i=1}^n d_i^{- \downarrow} \qquad\text{for all }k\in \N,$

But it is not true that

$\sum_{i=1}^\infty d_{i}^{+} = \sum_{i=1}^{\infty}\lambda_{i}^{+}\quad\text{and}\quad \sum_{i=1}^\infty d_{i}^{-} = \sum_{i=1}^{\infty}\lambda_{i}^{-}$

Indeed, $$(0,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots)$$ is a diagonal of an operator with eigenvalue list $$(-1,1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots)$$.

We need to keep track of how much "mass" was moved across zero:

$\sigma_{+} = \liminf_{n\to\infty} \sum_{i=1}^n (\lambda_i^{+ \downarrow} - d_i^{+ \downarrow})\quad\text{and}\quad \sigma_{-} = \liminf_{n\to\infty} \sum_{i=1}^n (\lambda_i^{- \downarrow} - d_i^{- \downarrow})$

$\sigma_{+} = \sum_{i=1}^n (\lambda_i^{+ \downarrow} - d_i^{+ \downarrow})\quad\text{and}\quad \sigma_{-} = \sum_{i=1}^n (\lambda_i^{- \downarrow} - d_i^{- \downarrow})$

Theorem (Bownik, J '22) Let $$\boldsymbol\lambda,\boldsymbol{d}\in c_{0}$$. Set

$\sigma_{+} = \liminf_{n\to\infty} \sum_{i=1}^n (\lambda_i^{+ \downarrow} - d_i^{+ \downarrow})\quad\text{and}\quad \sigma_{-} = \liminf_{n\to\infty} \sum_{i=1}^n (\lambda_i^{- \downarrow} - d_i^{- \downarrow})$

Let $$A$$ be a compact self-adjoint operator with eigenvalue list $$\boldsymbol\lambda.$$

The sequence $$\boldsymbol d$$ is a diagonal of an operator $$B$$ such that $$A\oplus \boldsymbol 0$$ and

$$B \oplus \boldsymbol 0$$ are unitarily equivalent, where $$\boldsymbol 0$$ denotes the zero operator on an infinite dimensional Hilbert space, if and only if the following four conditions hold:

$\sum_{i=1}^n \lambda_i^{+ \downarrow} \geq \sum_{i=1}^n d_i^{+ \downarrow} \quad\text{and }\quad \sum_{i=1}^n \lambda_i^{- \downarrow} \ge \sum_{i=1}^n d_i^{- \downarrow} \qquad\text{for all }k\in \N,$

$\boldsymbol d_+ \in \ell^1 \quad \implies\quad \sigma_{-}\geq \sigma_{+}.$

$\boldsymbol d_- \in \ell^1 \quad \implies\quad \sigma_{+}\geq \sigma_{-}$

Note: If $$\boldsymbol d\in\ell^{1}$$ ($$A$$ is trace class), then $$\sigma_{-} = \sigma_{+}$$. That is $$\sum d_{i} = \sum\lambda_{i}$$.

Theorem (Bownik, J '22) Let $$\boldsymbol\lambda,\boldsymbol{d}\in c_{0}$$.  Let $$A$$ be a compact self-adjoint operator with eigenvalue list $$\boldsymbol\lambda.$$ If

$\sum_{i=1}^n \lambda_i^{+ \downarrow} \geq \sum_{i=1}^n d_i^{+ \downarrow} \quad\text{and }\quad \sum_{i=1}^n \lambda_i^{- \downarrow} \ge \sum_{i=1}^n d_i^{- \downarrow} \qquad\text{for all }k\in \N,$

and

$\sigma_{+}=\sigma_{-}\in(0,\infty).$

then, the sequence $$\boldsymbol d$$ is a diagonal of $$A$$.

In some special cases we can get a stronger sufficiency result:

Not some operator with a different sized kernel.

Where did it go?

Example. Set $\boldsymbol\lambda = \left(-1,1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots\right)$ and $\boldsymbol{d} =\left (\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots\right)$

Then $\sigma_{+}=\sigma_{-}=1\in(0,\infty),$ and clearly

$\sum_{i=1}^n \lambda_i^{+ \downarrow} \geq \sum_{i=1}^n d_i^{+ \downarrow} \quad\text{and }\quad \sum_{i=1}^n \lambda_i^{- \downarrow} \ge \sum_{i=1}^n d_i^{- \downarrow} \qquad\text{for all }k\in \N,$

So, $$(0,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots)$$ and $$(\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots)$$ are diagonals of $$E=\operatorname{diag}(\boldsymbol\lambda)$$.

Thanks!

• This work was partially supported by NSF #1830066