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Min-max theorem

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Variational characterization of eigenvalues of compact Hermitian operators on Hilbert spaces

Not to be confused withMinimax theorem.

"Variational theorem" redirects here; not to be confused withvariational principle.

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Inlinear algebra andfunctional analysis, themin-max theorem, orvariational theorem, orCourant–Fischer–Weyl min-max principle, is a result that gives a variational characterization ofeigenvalues ofcompact Hermitian operators onHilbert spaces. It can be viewed as the starting point of many results of similar nature.

This article first discusses the finite-dimensional case and its applications before considering compact operators on infinite-dimensional Hilbert spaces. We will see that for compact operators, the proof of the main theorem uses essentially the same idea from the finite-dimensional argument.

In the case that the operator is non-Hermitian, the theorem provides an equivalent characterization of the associatedsingular values. The min-max theorem can be extended toself-adjoint operators that are bounded below.

Matrices

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LetA be an ×nHermitian matrix. As with many other variational results on eigenvalues, one considers theRayleigh–Ritz quotientR_A :Cⁿ \ {0} →R defined by

R_{A}(x)={\frac {(Ax,x)}{(x,x)}}

where(⋅, ⋅) denotes theEuclidean inner product onCⁿ. Equivalently, the Rayleigh–Ritz quotient can be replaced by

f(x)=(Ax,x),\;\|x\|=1.

The Rayleigh quotient of an eigenvector $v {\displaystyle v}$ is its associated eigenvalue $\lambda$ because $R_{A}(v)=(\lambda x,x)/(x,x)=\lambda$ . For a Hermitian matrixA, the range of the continuous functionsR_A(x) andf(x) is a compact interval [a,b] of the real line. The maximumb and the minimuma are the largest and smallest eigenvalue ofA, respectively. The min-max theorem is a refinement of this fact.

Min-max theorem

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Let ${\textstyle A}$ be Hermitian on an inner product space ${\textstyle V}$ with dimension ${\textstyle n}$ , with spectrum ordered in descending order ${\textstyle \lambda _{1}\geq ...\geq \lambda _{n}}$ .

Let ${\textstyle v_{1},...,v_{n}}$ be the corresponding unit-length orthogonal eigenvectors.

Reverse the spectrum ordering, so that ${\textstyle \xi _{1}=\lambda _{n},...,\xi _{n}=\lambda _{1}}$ .

(Poincaré’s inequality)—Let ${\textstyle M}$ be a subspace of ${\textstyle V}$ with dimension ${\textstyle k}$ , then there exists unit vectors ${\textstyle x,y\in M}$ , such that

${\textstyle \langle x,Ax\rangle \leq \lambda _{k}}$ , and ${\textstyle \langle y,Ay\rangle \geq \xi _{k}}$ .

Proof

Part 2 is a corollary, using ${\textstyle -A}$ .

${\textstyle M}$ is a ${\textstyle k}$ dimensional subspace, so if we pick any list of ${\textstyle n-k+1}$ vectors, their span ${\textstyle N:=span(v_{k},...v_{n})}$ must intersect ${\textstyle M}$ on at least a single line.

Take unit ${\textstyle x\in M\cap N}$ . That’s what we need.

{\textstyle x=\sum _{i=k}^{n}a_{i}v_{i}}

, since

{\textstyle x\in N}

Since

{\textstyle \sum _{i=k}^{n}|a_{i}|^{2}=1}

, we find

{\textstyle \langle x,Ax\rangle =\sum _{i=k}^{n}|a_{i}|^{2}\lambda _{i}\leq \lambda _{k}}

min-max theorem— ${\begin{aligned}\lambda _{k}&=\max _{\begin{array}{c}{\mathcal {M}}\subset V\\\operatorname {dim} ({\mathcal {M}})=k\end{array}}\min _{\begin{array}{c}x\in {\mathcal {M}}\\\|x\|=1\end{array}}\langle x,Ax\rangle \\&=\min _{\begin{array}{c}{\mathcal {M}}\subset V\\\operatorname {dim} ({\mathcal {M}})=n-k+1\end{array}}\max _{\begin{array}{c}x\in {\mathcal {M}}\\\|x\|=1\end{array}}\langle x,Ax\rangle {\text{. }}\end{aligned}}$

Proof

Part 2 is a corollary of part 1, by using ${\textstyle -A}$ .

By Poincare’s inequality, ${\textstyle \lambda _{k}}$ is an upper bound to the right side.

By setting ${\textstyle {\mathcal {M}}=span(v_{1},...v_{k})}$ , the upper bound is achieved.

Define thepartial trace ${\textstyle tr_{V}(A)}$ to be the trace of projection of ${\textstyle A}$ to ${\textstyle V}$ . It is equal to ${\textstyle \sum _{i}v_{i}^{*}Av_{i}}$ given an orthonormal basis of ${\textstyle V}$ .

Wielandt minimax formula (^[1]^: 44)—Let ${\textstyle 1\leq i_{1}<\cdots <i_{k}\leq n}$ be integers. Define a partial flag to be a nested collection ${\textstyle V_{1}\subset \cdots \subset V_{k}}$ of subspaces of ${\textstyle \mathbb {C} ^{n}}$ such that ${\textstyle \operatorname {dim} \left(V_{j}\right)=i_{j}}$ for all ${\textstyle 1\leq j\leq k}$ .

Define the associated Schubert variety ${\textstyle X\left(V_{1},\ldots ,V_{k}\right)}$ to be the collection of all ${\textstyle k}$ dimensional subspaces ${\textstyle W}$ such that ${\textstyle \operatorname {dim} \left(W\cap V_{j}\right)\geq j}$ .

$\lambda _{i_{1}}(A)+\cdots +\lambda _{i_{k}}(A)=\sup _{V_{1},\ldots ,V_{k}}\inf _{W\in X\left(V_{1},\ldots ,V_{k}\right)}tr_{W}(A)$

Proof

Proof

The ${\textstyle \leq }$ case.

Let ${\textstyle V_{j}=span(e_{1},\dots ,e_{i_{j}})}$ , and any ${\textstyle W\in X\left(V_{1},\ldots ,V_{k}\right)}$ , it remains to show that $\lambda _{i_{1}}(A)+\cdots +\lambda _{i_{k}}(A)\leq tr_{W}(A)$

To show this, we construct an orthonormal set of vectors ${\textstyle v_{1},\dots ,v_{k}}$ such that ${\textstyle v_{j}\in V_{j}\cap W}$ . Then ${\textstyle tr_{W}(A)\geq \sum _{j}\langle v_{j},Av_{j}\rangle \geq \lambda _{i_{j}}(A)}$

Since ${\textstyle dim(V_{1}\cap W)\geq 1}$ , we pick any unit ${\textstyle v_{1}\in V_{1}\cap W}$ . Next, since ${\textstyle dim(V_{2}\cap W)\geq 2}$ , we pick any unit ${\textstyle v_{2}\in (V_{2}\cap W)}$ that is perpendicular to ${\textstyle v_{1}}$ , and so on.

The ${\textstyle \geq }$ case.

For any such sequence of subspaces ${\textstyle V_{i}}$ , we must find some ${\textstyle W\in X\left(V_{1},\ldots ,V_{k}\right)}$ such that $\lambda _{i_{1}}(A)+\cdots +\lambda _{i_{k}}(A)\geq tr_{W}(A)$

Now we prove this by induction.

The ${\textstyle n=1}$ case is the Courant-Fischer theorem. Assume now ${\textstyle n\geq 2}$ .

If ${\textstyle i_{1}\geq 2}$ , then we can apply induction. Let ${\textstyle E=span(e_{i_{1}},\dots ,e_{n})}$ . We construct a partial flag within ${\textstyle E}$ from the intersection of ${\textstyle E}$ with ${\textstyle V_{1},\dots ,V_{k}}$ .

We begin by picking a ${\textstyle (i_{k}-(i_{1}-1))}$ -dimensional subspace ${\textstyle W_{k}'\subset E\cap V_{i_{k}}}$ , which exists by counting dimensions. This has codimension ${\textstyle (i_{1}-1)}$ within ${\textstyle V_{i_{k}}}$ .

Then we go down by one space, to pick a ${\textstyle (i_{k-1}-(i_{1}-1))}$ -dimensional subspace ${\textstyle W_{k-1}'\subset W_{k}\cap V_{i_{k-1}}}$ . This still exists. Etc. Now since ${\textstyle dim(E)\leq n-1}$ , apply the induction hypothesis, there exists some ${\textstyle W\in X(W_{1},\dots ,W_{k})}$ such that $\lambda _{i_{1}-(i_{1}-1)}(A|E)+\cdots +\lambda _{i_{k}-(i_{1}-1)}(A|E)\geq tr_{W}(A)$ Now ${\textstyle \lambda _{i_{j}-(i_{1}-1)}(A|E)}$ is the ${\textstyle (i_{j}-(i_{1}-1))}$ -th eigenvalue of ${\textstyle A}$ orthogonally projected down to ${\textstyle E}$ . By Cauchy interlacing theorem, ${\textstyle \lambda _{i_{j}-(i_{1}-1)}(A|E)\leq \lambda _{i_{j}}(A)}$ . Since ${\textstyle X(W_{1},\dots ,W_{k})\subset X(V_{1},\dots ,V_{k})}$ , we’re done.

If ${\textstyle i_{1}=1}$ , then we perform a similar construction. Let ${\textstyle E=span(e_{2},\dots ,e_{n})}$ . If ${\textstyle V_{k}\subset E}$ , then we can induct. Otherwise, we construct a partial flag sequence ${\textstyle W_{2},\dots ,W_{k}}$ By induction, there exists some ${\textstyle W'\in X(W_{2},\dots ,W_{k})\subset X(V_{2},\dots ,V_{k})}$ , such that $\lambda _{i_{2}-1}(A|E)+\cdots +\lambda _{i_{k}-1}(A|E)\geq tr_{W'}(A)$ thus
$\lambda _{i_{2}}(A)+\cdots +\lambda _{i_{k}}(A)\geq tr_{W'}(A)$ And it remains to find some ${\textstyle v}$ such that ${\textstyle W'\oplus v\in X(V_{1},\dots ,V_{k})}$ .

If ${\textstyle V_{1}\not \subset W'}$ , then any ${\textstyle v\in V_{1}\setminus W'}$ would work. Otherwise, if ${\textstyle V_{2}\not \subset W'}$ , then any ${\textstyle v\in V_{2}\setminus W'}$ would work, and so on. If none of these work, then it means ${\textstyle V_{k}\subset E}$ , contradiction.

This has some corollaries:^[1]^: 44

Extremal partial trace— $\lambda _{1}(A)+\dots +\lambda _{k}(A)=\sup _{\operatorname {dim} (V)=k}tr_{V}(A)$

$\xi _{1}(A)+\dots +\xi _{k}(A)=\inf _{\operatorname {dim} (V)=k}tr_{V}(A)$

Corollary—The sum ${\textstyle \lambda _{1}(A)+\dots +\lambda _{k}(A)}$ is a convex function, and ${\textstyle \xi _{1}(A)+\dots +\xi _{k}(A)}$ is concave.

(Schur-Horn inequality) $\xi _{1}(A)+\dots +\xi _{k}(A)\leq a_{i_{1},i_{1}}+\dots +a_{i_{k},i_{k}}\leq \lambda _{1}(A)+\dots +\lambda _{k}(A)$ for any subset of indices.

Equivalently, this states that the diagonal vector of ${\textstyle A}$ is majorized by its eigenspectrum.

Schatten-norm Hölder inequality—Given Hermitian ${\textstyle A,B}$ and Hölder pair ${\textstyle 1/p+1/q=1}$ , $|\operatorname {tr} (AB)|\leq \|A\|_{S^{p}}\|B\|_{S^{q}}$

Proof

Proof

WLOG, ${\textstyle B}$ is diagonalized, then we need to show ${\textstyle |\sum _{i}B_{ii}A_{ii}|\leq \|A\|_{S^{p}}\|(B_{ii})\|_{l^{q}}}$

By the standard Hölder inequality, it suffices to show ${\textstyle \|(A_{ii})\|_{l^{p}}\leq \|A\|_{S^{p}}}$

By the Schur-Horn inequality, the diagonals of ${\textstyle A}$ are majorized by the eigenspectrum of ${\textstyle A}$ , and since the map ${\textstyle f(x_{1},\dots ,x_{n})=\|x\|_{p}}$ is symmetric and convex, it is Schur-convex.

Counterexample in the non-Hermitian case

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LetN be the nilpotent matrix

{\begin{bmatrix}0&1\\0&0\end{bmatrix}}.

Define the Rayleigh quotient $R_{N}(x)$ exactly as above in the Hermitian case. Then it is easy to see that the only eigenvalue ofN is zero, while the maximum value of the Rayleigh quotient is⁠1/2⁠. That is, the maximum value of the Rayleigh quotient is larger than the maximum eigenvalue.

Applications

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Min-max principle for singular values

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Thesingular values {σ_k} of a square matrixM are the square roots of the eigenvalues ofM*M (equivalentlyMM*). An immediate consequence^{[citation needed]} of the first equality in the min-max theorem is:

\sigma _{k}^{\downarrow }=\max _{S:\dim(S)=k}\min _{x\in S,\|x\|=1}(M^{*}Mx,x)^{\frac {1}{2}}=\max _{S:\dim(S)=k}\min _{x\in S,\|x\|=1}\|Mx\|.

Similarly,

\sigma _{k}^{\downarrow }=\min _{S:\dim(S)=n-k+1}\max _{x\in S,\|x\|=1}\|Mx\|.

Here $\sigma _{k}^{\downarrow }$ denotes thek^th entry in the decreasing sequence of the singular values, so that $\sigma _{1}^{\downarrow }\geq \sigma _{2}^{\downarrow }\geq \cdots$ .

Cauchy interlacing theorem

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Main article:Poincaré separation theorem

LetA be a symmetricn ×n matrix. Them ×m matrixB, wherem ≤n, is called acompression ofA if there exists anorthogonal projectionP onto a subspace of dimensionm such thatPAP* =B. The Cauchy interlacing theorem states:

Theorem. If the eigenvalues ofA areα₁ ≤ ... ≤α_n, and those ofB areβ₁ ≤ ... ≤β_j ≤ ... ≤β_m, then for allj ≤m,

\alpha _{j}\leq \beta _{j}\leq \alpha _{n-m+j}.

This can be proven using the min-max principle. Letβ_i have corresponding eigenvectorb_i andS_j be thej dimensional subspaceS_j = span{b₁, ...,b_j}, then

\beta _{j}=\max _{x\in S_{j},\|x\|=1}(Bx,x)=\max _{x\in S_{j},\|x\|=1}(PAP^{*}x,x)\geq \min _{S_{j}}\max _{x\in S_{j},\|x\|=1}(A(P^{*}x),P^{*}x)=\alpha _{j}.

According to first part of min-max,α_j ≤β_j. On the other hand, if we defineS_m−j+1 = span{b_j, ...,b_m}, then

\beta _{j}=\min _{x\in S_{m-j+1},\|x\|=1}(Bx,x)=\min _{x\in S_{m-j+1},\|x\|=1}(PAP^{*}x,x)=\min _{x\in S_{m-j+1},\|x\|=1}(A(P^{*}x),P^{*}x)\leq \alpha _{n-m+j},

where the last inequality is given by the second part of min-max.

Whenn −m = 1, we haveα_j ≤β_j ≤α_j+1, hence the nameinterlacing theorem.

Lidskii's inequality

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Main article:Trace class § Lidskii's theorem

Lidskii inequality—If ${\textstyle 1\leq i_{1}<\cdots <i_{k}\leq n}$ then ${\begin{aligned}&\lambda _{i_{1}}(A+B)+\cdots +\lambda _{i_{k}}(A+B)\\&\quad \leq \lambda _{i_{1}}(A)+\cdots +\lambda _{i_{k}}(A)+\lambda _{1}(B)+\cdots +\lambda _{k}(B)\end{aligned}}$

${\begin{aligned}&\lambda _{i_{1}}(A+B)+\cdots +\lambda _{i_{k}}(A+B)\\&\quad \geq \lambda _{i_{1}}(A)+\cdots +\lambda _{i_{k}}(A)+\xi _{1}(B)+\cdots +\xi _{k}(B)\end{aligned}}$

Proof

Proof

The second is the negative of the first. The first is by Wielandt minimax.

${\begin{aligned}&\lambda _{i_{1}}(A+B)+\cdots +\lambda _{i_{k}}(A+B)\\=&\sup _{V_{1},\dots ,V_{k}}\inf _{W\in X(V_{1},\dots ,V_{k})}(tr_{W}(A)+tr_{W}(B))\\=&\sup _{V_{1},\dots ,V_{k}}(\inf _{W\in X(V_{1},\dots ,V_{k})}tr_{W}(A)+tr_{W}(B))\\\leq &\sup _{V_{1},\dots ,V_{k}}(\inf _{W\in X(V_{1},\dots ,V_{k})}tr_{W}(A)+(\lambda _{1}(B)+\cdots +\lambda _{k}(B)))\\=&\lambda _{i_{1}}(A)+\cdots +\lambda _{i_{k}}(A)+\lambda _{1}(B)+\cdots +\lambda _{k}(B)\end{aligned}}$

Note that $\sum _{i}\lambda _{i}(A+B)=tr(A+B)=\sum _{i}\lambda _{i}(A)+\lambda _{i}(B)$ . In other words, $\lambda (A+B)-\lambda (A)\preceq \lambda (B)$ where $\preceq$ meansmajorization. By the Schur convexity theorem, we then have

p-Wielandt-Hoffman inequality— ${\textstyle \|\lambda (A+B)-\lambda (A)\|_{\ell ^{p}}\leq \|B\|_{S^{p}}}$ where ${\textstyle \|\cdot \|_{S^{p}}}$ stands for the p-Schatten norm.

Compact operators

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LetA be acompact,Hermitian operator on a Hilbert spaceH. Recall that the non-zerospectrum of such an operator consists of real eigenvalues with finite multiplicities whose only possiblecluster point is zero. IfA has infinitely many positive eigenvalues, they accumulate at zero. In this case, we list the positive eigenvalues ofA as

\cdots \leq \lambda _{k}\leq \cdots \leq \lambda _{1},

where entries are repeated withmultiplicity, as in the matrix case. (To emphasize that the sequence is decreasing, we may write $\lambda _{k}=\lambda _{k}^{\downarrow }$ .) We now apply the same reasoning as in the matrix case. LettingS_k ⊂H be ak dimensional subspace, we can obtain the following theorem.

Theorem (Min-Max). LetA be a compact, self-adjoint operator on a Hilbert spaceH, whose positive eigenvalues are listed in decreasing order... ≤λ_k ≤ ... ≤λ₁. Then:

{\begin{aligned}\max _{S_{k}}\min _{x\in S_{k},\|x\|=1}(Ax,x)&=\lambda _{k}^{\downarrow },\\\min _{S_{k-1}}\max _{x\in S_{k-1}^{\perp },\|x\|=1}(Ax,x)&=\lambda _{k}^{\downarrow }.\end{aligned}}

A similar pair of equalities hold for negative eigenvalues.

Proof

LetS' be the closure of the linear span $S'=\operatorname {span} \{u_{k},u_{k+1},\ldots \}$ .The subspaceS' has codimensionk − 1. By the same dimension count argument as in the matrix case,S' ∩S_k has positive dimension. So there existsx ∈S' ∩S_k with $\|x\|=1$ . Since it is an element ofS', such anx necessarily satisfy

(Ax,x)\leq \lambda _{k}.

Therefore, for allS_k

\inf _{x\in S_{k},\|x\|=1}(Ax,x)\leq \lambda _{k}

ButA is compact, therefore the functionf(x) = (Ax,x) is weakly continuous. Furthermore, any bounded set inH is weakly compact. This lets us replace the infimum by minimum:

\min _{x\in S_{k},\|x\|=1}(Ax,x)\leq \lambda _{k}.

\sup _{S_{k}}\min _{x\in S_{k},\|x\|=1}(Ax,x)\leq \lambda _{k}.

Because equality is achieved when $S_{k}=\operatorname {span} \{u_{1},\ldots ,u_{k}\}$ ,

\max _{S_{k}}\min _{x\in S_{k},\|x\|=1}(Ax,x)=\lambda _{k}.

This is the first part of min-max theorem for compact self-adjoint operators.

Analogously, consider now a(k − 1)-dimensional subspaceS_k−1, whose the orthogonal complement is denoted byS_k−1^⊥. IfS' = span{u₁...u_k},

S'\cap S_{k-1}^{\perp }\neq {0}.

\exists x\in S_{k-1}^{\perp }\,\|x\|=1,(Ax,x)\geq \lambda _{k}.

This implies

\max _{x\in S_{k-1}^{\perp },\|x\|=1}(Ax,x)\geq \lambda _{k}

where the compactness ofA was applied. Index the above by the collection ofk-1-dimensional subspaces gives

\inf _{S_{k-1}}\max _{x\in S_{k-1}^{\perp },\|x\|=1}(Ax,x)\geq \lambda _{k}.

PickS_k−1 = span{u₁, ...,u_k−1} and we deduce

\min _{S_{k-1}}\max _{x\in S_{k-1}^{\perp },\|x\|=1}(Ax,x)=\lambda _{k}.

Self-adjoint operators

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The min-max theorem also applies to (possibly unbounded) self-adjoint operators.^[2]^[3] Recall theessential spectrum is the spectrum without isolated eigenvalues of finite multiplicity. Sometimes we have some eigenvalues below the essential spectrum, and we would like to approximate the eigenvalues and eigenfunctions.

Theorem (Min-Max). LetA be self-adjoint, and let

E_{1}\leq E_{2}\leq E_{3}\leq \cdots

be the eigenvalues ofA below the essential spectrum. Then

$E_{n}=\min _{\psi _{1},\ldots ,\psi _{n}}\max\{\langle \psi ,A\psi \rangle :\psi \in \operatorname {span} (\psi _{1},\ldots ,\psi _{n}),\,\|\psi \|=1\}$ .

If we only haveN eigenvalues and hence run out of eigenvalues, then we let $E_{n}:=\inf \sigma _{ess}(A)$ (the bottom of the essential spectrum) forn>N, and the above statement holds after replacing min-max with inf-sup.

Theorem (Max-Min). LetA be self-adjoint, and let

E_{1}\leq E_{2}\leq E_{3}\leq \cdots

be the eigenvalues ofA below the essential spectrum. Then

$E_{n}=\max _{\psi _{1},\ldots ,\psi _{n-1}}\min\{\langle \psi ,A\psi \rangle :\psi \perp \psi _{1},\ldots ,\psi _{n-1},\,\|\psi \|=1\}$ .

If we only haveN eigenvalues and hence run out of eigenvalues, then we let $E_{n}:=\inf \sigma _{ess}(A)$ (the bottom of the essential spectrum) forn > N, and the above statement holds after replacing max-min with sup-inf.

The proofs^[2]^[3] use the following results about self-adjoint operators:

Theorem. LetA be self-adjoint. Then

(A-E)\geq 0

for

E\in \mathbb {R}

if and only if

\sigma (A)\subseteq [E,\infty )

.^[2]^: 77

Theorem. IfA is self-adjoint, then

$\inf \sigma (A)=\inf _{\psi \in {\mathfrak {D}}(A),\|\psi \|=1}\langle \psi ,A\psi \rangle$

and

$\sup \sigma (A)=\sup _{\psi \in {\mathfrak {D}}(A),\|\psi \|=1}\langle \psi ,A\psi \rangle$ .^[2]^: 77

References

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^^a ^bTao, Terence (2012).Topics in random matrix theory. Graduate studies in mathematics. Providence, R.I: American Mathematical Society.ISBN 978-0-8218-7430-1.
^^a ^b ^c ^dG. Teschl, Mathematical Methods in Quantum Mechanics (GSM 99)https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe.pdf
^^a ^bLieb; Loss (2001).Analysis. GSM. Vol. 14 (2nd ed.). Providence: American Mathematical Society.ISBN 0-8218-2783-9.

External links and citations to related work

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Fisk, Steve (2005). "A very short proof of Cauchy's interlace theorem for eigenvalues of Hermitian matrices".arXiv:math/0502408.
Hwang, Suk-Geun (2004)."Cauchy's Interlace Theorem for Eigenvalues of Hermitian Matrices".The American Mathematical Monthly.111 (2):157–159.doi:10.2307/4145217.JSTOR 4145217.
Kline, Jeffery (2020)."Bordered Hermitian matrices and sums of the Möbius function".Linear Algebra and Its Applications.588:224–237.doi:10.1016/j.laa.2019.12.004.
Reed, Michael; Simon, Barry (1978).Methods of Modern Mathematical Physics IV: Analysis of Operators. Academic Press.ISBN 978-0-08-057045-7.
Edmunds, D. E.; Evans, W. D. (2018)."11.1 The Max–Min Principle for Semi-Bounded, Self-Adjoint Operators".Spectral theory and differential operators. Oxford science publications (2 ed.). Oxford: Oxford University Press.ISBN 978-0-19-881205-0.

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