Connective constant of a hexagonal lattice

reference request – Eigenvalues of the product of two symmetric matrices Answer

Hello pricey customer to our community We will proffer you an answer to this query reference request – Eigenvalues of the product of two symmetric matrices ,and the respond will breathe typical via documented data sources, We welcome you and proffer you fresh questions and solutions, Many customer are questioning concerning the respond to this query.

reference request – Eigenvalues of the product of two symmetric matrices

Let $A,B$ breathe $n instances n$ Hermitian matrices whose eigen values are non-negatives. Let $lambda_1(X) geq
lambda_1(X) geq lambda_2(X) geq cdots geq lambda_n(X)$
denote
ordered eigenvalues of $n instances n$ Hermitian matrix $X$.

Then, $lambda_i(AB) leq lambda_i(A)lambda_{j-i+1}(B),$
for any $i leq j$.

Proof.
Let $a_i,b_i$ and $c_i$ breathe the unitary eigenvectors of $A,B$ and $A+B$, respectively. Namely, $A a_i = lambda_i(A)a_i$ and so forth.

Let $V_a,V_b$ and $V_c$ are vector areas spanned by the vectors ${a_i,…,a_n}$ and ${b_{j-i+1},…,b_n}$ and ${c_1,…,c_j}$, respectively. Then, $V_a cap V_b cap V_c neq 0$. In truth, for capricious vector areas $X,Y$, we will say $dim (X cap Y) = dim(X)+dim(X) -dim(X+Y )$ and utilizing this twice, it follows that

$dim(V_a cap V_b cap V_c ) =dim(V_a cap( V_b cap V_c) )$

$=dim(V_a) + dim( V_b cap V_c) – dim(V_a + ( V_b cap V_c)) $

$=dim(V_a) + dim( V_b ) + dim( V_c) – dim( V_b +V_c) – dim(V_a + ( V_b cap V_c)) $

$geqdim(V_a) + dim( V_b ) + dim( V_c) – n -n $

$= (n-i+1) + (n-j+i-1)+1 + j – n -n$

$= 1,$

and thus there’s a nonzero unit vector $x in V_a cap V_b cap V_c.$

Because $x in V_c$, we will write $x= sum_{nu =1}^j x^nu c_nu$ after which,

$<x,(AB)x>= < sum_{nu =1}^j x^nu c_nu , (AB) sum_{nu =1}^j x^nu c_nu >$

$ = < sum_{nu =1}^j x^nu c_nu , sum_{nu =1}^j x^nu lambda_nu(AB)c_nu >$

$ = sum_{nu =1}^j | x^nu |^2 lambda_nu(AB)$

$ geq sum_{nu =1}^j | x^nu |^2 lambda_j(AB)$

$ = lambda_j(AB) cdots cdots cdots (1).$

Similarly, as a result of $x in V_a$, we will write $x= sum_{nu =i}^n x^nu a_nu$ after which,

$<x,Ax>= < sum_{nu =i}^n x^nu a_nu , Asum_{nu =i}^n x^nu a_nu >$

$ = < sum_{nu =i}^n x^nu a_nu , Asum_{nu =i}^n x^nu a_nu >$

$ = sum_{nu =i}^n | x^nu |^2 lambda_nu(A )$

$ leq sum_{nu =i}^j | x^nu |^2 lambda_i(A )$

$ = lambda_i(A ) cdots cdots cdots (2).$

Similarly,
$<x,Bx> leq lambda_{j-i+1}(B ) cdots cdots cdots (3).$

Combining the inequalities (1),(2),(3) and Cauchy Schwartz inequality, we get

$lambda_j(AB) leq <x,(AB)x> leq |<A^*x , Bx>| leq |A^*x| dot | Bx| leq lambda_i(A ) lambda_{j-i+1}(B ). $

This completes the proof.

we are going to proffer you the answer to reference request – Eigenvalues of the product of two symmetric matrices query by way of our community which brings all of the solutions from a number of dependable sources.

Add comment