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

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## 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,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

This completes the proof.

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