It is unique if you require $diag(L) = (1, \dots 1)$. Because, assume that the LU-decomposition is not unique. Any other LU-decomp would be of the form: $$A=LU=LI_nU=LDD^{-1}U=(LD)(D^{-1}U)$$ with $... View answer Accepted answer 1 votes The way to prove$A_{i,i}$is positive is as follows: We know 2 things because$A$is an M-matrix:$A^{-1}_{i,j} \geq 0 \quad \forall i,jA_{i,j} \leq 0 \quad \forall i\neq j$Then$0 \leq 1 = (A*...

(1) First, we prove $F_{2,2d}^a$ is the sum of two squares: So $F_{2,2d}^a=a_1x_1^{2d} +a_2x_2^{2d} -2dx_1^{a_1}x_2^{a_2}$ We can take the homonization $P(x) = a_1x^{2d} + a_2 -2dx^{a_1}$, which is ...

I can only answer the (1) versions: So, assuming $f\geq 0$ on [-1,1] $\Rightarrow^{Q2.1}$ $G(f)(x) = (1+x)^df\big(\frac{1-x}{1+x}\big)\geq0$ on $\mathbb{R}_+$. Then from Polya's theorem, we know we ...
Say $S\subset V$ is the maximum stable set of $G$. When we can form an orthogonal representation in the following way: for all $i,j\in S$ we need $u_i^Tu_j =0$ because $\{i,j\}\in \bar{E}$. Say \$S = \{...