Proof of Schur Product Theorem Does anyone know how I can find a proof using operators of Schur's Product Theorem? Most proofs I have seen are very terse.
Is there a way to prove it with operators and their matrices? Thanks!
 A: If the basis $\{v_1, . . . , v_n\}$ is orthonormal, then for any two vectors $v, w$ ∈ $V$
with $v = ∑ a_iv_i$ and $w = ∑ b_jv_j$ , then
$$
⟨v, w⟩ =∑a_ib_j ⟨v_i, v_j ⟩ =∑a_ib_jδ_{ij} = ∑a_ib_i
$$
so if we identify $V$ with $\mathbb{R}^n$ via the basis (which we will do from now on), the inner product becomes the usual dot product. This is not true if the basis is not orthonormal, and in fact the statement of the problem is false in that case.
If $A$ is positive, by the Spectral Theorem, there is an orthonormal eigenbasis $\{e_1, . . . , e_n\}$ with nonnegative eigenvalues. 
In terms of matrices,
$M =∑λ_ie_ie^{T}_i$
with $λ_i ≥ 0$. Similarly, $B$ also has an orthonormal eigenbasis $\{f_1, . . . , f_n\}$ with eigenvalues $ν_j ≥ 0$, so in terms of matrices, $N = ∑ν_jf_jf^{T}_j$.
If we denote the componentwise product of matrices by ○, then
$$L = M ○ N = (∑λ_ie_ie^{T}_i) ○(∑ν_jf_jf^{T}_j)=∑λ_iν_j (e_ie^{T}_i) ○ (f_jf^{T}_j)=
∑
λ_iν_j (e_i ○ f_j)(e_i ○ f_j)^T
$$
Now let the latter matrix $L$ define a positive operator, $R$. If $v ∈ V$ is an arbitrary
vector, then
$$⟨v, Rv⟩ = v^{T}Lv=
∑
λ_iν_jv^T(e_i ○ f_j)(e_i ○ f_j)^{T}v=∑λ_iν_jv^{T}(e_i ○ f_j))^2 ≥ 0
$$
because each term in the sum is nonnegative.
