Existence and Uniqueness of the Positive Definite Solution for the Matrix Equation $$

Theorem 1. Equation (6) has a positive semidefinite solution for any A ∈ 𝒞(n).

Proof. A matrix Y is a solution of (6) if and only if it is a fixed point of the map F defined by $$. Note that F maps [0, F(0)] into itself, because F is order reversing. Hence it has a fixed point in the [0, F(0)]. That is to says (6) has a positive semidefinite solution.

Theorem 2. If Y is positive semidefinite solution of (6), then Y ∈ [0, F(0)].

Proof. By Y ≥ 0, and C₁ > 0, we know that $$ and $$; then $$. By Y ≥ 0, we know that $$. That is, Y ∈ [0, F(0)].

Lemma 3. Let $$. For any Y ∈ [0, F(0)] and 0 < t < 1, one has

Proof. By Theorem 2, we know F(Y)∈[0, F(0)] for any Y ∈ [0, F(0)]. Then F²(Y)∈[0, F(0)] for any Y ∈ [0, F(0)]. Let $$, and $$. For any Y ∈ [0, F(0)] and 0 < t < 1, we have

Theorem 4. Equation (6) has a unique positive definite solution Y ∈ [0, F(0)], and for any Y₀ ∈ [0, F(0)], the iteration

converges to Y; that is, lim _n→∞Y_n = Y.

Proof. Consider the matrix sequence (9). Let $$. We first suppose that Y₀ = 0. Then we get

Analogously one can prove that Hence, the sequence {F^2k(0),  k ≥ 0} is monotone increasing and is bounded from above by F(0). Thus, the sequence {F^2k(0),  k ≥ 0} has a finite positive definite limit. Moreover, the sequence {F^2k+1(0),  k ≥ 0} is monotone decreasing and is bounded from below by 0. Thus, the sequence {F^2k+1(0),  k ≥ 0} has a finite positive definite limit. Let then 0 ≤ Y⁽ⁱ⁾ ≤ F(0) for i = 1,2. Clearly, both Y⁽¹⁾ and Y⁽²⁾ are the positive fixed points of F²(Y). Then Let t₀ = sup   {t∣Y⁽¹⁾ ≥ tY⁽²⁾}; then 0 < t₀ < +∞. Now we prove that t₀ ≥ 1. Assume, on the contrary, that 0 < t₀ < 1; then Y⁽¹⁾ ≥ t₀Y⁽²⁾. By Lemma 3 and monotonicity of F²(Y), we have Since (1 + η(t₀))t₀ > t₀, then it is a contradiction to the definition of t₀. Hence we have t₀ ≥ 1; therefore Y⁽¹⁾ ≥ Y⁽²⁾. Similarly, we can get Y⁽¹⁾ ≤ Y⁽²⁾. Therefore, Y⁽¹⁾ = Y⁽²⁾. Let Ω = [0, F(0)]. By Theorem 1, we know F(Ω)⊆Ω. Hence F²(Ω)⊆Ω. That is, F² has only one positive fixed point in Ω. In other words, the equation Y = F²(Y) has only one positive definite solution. Hence Y⁽¹⁾ = Y⁽²⁾. It follows that lim _n→∞Fⁿ(0) is a fixed point of F². Since the positive definite solution of F(Y) = Y solves Y = F²(Y), then F²(Y) = Y has only one positive definite solution. Hence lim _n→∞F⁽ⁿ⁾(0) is the unique fixed point of F.

For any Y₀ > 0, we get

By induction, we have for any k = 1,2, … Now letting k → ∞ on both sides of the above inequalities, we can get

Since (1) is equal to (6) when X = Y + Q, then we know that X is a positive definite solution of (1) if and only if Y is a positive semidefinite solution of (6). And furthermore, Y ∈ [0, F(0)] if and only if X ∈ [Q, G₁(Q)]. Thus, we can get the following conclusions about (1).

Theorem 5. Equation (1) with $$ has a positive definite solution for any A ∈ 𝒞(n).

Theorem 6. If X is positive definite solution of (1) with $$, then X ∈ [Q, G₁(Q)] ⊂ S₁(n).

Theorem 7. Let Q ∈ 𝒫(n) and $$ such that $$. Then (1) has a unique solution X in S₁(n) and for any X₀ ∈ [Q, G₁(Q)], the iteration

converges to X; that is, lim _n→∞X_n = X.

Theorem 8. Equation (1) with $$ and $$ has a positive definite solution for any A ∈ 𝒞(n).

Theorem 9. If X is a positive definite solution of (1) with $$ and $$, then X ∈ [G₂(Q), Q] ⊂ S₂(n).

Theorem 10. Let Q ∈ 𝒫(n) and $$ such that $$. Then (1) has a unique solution X in S₂(n), and for any X₀ ∈ [G₂(Q), Q], the iteration

converges to X; that is, lim _n→∞X_n = X.