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For any vectors $x$ and $y$ in an inner product space, $|\langle x, y \rangle| \leq \|x\| \cdot \|y\|$. Example: For $x = (1,2)$ and $y = (3,4)$, $|1 \cdot 3 + 2 \cdot 4| = 11 \leq \sqrt{1^2 + 2^2} \cdot \sqrt{3^2 + 4^2}$ which simplifies to $11 \leq \sqrt{5} \cdot \sqrt{25} = 11$.

For any matrices $A$ and $B$ of the same size, $\|A + B\| \leq \|A\| + \|B\|$. Example: If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$, then $\|A+B\|$ should be less than or equal to $\|A\|+\|B\|$ for any chosen matrix norm.

A symmetric matrix $A$ is positive definite if and only if all leading principal minors of $A$ are positive. Example: The matrix $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ is positive definite because all its principal minors (2, 3) are positive.

For any matrices $A$ and $B$, $\|A \cdot B\|_F \leq \|A\|_F \cdot \|B\|_F$. Example: If $A$ and $B$ are $2 \times 2$ matrices, then the Frobenius norm of their product should not exceed the product of their Frobenius norms.

For any two positive semidefinite matrices $A$ and $B$, their Schur (or Hadamard) product $A \circ B$ is also positive semidefinite. Example: Let $A$ and $B$ be $2 \times 2$ positive semidefinite matrices; their Schur product will have non-negative eigenvalues.

If $A$ is a square matrix and $B$ is a positive semidefinite matrix, then $\text{tr}(AB) \geq 0$. Example: For $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B$ being any positive semidefinite matrix, the trace of their product will be non-negative.

For any matrix $A$ with rows $r_1, r_2, \ldots, r_n$, the determinant satisfies $|\text{det}(A)| \leq \|r_1\| \cdot \|r_2\| \cdots \|r_n\|$. Example: For matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, $|\text{det}(A)| = |(1)(4) - (2)(3)| = |4 - 6| = 2 \leq \sqrt{1^2 + 2^2} \cdot \sqrt{3^2 + 4^2} = \sqrt{5} \cdot \sqrt{25} = 11$.

For any matrices $A$ and $B$ of the same size with singular values $\sigma_1 \geq \sigma_2 \geq \ldots \geq \sigma_n$ and $\tau_1 \geq \tau_2 \geq \ldots \geq \tau_n$ respectively, $|\text{tr}(A^*B)| \leq \sum_{i=1}^{n}\sigma_i\tau_i$. Example: If $A = \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}$, then $|\text{tr}(A^*B)| = |\text{tr}(AB)| = |3 \cdot 1 + 1 \cdot 3| = 6 \leq (3 \cdot 1 + 1 \cdot 3) = 6$.

For Hermitian matrices $A$ and $B$ with eigenvalues $\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n$ and $\mu_1 \geq \mu_2 \geq \ldots \geq \mu_n$ respectively, $\text{tr}(AB) \leq \sum_{i=1}^{n}\lambda_i\mu_i$. Example: If $A = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix}$, then $\text{tr}(AB) = 4 \cdot 3 + 2 \cdot 1 = 14 \leq (4 \cdot 3 + 2 \cdot 1) = 14$.

If $A$ and $B$ are positive semidefinite matrices, then $A + B$ is also positive semidefinite. Example: Given that $A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ are positive semidefinite, their sum $\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$ is also positive semidefinite.

Given any $p, q, r > 0$ such that $\frac{1}{r} = \frac{1}{p} + \frac{1}{q}$, for matrices $A$ and $B$ we have $\|AB\|_r \leq \|A\|_p \cdot \|B\|_q$. Example: If $A$ and $B$ are $2 \times 2$ matrices and $p = q = r = 2$ (Frobenius norm), then $\|AB\|_2 \leq \|A\|_2 \cdot \|B\|_2$.

For Hermitian matrices $A$ and $B$, and their sum $A+B$, the eigenvalues satisfy $\lambda_{i+j-1}(A+B) \leq \lambda_i(A) + \lambda_j(B)$ for all $i, j$. Example: If $A$ and $B$ are $2 \times 2$ Hermitian matrices with eigenvalues $(\lambda_1, \lambda_2)$ and $(\mu_1, \mu_2)$ respectively, then $\lambda_2(A+B) \leq \lambda_1(A) + \mu_2(B)$.

For any matrices $A$ and $B$, and $p, q > 1$ with $\frac{1}{p} + \frac{1}{q} = 1$, we have $|\text{tr}(AB^*)| \leq \|A\|_p \cdot \|B\|_q$. Example: For $p = q = 2$ (Frobenius norms), $|\text{tr}(AB^*)| \leq \|A\|_F \cdot \|B\|_F$ for any $A, B$.

For positive semidefinite matrices $A$ and $B$ such that $B$ is invertible, $\text{det}(A+B) \geq \text{det}(A) + \text{det}(B)$. Example: Let $A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$ then $\text{det}(A+B) = \text{det}\begin{bmatrix} 3 & 0 \\ 0 & 6 \end{bmatrix} = 18 \geq \text{det}(A) + \text{det}(B) = 4 + 4 = 8$.