Positive semidefiniteness of a block matrix of positive semidefinite matrices

Given any symmetric matrix $\mathbf{M} = \begin{pmatrix} \mathbf{A} & \mathbf{B}\\ \mathbf{B}^\mathrm{T}& \mathbf{C} \end{pmatrix}$, the following conditions are equivalent:

(1) $\mathbf{M}\succeq 0$ ($\mathbf{M}$ is positive semidefinite)

(2) $\mathbf{A}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;\mathbf{C} - \mathbf{B}^T\mathbf{A}^{\dagger }\mathbf{B} \succeq 0$.

(3) $\mathbf{C}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, $\;\;\mathbf{A} - \mathbf{B}\mathbf{C}^{\dagger }\mathbf{B}^T \succeq 0$.

Then, if we are given that $\mathbf{A}\succeq 0$, $\mathbf{B}\succeq 0$ and $\mathbf{C}\succeq 0$, is there any way to verify (2) or (3) and therefore conclude that $\mathbf{M}$ will be positive semidefinite as well?

The conditions$\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, and $\mathbf{B} \succeq0$, do not follow from anything. W/out them the equivalence follows from the Sylvester criteria for positive definiteness and determinant formula for Schur complement. What is the source of the problem?

Hint By definition a matrix $M\in\mathbb{R}^{n\times n}$ is positive semidefinite if $v^TMv\geq 0$ for all $v\in\mathbb{R}^n$. Use the fact $ \det M= \det\Big(\mathbf{C} - \mathbf{B}^T\mathbf{A}^{-1}\mathbf{B} \Big) $

and proof by inductin in $n\in\mathbb{N}$ for $n\geq 1$.