Equivalence of two invertibility criteria. I would like to prove that for two right-invertible matrices $A, B \in \mathbb{R}^{m \times n}$, the following two statements are equivalent
(a) $(I - A^+ A) + (I - B^+ B) $ is invertible.
(b) $\begin{bmatrix} A \\ B \end{bmatrix}$ is right-invertible.
With $A^+$ and $B^+$, I mean the pseudoinverse of $A$ and $B$. I was able to prove $(a) \Rightarrow (b)$. However, I can't prove $(b) \Rightarrow (a)$. 
My results so far:
Multiplying the matrix in (a) with $\frac{1}{2}$ does not change its invertibility and, hence, I can analyze
$$
 I - \frac{1}{2} (A^+ A + B^+ B) 
$$
instead. I know that for
$$
 \| \frac{1}{2} (A^+ A + B^+ B) \|_2 < 1
$$
this new matrix is invertible (by using the Neumann series).
By using the Penrose properties, I can show that $A^+A$ and $B^+B$ are idempotent matrices for which it is known that all Eigenvalues are either 0 or 1. And since $A^+A$ and $B^+B$ are symmetric matrices, the 2-norm is equal to the largest Eigenvalue. With all that, I can show $$ \| \frac{1}{2} (A^+ A + B^+ B) \|_2 \leq \| \frac{1}{2} (A^+ A) \|_2 + \| \frac{1}{2} (B^+ B) \|_2 = 1$$
My problem is that I need to show that the 2-norm is strictly less than 1.
I would be happy for any hints.
 A: We use the following fact below, the matrix $A_{m \times n}$ has a right inverse ( i.e., a matrix $C_{n \times m}$ such that $AC$ is the identity matrix exists) iff the rows of $A$ are linearly independent. In this case $AA^\top$ is invertible and the Moore-Penrose inverse of $A$ is $A^\top(AA^\top)^{-1}$.
To show $(b) \implies (a)$ the following argument works.
Since $C  = \begin{bmatrix} A \\ B \end{bmatrix}$ has a right inverse, the rows of $C$ must be linearly independent, so in particular we must have, $A$ and $B$ have linearly independent rows and
$$ \mathcal{C}(A^\top) \cap \mathcal{C}(B^\top) = \{ 0 \},$$ here $\mathcal{C}(X)$ denotes the column space of an arbitrary matrix $X$.
The linear independence of the rows of $A$ and $B$ imply they are right invertible and $A^+ = A^\top (AA^\top)^{-1}$ and $B^+ = B^\top (BB^\top)^{-1}$.
Let $P_1 = I - A^+A$ and $P_2 = I - B^+B$. We need to show $P_1 + P_2$ is invertible.
From the usual properties of the Moore-Penrose pseudoinverse, we have that both $P_1$ and $P_2$ are symmetric and idempotent. So $P_1 = P_1^2 = P_1^\top P_1$ and $P_2 = P_2^2 = P_2^\top P_2$, so $P_1$ and $P_2$ and hence $P_1 + P_2$ is positive semi-definite. 
To show $P_1 + P_2$ is invertible it is sufficient to show it is strictly positive definite. 
Let $x$ be a vector such that $x^\top(P_1 + P_2)x = 0$. For this $x$ we have $x^T P_1^\top P_1 x + x^\top P_2^\top P_2 x =  \|P_1 x \|^2 + \| P_2 x\|^2 =0$, so $P_1x = P_2x=0$, so $x=A^+Ax = A^\top (AA^\top)^{-1}Ax$ and $x=B^+Bx = B^\top (BB^\top)^{-1}Bx$ so $x \in \mathcal{C}(A^\top) \cap \mathcal{C}(B^\top)$ and so $ x = 0$.
