If $A,B$ are PSD matrices, $V^{\top}(A+B)V = 0$ implies $V^{\top}AV=0$ and $V^{\top}BV=0$ Let $A,B$ be two Positive Semi Definite matrices of dimensions $n \times n$ such that $rank(A+B)= r$. Let $V \in \Bbb{R}^{n\times(n-r)}$ be matrix with $\mathcal{R}(V) = \mathcal{N}(A+B)$.
If
$
V^{T}(A+B) V=V^{T} A V+V^{T} B V=0
$, then does it imply that  $V^{T} A V =0$ and $V^{T} B V=0$?
Here- $\mathcal{R}$ is the range and $\mathcal{N}$ is the Nullity. 
This question says that $V^{T} A V$ and $V^{T} B V$ will be positive semi definite but I am unable to see how if sum of two postive semi definite matrices is a zero matrix and then the individual matrices are essentially zero matrices. 
 A: We could say:
$$
V^T(A+B)V = V^TAV + V^TBV = \mathbf0_{r \times r}
$$
and for every $x$, $x^T\mathbf0 x = 0 $. if $A$ and $B$ are PSD, then $V^TAV$ and $V^TBV$ are also PSD (it's easy to prove). So we have: 
$$
z^T(V^T(A+B)V)z = 0  \to z_1^TAz_1 +  z_1^TBz_1 = 0      
$$
On the other hand, for both $A$ and $B$ we have $z_1^TAz_1 \geq 0$ and $z_1^TBz_1\geq 0$, so according to above equality, they must be zero.  
A: Given that
$V^T(A + B)V = 0, \tag 1$
for any suitable vector $x$ we have
$x^TV^T(A + B)Vx = 0, \tag 2$
whence
$(Vx)^TA(Vx) + (Vx)^TB(Vx) = (Vx)^T(A + B)(Vx)$
$= x^TV^T(A + B)Vx = 0; \tag 3$
$A$ and $B$ are positive semidefinite, so
$(Vx)^TA(Vx) \ge 0, \; (Vx)^TB(Vx) \ge 0; \tag 4$
but the sum of two non-negative real numbers is zero if and only if they vanish individually; thus in light of (3),
$(Vx)^TA(Vx) = 0 = (Vx)^TB(Vx); \tag 5$
and thus
$x^T(V^TAV)x = (x^TV^T)A(Vx) = (Vx)^TA(Vx) = 0; \tag 6$
we further recall that positive semidefinite matrices are symmetric, that is
$A^T = A, \; B^T = B; \tag{6.1}$
this implies that
$(V^TAV)^T = V^TA^T(V^T)^T = V^TAV,$
$(V^TBV)^T = V^TB^T(V^T)^T = V^TBV, \tag{6.2}$
i.e., $V^TAV$ and $V^TBV$ are also symmetric; by virtue of this fact, it follows from (6) that
$V^TAV = 0, \tag 7$
with the same argument applying to show that
$V^TBV = 0 \tag 8$
as well.
Nota Bene:  In the above we have called upon the well-known fact that the only symmetric matrix $C$,
$C^T = C \tag 9$
such that
$\forall \; \text{vectors} \; z, \; z^TCz = 0 \tag{10}$
is in fact the zero matrix:
$C = 0; \tag{11}$
we flesh out the preceding argument by demonstrating this useful observation:  taking
$z = x + y, \tag{12}$
we may write
$(x + y)^TC(x + y) = 0, \tag{13}$
that is, expanding,
$x^TCx + x^TCy + y^TCx + y^TCy = 0; \tag{14}$
in light of (10) this becomes
$x^TCy + y^TCx = 0; \tag{15}$
we note that $x^TCy$, $y^TCx$ are scalars, whence
$x^TCy = (x^TCy)^T = y^TC^Tx = y^TCx; \tag{16}$
thus, from (15),
$2x^TCy = 0 \Longrightarrow x^TCy = 0; \tag{17}$
since this holds for all vectors $x$, $y$ we have
$C = 0. \tag{18}$
End of Note.
