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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.

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  • $\begingroup$ Hint: all the diagonal entries of a psd matrix are non-negative. $\endgroup$ – kimchi lover Oct 9 at 15:53
  • $\begingroup$ @kimchilover: Thank you, I understand that diagonal elements of PSD matrix are non-negative hence, the diagonal elements of both $V^TAV$ and $V^TBV$ are zeros. So, can we say all the non-diagonal elements are zeros as well? $\endgroup$ – kasa Oct 9 at 15:58
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    $\begingroup$ All the 2 by 2 principle submatrices of a psd matrix are psd. Ask yourself, for which $a$ is $\pmatrix{0&a\\a&0}$ positive semidefinite? $\endgroup$ – kimchi lover Oct 9 at 18:10
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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.

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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.

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