# 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.

• Hint: all the diagonal entries of a psd matrix are non-negative. – kimchi lover Oct 9 at 15:53
• @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? – kasa Oct 9 at 15:58
• 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? – kimchi lover Oct 9 at 18:10

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.

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.