How to prove Rank(A+B)$\ge$max{Rank(A),Rank(B)} for positive semi-definite A and B

Question

How to prove $\text{Rank}(A+B)\ge\max(\text{Rank}(A),\text{Rank}(B))$ if $A\in S_+^n$ and $B\in S_+^n$ (i.e. $A$ and $B$ are both $n\times n$ symmetry positive semi-positive matrices)?

Answer 1

Hint Since $A+B$ is symmetric, it is orthogonally diagonalizable.$\newcommand{\rank}{\operatorname{rank}}$

If $v_1,..,v_k$ is a basis for the eigenspace corresponding to $\lambda=0$, then, since $A+B$ is diagonalizable, you have $\rank(A+B)=n-k$.

Now, use $$v_j^T (A+B) v_j=0$$ and the fact that $A,B$ are positive semi-definite, to deduce that $$v_j^TAv_j=v_j^TBv_j=0$$

Answer 2

HINT:

The main statement is $$\operatorname{Image}(A+B) = \operatorname{Image}A + \operatorname{Image} B$$ from which the inequality $$\operatorname{rank} (A+B) \ge \max (\operatorname{rank} A, \operatorname{rank} B)$$ readily follows.

To see the first inequality, notice that for $A$ positive semi-définite we have $$(\operatorname{Image} A) ^{\perp} = \{x \ | \ \langle A x, x \rangle = 0 \}$$ and so $$(\operatorname{Image} (A+ B) )^{\perp} = (\operatorname{Image} A) ^{\perp} \cap (\operatorname{Image} B) ^{\perp} $$

Answer 3

Suppose $v_1,v_2,...v_k$ forms a basis for $nullspace$ of $A+B$(i.e. eigenspace corresponding to eigenvalue $0$), now we have $Rank(A+B)=n-k$ and $v_i^T(A+B)v_i=0$ for i=1,2,3,...k.

Due to the fact that $A$ and $B$ are positive semi-definite matrices, we have:

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space v_i^TAv_i=0$ and $v_i^TBv_i=0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space$ Then let $A=A^{1/2}A^{1/2}$, due to $A^{1/2}$ is also positive semi-definite we can get

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space v_i^TA^{1/2}A^{1/2}v_i=(A^{1/2}v_i)^TA^{1/2}v_i=0$

Now I achieve a key intermediate result: $A^{1/2}v_i=0$. And left multiplying $A^{1/2}$ on both sides to get

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space Av_i=0$

Now,we can say $v_1,v_2,...v_k$ spans a k-dimension subspace belonging to $A$'s $nullspace$. So we have $k{\le}n-Rank(A)\space i.e. Rank(A){\le}n-k=Rank(A+B)$.

Similarly, we can get $Rank(B){\le}Rank(A+B)$.