Prove $\ker(A+T(A))\subseteq \ker(A)$ for $A\geq 0$ and $T$ positive linear map

Question

Let $A\geq 0$ be a positive semi-definite complex matrix in $M_d(\mathbb{C})$. Let $T:M_d(\mathbb{C})\to M_d(\mathbb{C})$ be a positive linear map between $d\times d$ complex matrices, i.e., $A\geq 0\Rightarrow T(A)\geq 0$.

I want to prove that $\ker(A+T(A))\subseteq \ker(A)$. Intuitively this should be easy in the following sense: on one extreme, consider $T(A) \propto I_d$ (identity matrix), in which case the sum will give trivial kernel (essentially because in the spectral decomposition of $A$, all the zero eigenvalues are shifted to positive value), hence $A+T(A)$ can remove all vectors in the kernel of $A$. On the opposite extreme, $T(A)=A$ will make their kernels equal. There are easy intermediate cases, namely $[T(A),A]=0$, in which case we can diagonalize both simultaneously, and the inclusion follows by analysing whether $T(A)$ shrinks or expands the support of the nonzero block diagonal.

However, when they do not commute, I am not sure if there is an easy way to do this. The basic idea is still the same in that $T$ should either increase or decrease the rank of $A$, but it is not clear where $\ker(T(A))$ (and hence $\ker(A+T(A))$) is "pointing at": my intuition does not exclude the possibility that $\ker(A+T(A))\not\subseteq \ker(A)$.

I would imagine the positivity of the map $T$ is essential. Is there a transparent way of seeing this?

Answer 1

$A=B^*B$ and $T(A)=C^*C$ for some $B,C\in M_d(\Bbb C)$. $\forall x\in\Bbb C^d$, if $(A+T(A))x=0$ then $$0=x^*(A+T(A))x=\|Bx\|^2+\|Cx\|^2$$ hence $Ax=B^*Bx=B^*0=0$.

Answer 2

The $T$ here is just a distraction. In general, given any two positive semi-definite matrices $A$ and $B$, we have $\ker(A+B)=\ker(A)\cap\ker(B)$. Note that $\langle Ax,x\rangle=0$ only if $Ax=0$ (and similarly for $Bx$), because $$ 0\le\langle A(Ax-tx),Ax-tx\rangle =\langle A(Ax),Ax\rangle-2t\langle Ax,Ax\rangle $$ even if the real number $t$ approaches positive or negative infinity. It follows that \begin{align*} x\in\ker(A+B) &\iff (A+B)x=0\\ &\iff \langle (A+B)x,x\rangle=0\\ &\iff \langle Ax,x\rangle+\langle Bx,x\rangle=0\\ &\iff \langle Ax,x\rangle=\langle Bx,x\rangle=0\\ &\iff Ax=Bx=0\\ &\iff x\in\ker(A)\cap\ker(B).\\ \end{align*}