Let $A \in\mathbb(C)^{p\times n}, C \in \mathbb(C)^{p\times q}$. Show that you cannot have more than one matrix $B\in\mathbb(C)^{n\times q}$ s.t. both of the properties hold:
1) $AB = C$ and
2) $B^{H}\vec{v}=0$ for every vector $\vec{v}\in N_{A}$.
Directions:
a) Assume there are two matrices B1 and B2 such that properties 1 and 2 hold. Show that the columns of D=B1−B2 are in $N_{A}$ because of property 1.
b) Show that $D^{H}D=0$ because of property 2.
a) Assume that there is indeed two matrices B1 and B2 such that property 1 holds.
Further consider,
$AB_{1} - AB_{2} = C-C$
$\implies$ $A(B_{1}-B_{2}) = 0$
let $D = (B_{1}-B_{2})$, such that
$AD = 0$
Note that A $\in$ $\mathbb(C)^{pxn}$ and $B_{1},B_{2}$ $\in$ $\mathbb(C)^{nxq}$, since subtraction does not change the size of the resulting matrix, then D $\in$ $\mathbb(C)^{nxq}$, thus by matrix multiplication O $\in$ $\mathbb(C)^{pxq}$ matrix. Now I don't know how to show that the columns of D are in $N_{A}$ because of property 1; shouldn't they automatically be because matrix multiplication gives me the zero matrix?
b) We take a similar approach such that
$(B_{1}^{H}-B_{2}^{H})\vec{v}=0$
$\implies(B_{1}-B_{2})^{H}\vec{v}=0$
$\implies(D)^{H}\vec{v}=0$
Now, can't I just simply multiply by D from the right such that I obtain,
$\implies(D)^{H}\vec{v}D=0D$
$\implies(D)^{H}D\vec{v}=0$
Then since $D\in N_{A}$, shouldn't I just get that
$(D)^{H}D=0$
Any help would be greatly appreciated.
Suppose there exist $B_1$ and $B_2$ satisfying the two properties. Set $D=B_1-B_2$. Then $AD=0$. Thus, for every $v\in\mathbb{C}^q$, $Dv\in N(A)$. Note that $D$ satisfies property 2, which implies $D^H(Dv)=0$, for every $v$.
Can you finish?