Let A be a $m*n$ real valued matrix, $r$ the rank of A and let $S$ be the set of all real $n*k$ matrices such that AB=0 (the $m*k$ zero matrix).
Show that S is a vector subspace of $\mathbb{R^{n,k}}$ and find the dimension of $S$
I am stuck with the second question.
My reasoning has been that I can see both B and A as represenations of linear functions, $B$ being a function f from $\mathbb{R}^{k}$ to $\mathbb{R}^{n}$ and $A$ being a function g from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$, $AB$ being the fucntion composition, therefore if $AB$=0
Then g(f(v))=0 for every vector v, thus $im(f) \subseteq ker(g)$.
By the Rank–nullity Theorem, $dim(ker(g))$=n-r, and $dim(im(f))$=$rank(B)$.
Therefore S is just the set of all matrices such that $rank(B) \leq n-r$ but I can't continue.
Am i on the right track?
Best Answer
$AB=0$ holds if and only if each column of $B$ is in the kernel of $A$.
This observation gives a vector space isomorphism $S\cong(\ker A)^k$.