If $B^T$ consists of a basis of $\mathrm{im} (A)^\perp$, then $\mathrm{im}(A)=\ker (B)$?

Question

Well basically, the question is in the title:

Suppose we have $A\in\mathbb{R}^{n\times m}$ with rank $d$ and we fix a basis $(b_1,\ldots,b_{n-d})$ of $\mathrm{im}(A)^\perp$. Let $B^T=(b_1,\ldots,b_{n-d})$ (so $b_i$ are the rows of $B$). Then the author states:

So we have $BA=0$, i.e. $\mathrm{im}(A)=\ker (B)$.

Once again, I feel like I'm completely missing out on sth. here. It's obvious that $BA=0$, but that only gives $\mathrm{im}(A)\subseteq \ker(B)$, doesn't it?

I have some intuition why this holds, but it's not very smooth: We have $$\ker(B)^\perp\cong\mathrm{im}(B)=\mathrm{span}\{b_1,\ldots,b_{n-d}\}=\mathrm{im}(A)^\perp,$$ because of the homomorphism theorem, but I feel like this is not very well thought out (plus, there's only "$\cong$" and no "$=$").

I've been away from linear algebra for quite some time, so any help would be appreciated!

Answer 1

Note that $dim(Ker(B))=d=dim(Im(A))$. Since you have the inclusion in one direction, and the dimension of the spaces are the same, so in fact these spaces are equal.

Answer 2

Rows of $B$ are orthogonal to the columns of $A$ and therefore $BA=0$ and hence $\mathrm{im}(A)\subset \ker(B)$ as you have mentioned. However whenever you have two finite-dimensional vector spaces $V$ and $W$ with $V\subset W$ and they have the same dimension you can conclude that they are equal.