Affine subspace of $\Bbb{R}^{27}$

Question

I have affine subspace $K$, $K \subset \mathbb R^{27}$. It's elements are solutions of system of linear equations $Ax=b, b \in R^{16}$. What are maximum and minimum dimensions of said subspace, if I know that rank of $A$ is equal to rank of $(A,b)$(i.e. said subspace isn't empty)? My intuition says to me that maximum is 16 and minimum is 0 (one solution), but I can't understand how to prove it from definition of affine subspace and it's direction.

Answer 1

Use Rank-nullity theorem. Specifically the equality $$\dim(\text{im}(A))+\dim(\ker(A))=\dim(\mathbb{R^{27}}).$$ Since $\text{rank}(A)=\text{rank}(A|b),$ we get that $\dim(\ker(A))=\dim(K).$ What more, by definition $\text{rank}(A)=\dim(\text{im}(A)).$ So the equality from Rank-nullity theorem changes to $$\dim(K)=27-\text{rank}(A)$$ Rank varies from $0$ to $16.$ So $$11\leq \dim(K)\leq 27.$$ The crutial assumption is the one that says $\text{rank}(A)=\text{rank}(A|b).$ It actually means the same as $b\in \text{im}(A).$ From that you get that $K$ is just translated $\ker(A).$

In fact, put $b=Ax_0$ for some $x_0\in\mathbb{R}^{27}.$ Then $$K=\{x\in\mathbb{R}^{27}:Ax=b\}=\{x\in\mathbb{R}^{27}:Ax=Ax_0\}=\{x\in\mathbb{R}^{27}:A(x-x_0)=0\}=\ker(A)+x_0.$$