Proof that if $A$ and $B$ are $n \times n$- matrices such that $Ker(A-B)= \mathbb R^n$, that $A=B$.
Definition : If $f: V \to W$ is a linear map, then the kernel of $f$ is
$Ker(f)= \{\overrightarrow v\in V: f(\overrightarrow v) = \overrightarrow 0_W\}$ a sub vector space of $V$.
I know that $A$ and $B$ are vector spaces so that a linear map can be constructed. And I know that by chosing a basis $\{\overrightarrow v_1, \ldots, \overrightarrow v_n\}$ for the $n$-dimensional vector space $V$ we can identify vectors $\overrightarrow v \in V$ with their column vector of coefficients with repect to the basis. So let's do the same for $W$ with basis $\{\overrightarrow w_1, \ldots, \overrightarrow w_n \}$ then a linear map is constructed as is mentioned in the definition $f: V \to W$ by representing an $m \times n$-matrix of which we take the i-th column as the column vector with respect to the basis $\{\overrightarrow w_1, \ldots, \overrightarrow w_n \}$ of the image $f(\overrightarrow v_k)\in W$.
Now I notice this is just a beginning but it seems by definition of the kernel the $A-B=0$. And $\overrightarrow 0 \in \mathbb R^n$.
What should I look for to complete this proof?
Thought: Can I use the direct sum in this question because it seems to me I can use it somehow?
If $v\in\mathbb{R}^n$, then $v\in\ker(A-B)$ and therefore $(A-B)v=0$. Since this occurs for each vector of $\mathbb{R}^n$, $A-B=0$. In other words, $A=B$.