TITLE: Is $U = \left\{A \in \mathbb{M}^ {n \times n}(\mathbb{R} ): \ker{A} \cap \text{Im}A = \{\vec{0}\} \right\}$ a vector space?
QUESTION [5 upvotes]: I recently tried to prove or reject the conjecture that the set of matrices
$$U = \left\{A \in \mathbb{M}^ {n \times n}(\mathbb{R} ): \ker{A} \cap \text{Im}A = \{\vec{0}\} \right\}$$
is a linear subspace of $\mathbb{M}^{n \times n}(\mathbb{R})$.

I didn't find it hard to prove that the set is multiplicative: Assume that $A \in U$. We know that $\ker{0}\cap \text{Im}0 = \vec{0}$. So in the case that $\lambda=0$ we are done. Suppose that $\lambda \neq 0$. It follows easily that $\ker{A}=\ker{\lambda A}$ and that $\text{Im}A = \text{Im}{\lambda A}$. 
I thought additivity was the hardest part. I don't have a good intition when it comes to " adding" linear maps. Can you help me with this?

REPLY [3 votes]: Let $A$ be a simple matrix $\notin U$ for dimension $2$, say $\ker A={\rm im\,} A=\langle e_1\rangle$ where $e_1=\pmatrix{1\\0}$, e.g. take
$$A:=\pmatrix{0&1\\0&0}$$
with this example, $A+I$ and $I$ both have trivial kernel, so $A+I$ and $I$ are in $U$ but $A\notin U$, so $U$ is not closed under addition.