# How to invert a matrix

I would like to disprove the following claim, that seems false to me, finding a counterexample.

Let $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{n \times k}$, for $k < n$. Let us assume that $rk(A) = rk (B) = k$, that is both matrices are full rank. Let us now consider a matrix $\mathbf{C} \in \mathbb{R}^{k \times k}$ such that $c_{i,j} = c$, where c is a real constant such that $(\mathbf{A}^\intercal\mathbf{B} + \mathbf{C})_{i,j} > 0$. Then $\mathbf{A}^\intercal\mathbf{B} + \mathbf{C}$ is nonsingular.

I think it should be false because $\mathbf{A}^\intercal\mathbf{B}$ could have a rank much smaller than $k$ and $\mathbf{C}$ is just a rank one update, but I'm not so sure as I haven't been able to find a counterexample. Can someone check my reasoning and, and in case propose a counterexample?

Thanks!

Let $n = 2k$. Then $X = \mathbb{R}^{n} = \mathbb{R}^{2k} = U \oplus V$, where $U = V = \mathbb{R}^k$. Let $A$ be an embedding of $U$ into $X$ $$A \,:\, U \to X \,:\, u \mapsto (u \oplus \underbrace{\mathbf{0}}_{\in V}) \text{,}$$ which in coordinates means $A((x_1,\ldots,x_k)) = (x_1,\ldots,x_k,\underbrace{0,\ldots,0}_{\textrm{$k$times}})$. Let $B$ be an embedding of $V$ into $X$ $$B \,:\, V \to X \,:\, v \mapsto (\underbrace{\mathbf{0}}_{\in U} \oplus v) \text{,}$$ which in coordinates means $B((x_1,\ldots,x_k)) = (\underbrace{0,\ldots,0}_{\textrm{$k$times}},x_1,\ldots,x_k)$. Then $A^T$ is the mapping $$A^T \,:\, X \to U \,:\, (u \oplus v) \mapsto u \text{,}$$ i.e. in coordinates $A((x_1,\ldots,x_n)) = (x_1,\ldots,x_k)$. Then, clearly $A^TB = 0$.
For the claim to hold, $C$ would thus need to have full rank, which is impossible if all it's coefficients are the same.
Geometrically, what happens here is that $B$ maps $\mathbb{R}^k$ to one $k$-dimensional subspace $V$ of $\mathbb{R}^n$, and $A^T$ projects $\mathbb{R}^n$ onto a $k$-dimensional subspace $U$ orthogonal to $V$, and thus projects the range of $B$ to the zero vector.