If $A \in \mathbb R^{n \times n}$ is singular and has non-zero row sums that are the same for every row, prove that:
$A + \lambda \, 1_n 1_n^\top$ is singular for any nonzero scalar $\lambda$.
$A + 1_n f^\top$ is also singular for any $f \in \mathbb R^n$.
Note that $1_n 1_n^\top$ is an $n \times n$ matrix of ones. Thank you so much in advance for all your help!
On
I'll try to solve the first problem. Since $A$ is singular, there exist $u\in \mathbb R^n$ s.t. $uA^T=0$. Thus, $$u^TA1_n=0=u^T\cdot (s1_n)=s(u^T\cdot 1_n)=0$$($s$ is the sum of elements in each row)
Hence, $$u^T(A+\lambda1_n1_n^T)=u^TA+\lambda u^T1_n1_n^T=0+\lambda(u^T\cdot1_n)1_n^T=0+0=0$$ So the resulting matrix is singular $\forall \lambda\in \mathbb R$
For the second problem, I think you will just need to replace $1_n^T$ with $f^T$ and use the same reasoning.
Hint for 1: Suppose that $Av = 0$. Then there necessarily exists a value of $t \in \Bbb R$ such that $$ (A + \lambda 1_n1_n^T)(v + t\,1_n) = 0 $$ Alternatively, the hint for 2 below also applies.
Hint for 2: Note that the column space of $1_n f^T$ is a subspace of the column space of $A$.