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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!

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    $\begingroup$ What is $\lambda11^T$? $\endgroup$ May 7 '18 at 11:24
  • $\begingroup$ @JoséCarlosSantos 1 is the vector of order n x 1 and 1^T is its transpose. Sorry for not being clear on that and thanks for your question, I will edit my post. $\endgroup$
    – Sophil
    May 7 '18 at 11:27
  • $\begingroup$ @RodrigodeAzevedo Thank you so much for your help on editing my question!!! I am new to MSE and I still learn to format my questions and the mathematical notations. $\endgroup$
    – Sophil
    May 7 '18 at 11:33
  • $\begingroup$ @lhf Thanks for your attention! \lambda is any non-zero scalar. $\endgroup$
    – Sophil
    May 7 '18 at 11:34
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    $\begingroup$ Since you're fairly new to MSE I thought I'd point out that you'll find that simple "Here's the statement of my exercise, solve it for me" posts may be poorly received. What is better is for you to edit the question and add context: What you understand about the problem, what you've tried so far, etc. $\endgroup$
    – John Doe
    May 7 '18 at 11:36
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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.

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  • $\begingroup$ You are welcome! :) $\endgroup$
    – Macrophage
    May 7 '18 at 14:11
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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 $A$.

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  • $\begingroup$ for your hints! I haven't learned column space yet but I can use your first hint to solve the second one. $\endgroup$
    – Sophil
    May 7 '18 at 14:10
  • $\begingroup$ @Sophil here's another approach, then. Since $A$ is singular, there is a vector $y$ such that $y^TA = 0$ (or if you prefer, $A^Ty = 0$). Try to show that $y^T 1_n = 0$, which means that $y^T(A + 1_n f^T) = 0$. $\endgroup$ May 7 '18 at 14:12
  • $\begingroup$ @Sophil actually, I just noticed that this is exactly Macrophage's approach $\endgroup$ May 7 '18 at 14:17
  • $\begingroup$ Thanks so much for your help! I've just noticed that my first response to you was somehow missing the first word but I cannot edit it, it was "Thanks for your hints!" :) Anw, thanks a lot! :) $\endgroup$
    – Sophil
    May 7 '18 at 14:21

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