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$\newcommand{\s}{\operatorname{sum}}$Problem: Let $E=[1]_{n\times n}$ and let $\s(X)$ be the sum of all elements of matrix $X$.

If $A$,$B$ $\in$ $M_{n \times n}$ $(\mathbb C)$ are regular matrices so that $A+B=\lambda E$, $\lambda \in \mathbb C$, prove the following: $$[1-\lambda \s(A^{-1})][1-\lambda \s(B^{-1})]=1$$

We did not say much about sums of all elements. I did some research and found this topic, but I do not know how to use said information here.

This is what I've done so far:

$$[1-\lambda \s(A^{-1})][1-\lambda \s(B^{-1})]=1$$

$$1-\lambda \s(A^{-1})-\lambda \s(B^{-1})+\lambda^2\s(A^{-1}\s(B^{-1})=1$$

$$-\lambda \s(A^{-1})-\lambda \s(B^{-1})+\lambda^2\s(A^{-1})\s(B^{-1})=0$$

$$\lambda [-\s(A^{-1})-\s(B^{-1})+\lambda \s(A^{-1})\s(B^{-1})]=0$$

If $\lambda=0$ this is true, but if $\lambda \not= 0$ I have:

$$-\s(A^{-1})-\s(B^{-1})+\lambda \s(A^{-1})\s(B^{-1})=0$$

I am not sure how to make connection between $A$, $B$ and their inverse matrices. I know that regular matrices $A$ and $B$ mean their inverse exists. Also $\det(A+B)=0$ which means $(A+B)^{-1}$ does not exist.

Any help is greatly appreciated.

Community
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Asleen
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1 Answers1

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Let us denote by $\sigma(M)$ the sum of the entries of matrix $M$. If we denote by $v^T=(1,1,\cdots, 1)$, then $\sigma(M)= v^TMv$. The way you defined it, $E=vv^T$. Now if $A+B=\lambda E$, you have $$ \begin{aligned} \lambda \sigma(A^{-1})\sigma(B^{-1})&= \lambda(v^TA^{-1}v)(v^TB^{-1}v)= v^TA^{-1}(\lambda E) B^{-1}v\\ &=v^TA^{-1}(A+B)B^{-1}v = v^TA^{-1}v+v^TB^{-1}v=\sigma(A^{-1})+\sigma(B^{-1}) \end{aligned} $$

Hamed
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