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Let $A$ be an invertible $10\times 10$ matrix with real entries such that the sum of each row is $1$. Then is the sum of the entries of each row of the inverse of $A$ also $1$?

I created some examples, and found the proposition to be true. I also proved that if two matrices with the property that the sum of the elements in each row is $1$ are multiplied, then the product also has the same property. Clearly, $I$ has this property. I think I have a proof running along the following lines: $$A^{-1}A=I$$ where $A$ and $I$ satisfy the aforementioned property. Also, if $A^{-1}$ did not satisfy this property, then neither would the product of $A$ and $A^{-1}$, which is a contradiction.

Is the proposition true, and if so, is my proof correct?

algebraically_speaking
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  • The proposition is true, the proof isn't correct imo: that the product of two matrices with the condition (let's call it QS matrix) is again a QS matrix doesn't mean, or at least requires proof, that if the product of two matrices is QS and one of them is, then also the other one. – Timbuc Dec 15 '14 at 14:47

1 Answers1

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Any square matrix $\;n\times n\;$ has rows sum equal to $\;1\;$ iff $\;u:=(1,1,\ldots,1)^t\;$ is an eigenvector with eigenvalue $\;1\;$, but then

$$Au=u\implies A^{-1}\left(Au\right)=A^{-1}u\iff u= A^{-1}u$$

so the answer is yes.

Timbuc
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  • may you explain why the first line of ur answer valid. How?@timbuc – David Dec 17 '14 at 17:39
  • @Gloom Write $;A=(a_{ij});,;;1\le i,j\le n;$ ,so that

    $$\sum_{j=1}^na_{ij}=1;,;;\forall i=1,2,...,n\iff A\begin{pmatrix}1\1\\ldots\1\end{pmatrix}=\begin{pmatrix}\sum_{j=1}^na_{1j}=1\\sum_{j=1}^na_{2j}=1\\ldots\\sum_{j=1}^na_{nj}=1\end{pmatrix}\iff\begin{pmatrix}1\1\ \ldots\1\end{pmatrix}$$is an eigenvector of $;A;$ with eigenvalue $;1;$

     – Timbuc Dec 17 '14 at 20:15