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Am trying to prove that the product of a row vector $$\kappa$$ size (1 x n) and $$Matrix B$$ where the sum of the rows equals 1 produces a matrix C (1 x n) where the sum of the row is also 1.

So far, have worked out that $$ \kappa B = \sum\limits_{i=1}^n \kappa_i B $$ where $$ B = row(r_j)_{j=1:n}$$

I postulate that because row vector k is 1 and that the sum of rows B is 1, then this vector * matrix should produce a row that equals 1. Is there a more elegant way to show this.

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  • $\begingroup$ do you have any other problems? If not, it is advisable to accept the answer that helped you. $\endgroup$ – Mr Tsjolder Feb 16 '15 at 8:08
  • $\begingroup$ Hi, how do I accept it? $\endgroup$ – liujm Feb 16 '15 at 12:14
  • $\begingroup$ Normally, there should be this nice and green V under the voting arrows of my answer. $\endgroup$ – Mr Tsjolder Feb 16 '15 at 12:16
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I'm afraid there's not really a more elegant way, but you could write it down more formally if you would write it as follows:

Let $\kappa = (\kappa_1, \kappa_2, \ldots, \kappa_n)$ and $B = (\beta_1, \beta_2, \ldots, \beta_n)^T$ where $\beta_i$ is the $i^{th}$ row of $B$. Using your arguments you now get $\kappa B = \sum_{i=1}^{n} \kappa_i\beta_i$. So if you want to get to know the sum of elements of $\kappa B$ now, you can calculate the sum of elements of each $\beta_i$ and add these up. Thus we get: $$\mathrm{sum}(\kappa B) = \sum_{i=1}^{n} \mathrm{sum}(\kappa_i\beta_i) = \sum_{i=1}^{n} \kappa_i\mathrm{sum}(\beta_i) = \sum_{i=1}^{n} \kappa_i = 1$$ where sum is the function defined as $\mathbb{R}^n \to \mathbb{R} : (x_1, \ldots x_n) \mapsto \mathrm{sum} (x_1, \ldots x_n) = \sum_{i=1}^{n} x_i$.

I hope this answers your question.

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