Vector Space Axiom: Can this be done easier?

  • Thread starter Dustinsfl
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  • #26
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All I have to do is delete the last two lines in the first part and I will x + y.
 
  • #27
vela
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Right, so now it's the abuse of notation problem. You know what you need to show; you just have to write it down properly. It's akin to knowing what you want to say in an essay, and knowing how to write grammatically correct sentences.

Let [itex]A, B \in S[/itex]. Let C=A+B. Then
[tex]A \in S \Rightarrow a=\sum_{i=1}^3 a_{i1} = \sum_{i=1}^3 a_{i2} = \sum_{i=1}^3 a_{i3} = \sum_{i=1}^3 a_{1i} = \sum_{i=1}^3 a_{2i} = \sum_{i=1}^3 a_{3i} = \sum_{i=1}^3 a_{ii}=\sum_{i=1}^3 a_{4-i,i}[/tex]

[tex]B \in S \Rightarrow b=\sum_{i=1}^3 b_{i1} = \sum_{i=1}^3 b_{i2} = \sum_{i=1}^3 b_{i3} = \sum_{i=1}^3 b_{1i} = \sum_{i=1}^3 b_{2i} = \sum_{i=1}^3 b_{3i} = \sum_{i=1}^3 b_{ii}=\sum_{i=1}^3 b_{4-i,i}[/tex]

The sum of the elements of the i-th row of C is

[tex]\sum_{j=1}^3 c_{ij} = \sum_{j=1}^3 (a_{ij}+b_{ij}) = \sum_{j=1}^3 a_{ij} + \sum_{j=1}^3 b_{ij} = a+b[/tex]

The first equality is from the definition of matrix addition; the second equality is from what you know about rewriting sums; and the third equality holds because A and B are in S. From this, you can see that all the rows sum to the same number, a+b.

I'll leave it to you to fill in the rest. Eventually, you'll be able to conclude the sums of the rows, columns, and diagonals are all equal to each other; therefore, C is an element of S, which is what you wanted to show.
 

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