Sum of vector subspaces | Mathematics | Page 1

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I suppose we have to use the double inclusion to show that the 2 sets, U + W and <U union W> are equal

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I am not sure if my proof is correct

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Let v be a vector from U+W then it is of the form v = u + w where u is a vector of U and w is a vector of W. But then v is a linear combination of the vector u and w which belong to the set U union W. So U + W is included in <U union W>.

Now the other inclusion:
Let v be a vector from <U union W> then it is of the form v = ax1 + ax2 + ax3 +... + axn where xk is a vector from U union W
We separate all the vectors from U from the ones that are in W and get that v = (linear combination of vectors of U) + (linear combination of ve tors from W) = vector from U + vector from W.
This means that <U union W> is included in U + W

By our double inclusion we get that the 2 sets are equal

#Sum of vector subspaces