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Mathematics
Linear and Abstract Algebra
Showing direct sum of subspaces equals vector space
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[QUOTE="fresh_42, post: 6234412, member: 572553"] You have to solve the equation ##\mathbb{R}^3 \ni (x_1,x_2,x_3)= \mathbf{u} + \mathbf{v}## where ##\mathbf{u} \in U, \mathbf{v} \in V## have to be expressed in terms of the ##x_i\,.## Yes the elements of ##U## look like ##(-r-3s,r,s)## and the elements of ##V## as ##(t,t,2t)##, but what are ##r,s,t## given an arbitrary point ##(x_1,x_2,x_3)\,?## This would show ##\mathbb{R}^3=U+V## and with ##U\cap V= \{\,0\,\}## we get ##\mathbb{R}^3 =U\oplus V##. Of course the dimension argument would give the result immediately, as I first said. The presentation of an arbitrary point as a sum of vectors ##\mathbf{u},\mathbf{v}## is just for practice. [/QUOTE]
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Mathematics
Linear and Abstract Algebra
Showing direct sum of subspaces equals vector space
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