Showing direct sum of subspaces equals vector space

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  • Thread starter schniefen
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  • #1
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Summary:

Consider ##U=\{\textbf{x} \in \mathbf{R}^3; x_1+x_2+3x_3=0\}## and ##U=\{t(1,1,2)\in\mathbf{R}^3; t \in \mathbf{R}\}##. Show that ##\mathbf{R}^3=U \bigoplus V##.

Main Question or Discussion Point

If one shows that ##U\cap V=\{\textbf{0}\}##, which is easily shown, would that also imply ##\mathbf{R}^3=U \bigoplus V##? Or does one need to show that ##\mathbf{R}^3=U+V##? If yes, how? By defining say ##x_1'=x_1+t,x_2'=x_2+t,x_3'=x_3+2t## and hence any ##\textbf{x}=(x_1',x_2',x_3') \in \mathbf{R}^3## can be expressed as a sum of ##\textbf{u}=(x_1,x_2,x_3) \in U## and ##\textbf{v}=(t,t,2t) \in V## for ##t \in \mathbf{R}##?
 

Answers and Replies

  • #2
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If ##U\cap V =\{\,0\,\}## then we have a plane and a straight outside the plane, so they have to span a three dimensional space. So whether this is sufficient or not has to be answered by your teacher. I guess you should show that the sum equals the total space for practicing reasons.

In your proof, I don't see why ##\mathbf{u} \in U##. You need to set ##x_3=-\dfrac{1}{3}(x_1+x_2)## and show that the system can be solved.
 
  • #3
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In your proof, I don't see why ##\mathbf{u} \in U##. You need to set ##x_3=-\dfrac{1}{3}(x_1+x_2)## and show that the system can be solved.
Solving the system one can write ##\mathbf{u}=(-r-3s, r, s) \in U## for ##r,s\in \mathbf{R}##.

I guess you should show that the sum equals the total space for practicing reasons.
How would one do that? Finding the basis vectors for ##U## and ##V## respectively and check if they are linearly independent? Or can one, less rigorously, state that ##\mathbf{u}=(-r-3s, r, s) \in U## and ##\mathbf{v}=(t,t,2t) \in V## for ##r,s,t\in\mathbf{R}##. Since ##\mathbf{u}+\mathbf{v}=(-r-3s+t,r+t,s+2t)##, every ##\mathbf{x}\in\mathbf{R}^3## can be expressed as a sum of ##\mathbf{u}+\mathbf{v}## by choosing appropriate constants.
 
  • #4
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How would one do that? Finding the basis vectors for ##U## and ##V## respectively and check if they are linearly independent? Or can one, less rigorously, state that ##\mathbf{u}=(-r-3s, r, s) \in U## and ##\mathbf{v}=(t,t,2t) \in V## for ##r,s,t\in\mathbf{R}##. Since ##\mathbf{u}+\mathbf{v}=(-r-3s+t,r+t,s+2t)##, every ##\mathbf{x}\in\mathbf{R}^3## can be expressed as a sum of ##\mathbf{u}+\mathbf{v}## by choosing appropriate constants.
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.
 

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