Showing direct sum of subspaces equals vector space

[schniefen]

TL;DR

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 $\mathbf{R}^3=U \bigoplus V$.

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 $t \in \mathbf{R}$?

 

[fresh_42]

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.

 

[schniefen]

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.

 

[fresh_42]

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.