Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to show that $U_1:=\langle(1,1,1)^T\rangle$ and $U_2:=\langle(1,0,0)^T,(0,0,1)^T\rangle$ are complementary subspaces and how to decompose $(1,2,3)^T$ regarding these ? Best Regards, Thomas

share|improve this question
    
What could you say about linearity independence? –  Sigur May 15 '13 at 0:26
    
$<U_1, U_2>$ is linear independent !? –  fast-forward May 15 '13 at 0:32
    
and $U_1 \cap U_2 = \{\}$ therefore they are complementary subspaces !? –  fast-forward May 15 '13 at 0:34
    
I changed $<(1,0,0)^T,(0,0,1)^T>$ to $\langle(1,0,0)^T,(0,0,1)^T\rangle$. That is standard. Notice the extra spacing you get with the first form. That's because "$<$" is a binary relation symbols, and a certain amount of blank space precedes and follows those, as in $3<5$. –  Michael Hardy May 15 '13 at 1:08
add comment

3 Answers 3

up vote 0 down vote accepted

Notice that $U_2$ has zero in the $y$ coordinate, so whatever part is due to $U_1$ must account for the 2 in your vector. Thus $(1,2,3) = (2,2,2)+(-1,0,1)$ is the place to start.

Edit: Whoops! For the first part, the easiest thing is to write the standard basis $(1,0,0) ,(0,1,0),(0,0,1)$ in terms of your vectors since they're so simple.

share|improve this answer
    
I guess I only have to show that $U_1 \cap U_2 = \{\}$ and <U1, U2> = V. $(1,2,3) = 2(1,1,1)^T + (-1)(1,0,0)^T + (0,0,1)^T$ thats all !? –  fast-forward May 15 '13 at 0:41
    
Yes, that's fine. (Although the intersection is $\{0\}$, not empty.) –  Sharkos May 15 '13 at 0:47
add comment

You need to show that every $x \in \mathtt{R}^3$ (or whatever vector space you are interested in, your question doesn't specify one) can be uniquely represented as $$ x = u_1 + u_2 \quad\text{where $u_1 \in U_1$ and $u_2 \in U_2$.} $$

Since you already have a basis for $U_1$ and $U_2$, this is equivalent to showing that combining those basis vectors produces a basis for $\mathbb{R}^3$, which in turn is equivalent to showing that the matrix built from those basis vectors has determinant $\neq 0$. In your case, that means showing that $$ \left|\begin{array}{lll} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 1 \end{array}\right| \neq 0 \text{.} $$

To decompose a vector into $u_1, u_2$, one way is to find it's coordinates relative to the basis created from the basis vectors of the subspaces. You can then easily combine one vector per subspace, whose sum will produce the original vector.

share|improve this answer
    
one vector per subspace means in this example the sum of $(1,1,1)^T + (0,0,1)^T$ ??? –  fast-forward May 15 '13 at 1:00
    
@fast-forward Where does it say "one vector per subspace"? –  fgp May 15 '13 at 7:53
add comment

The best way to do this is to solve the system of equations $Ax=b$, where $$A=\left[\begin{array}{ccc}1&1&0\\1&0&0\\1&0&1\end{array}\right]\quad\text{and}\quad b= \left[\begin{array}{c} 1\\2\\3\end{array}\right].$$ With the numbers you have, there are shortcuts, as others have suggested.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.