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I know there is lots of topics about intersection of two vector subspaces and basis but i still dont fully understand how we should handle these question. So this is my homework:

Suppose U and W are subspaces of $R^3:\\$ $U=[(1,0,-1),(0,1,1)]\\W=[(2,4,0),(0,0,\sqrt{3})]\\$

Find a basis of $U\cap W \\$

So i know $U \cap W$ -> $a_1*(1,0,-1)+a_2*(0,1,1)-b_1*(2,4,0)-b_2*(0,0,\sqrt{3})=0\\$ a possible combination of coefficients $a_1=1,a_2=2,b_1=\frac12,b_2=-\sqrt{3}\\$

then i put the $1*(1,0,-1)+2*(0,1,1)$ in a matrix $$ \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ -1 & 2 \\ \end{bmatrix} \\$$

then i bring it to row-achelon form

$$ \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ 0 & 0 \\ \end{bmatrix} \\$$

so Basis=$\{ (1,0,-1),(0,1,1) \}$ (i m not sure if the last vector should be $(0,1,1)$ or $(0,2,2)) $?

So is the solution correct ?

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No, your basis is not a member of $W$, so it can't be an intersection –  Tim Ratigan Nov 6 '13 at 12:02
    
I did not follow everything, but unless the subspaces $V,W$ are equal (and U,W are not parallel), their intersection is a line, so the basis should consist of just 1 vector. do you know how to recover the equation of a plane from its basis vectors? –  user99680 Nov 6 '13 at 12:05
    
Also think of the following, @nbdip : if the intersection of two 2-dimensional spaces is a 2-dimensional space, then the intersect is one of the two subspaces... –  DonAntonio Nov 6 '13 at 12:05
    
i m really confused right now @Tim.Ratigan What did you mean by "your basis is not a member of w" edit:ok i get it. But if i put the other vectors with coefficient b1 and b2 in to the matrix.should that work –  nbdip Nov 6 '13 at 12:10
    
@user99680 do you mean for example for R^2=(1,0)*a+(0,1)*b –  nbdip Nov 6 '13 at 12:11

3 Answers 3

up vote 1 down vote accepted

Let $U=\left\{\begin{pmatrix}1\\0\\-1\end{pmatrix}\lambda+\begin{pmatrix}0\\1\\1\end{pmatrix}\mu\mid \lambda,\mu\in\mathbb{R}\right\}$, $V=\left\{\begin{pmatrix}2\\4\\0\end{pmatrix}\lambda+\begin{pmatrix}0\\0\\\sqrt{3}\end{pmatrix}\mu\mid \lambda,\mu\in\mathbb{R}\right\}$

For $U\cap W,$ $\begin{bmatrix}1&0\\0&1\\-1&1\end{bmatrix}\cdot\begin{bmatrix}\lambda_1\\\mu_1\end{bmatrix}=\begin{bmatrix}2&0\\4&0\\0&\sqrt{3}\end{bmatrix}\cdot\begin{bmatrix}\lambda_2\\\mu_2\end{bmatrix}$ $$ \lambda_1=2\lambda_2 $$ $$ \mu_1=4\lambda_2 $$ $$ -\lambda_1+\mu_1=\sqrt{3}\mu_2 $$ $$ \Rightarrow\lambda_2=\frac{\sqrt{3}}{2}\mu_2 $$ Thus $V\cap W=\left\{\begin{pmatrix}2\\4\\0\end{pmatrix}\frac{\sqrt{3}}{2}\mu+\begin{pmatrix}0\\0\\\sqrt{3}\end{pmatrix}\mu\mid \mu\in\mathbb{R}\right\}=\left\{\begin{pmatrix}\sqrt{3}\\2\sqrt{3}\\\sqrt{3}\end{pmatrix}\mu\mid \mu\in\mathbb{R}\right\}$

Note that the magnitude of the basis vector is largely irrelevant, so you can normalize it to make it $\langle1,2,1\rangle$.

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then we could do (1,0,-1)*sqrt(3)m +(0,1,1)*2*(sqrt(3))m aswell right ? –  nbdip Nov 6 '13 at 12:27

Since $\{(1,0,-1),(0,1,1), (0,0,\sqrt 3\}$ is a basis for $\mathbb{R}^3$ then $U \cup W=\mathbb{R}^3$. So, $\dim (U \cap W)=1$. Moreover we can choose $\{(1,2,1)\}$ a basis of $U \cap W$ because $\{(1,2,1)\} \in U \cap W$.

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Uhh... $\langle 1,1,0\rangle\not\in W$... Try finding a linear combination of $\langle 2, 4, 0\rangle$ and $\langle 0,0,\sqrt{3}\rangle$ that makes $\langle 1,1,0\rangle$ –  Tim Ratigan Nov 6 '13 at 12:22
    
Thanks... I'll correct that –  Jlamprong Nov 6 '13 at 12:23
    
Yes, I did an error computation. Thanks –  Jlamprong Nov 6 '13 at 12:28

This is my idea: just like we can obtain a basis from an expression $ax+by+cz=0$ to a basis $(x_0,y_0,z_0),(x_1,y_1,z_1)$ for the plane described by that equation (by using the basic linear -algebra -fact that $ax+by+cz=0$ has 2 free variables; say $ y,z $), we can reverse the process to get an equation $ax+by+cz=0$ when we're given two basis vectors $(x_0,y_0,z_0),(x_1,y_1,z_1)$. Once we have the two planes associated to the given basis in the form $ax+by+cz=0$, we can easily find their intersection.

Then, the plane spanned by $(1,0,-1),(0,1,1)$ is $x-y+z=0$ , and the plane spanned by $(2,4,0),(0,0,\sqrt 3)$ is $4\sqrt 3x-2\sqrt3 y$. It then just comes down to solving the simple system:

$i)x-y+z=0$

$ii)4\sqrt3x-2\sqrt3y=0$

Whose solution is $y=2z$ ; back-substitution in $i$ gives us $x-2z+z=0$ , so $x=z$, and $y=2z$ gives us the line $(x,2x,x)$, spanned by, e.g. $(1,2,1)$

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