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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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  • $\begingroup$ No, your basis is not a member of $W$, so it can't be an intersection $\endgroup$ Nov 6, 2013 at 12:02
  • $\begingroup$ 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? $\endgroup$
    – user99680
    Nov 6, 2013 at 12:05
  • $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Nov 6, 2013 at 12:05
  • $\begingroup$ 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 $\endgroup$
    – nbdip
    Nov 6, 2013 at 12:10
  • $\begingroup$ @user99680 do you mean for example for R^2=(1,0)*a+(0,1)*b $\endgroup$
    – nbdip
    Nov 6, 2013 at 12:11

4 Answers 4

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

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 $U\cap W=\left\{\begin{pmatrix}2\\4\\0\end{pmatrix}\frac{\sqrt{3}}{2}\mu+\begin{pmatrix}0\\0\\\sqrt{3}\end{pmatrix}\mu\mid \mu\in\mathbb{R}\right\}=\left\{\begin{pmatrix}\sqrt{3}\\2\sqrt{3}\\\sqrt{3}\end{pmatrix}\mu\mid \mu\in\mathbb{R}\right\}$

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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  • $\begingroup$ then we could do (1,0,-1)*sqrt(3)m +(0,1,1)*2*(sqrt(3))m aswell right ? $\endgroup$
    – nbdip
    Nov 6, 2013 at 12:27
  • $\begingroup$ Why did you equalize them? $\endgroup$
    – shinzou
    May 31, 2015 at 17:52
  • $\begingroup$ Did you mean W instead of V? $\endgroup$
    – blueether
    Oct 24, 2018 at 22:09
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Let me try a comprehensive answer. The Zassenhaus algorithm does this job, but it is somehow difficult to explain. Another approach is to remember that elementary row operations preserve linear dependence between columns of a matrix. Suppose that $$U=[u_1=(1,0,0,-1),u_2=(0,1,0,-1),u_3=(0,0,1,1)]$$ and $$W=[w_1=(1,0,-1,0),w_2=(0,1,-1,0),w_3=(0,0,0,1)].$$ Note that the generators of $U$ and $W$ are linearly independent, otherwise we could exclude the dependent vectors. Now, form a matrix whose columns are the vectors of U followed by the vectors of W: $$M=\left[\begin{matrix} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 & -1 & 0 \\ -1 & -1 & 1 & 0 & 0 & 1 \\ \end{matrix}\right]. $$ Computing the reduced row echelon form of $M$ we obtain that $$\text{rref}(M)=\left[\begin{matrix} 1 & 0 & 0 & 0 & -1 & -\frac{1}{2} \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 1 & 1 & \frac{1}{2} \\ \end{matrix}\right]. $$ Therefore we can conclude from the aforementioned fact, that $$\begin{aligned} w_2&=-u_1+u_2+w_1 \\ w_3&=-\frac{1}{2}u_1+\frac{1}{2}u_3+\frac{1}{2}w_1 \end{aligned}$$ and that $u_1$, $u_2$, $u_3$ and $w_1$ form a basis of $U+W$. We cannot say that $w_2$ and $w_3$ form a basis of $U\cap W$, because they are not linear combinations of vectors of $U$, but defining $$\begin{aligned} w_2'&:=w_2-w_1 = -u_1+u_2 \\ w_3'&:=w_3-\frac{1}{2}w_1=-\frac{1}{2}u_1+\frac{1}{2}u_3 \end{aligned}$$ we see that $w_2'\in U\cap W$ and $w_3'\in U\cap W$. It is easy to see from general considerations that $w_2'$ and $w_3'$ are linearly independent and from $$\text{dim}(U+W)+\text{dim}(U\cap W)=\text{dim}(U)+\text{dim}(W),$$ it turns out that they form a basis of $U\cap W$. This procedure can clearly be generalized and even transformed into a proof of the theorem about the dimension of the sum of subspaces.

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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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  • $\begingroup$ 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 $\langle 1,1,0\rangle$ $\endgroup$ Nov 6, 2013 at 12:22
  • $\begingroup$ Thanks... I'll correct that $\endgroup$
    – Jlamprong
    Nov 6, 2013 at 12:23
  • $\begingroup$ Yes, I did an error computation. Thanks $\endgroup$
    – Jlamprong
    Nov 6, 2013 at 12:28
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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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