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Problem:

Let $A = \begin{bmatrix}5& 2& -1\\3& 1& 0\\ -1& 0& -1 \end{bmatrix}$, $B = \begin{bmatrix} 4& -3\\ -2& 3\\ 1& -2\end{bmatrix}$, $U = C(A)$ and $V = C(B)$ where U and V are the column space of $A$ and $B$.
Find a basis of $\ U \cap V$.

What I have done:
Firstly, I calculate the bases of U and V, which are $\{ [ 5, 3, 1 ]^T, [2, 1, 0]^T \}$ and $\{[4, -2, 1]^T, [-3, 3, -2]^T\}$.
Then I want to get the intersection of it. But I find what I have done seems like out of logic, which makes me puzzle at how to solve it.
SO I check the answer:

$C(A)\cap C(B) = span{(1,1,−1)}$
Hint: solve the system $\begin{bmatrix}A& B \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} = 0$ and use the fact that any nonzero solution will give an element in the intersection, namely Ax or By. Now just look for the right number of linearly independent elements in the intersection.

I think the hint want me to solve a linear system and get a nonzero solution that is the answer.
Do I think it right?
And I am wondering if there is a better way to get a basis of $C(A) \cap C(B)$?
If not mind, could anyone help me and give some inspirations?
Thanks in advance.

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  • $\begingroup$ The solutions $(x,y)^T$ of the equation $[A \,| -B](x,y)^T = 0$ satisfy $Ax = By$, so $Ax = By$ is in the intersection of the column spaces of $A$ and $B$. So, first find the solutions $(x,y,z,a,b)^T$ of the larger system and then apply $B$ to all possible $(a,b)^T$ to get the intersection $C(A)\cap C(B)$. $\endgroup$ – amsmath Apr 17 '19 at 13:41
  • $\begingroup$ @amsmath Thanks for your help. And I get general solution of $\begin{bmatrix}A& B \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} = 0$ is $\begin{bmatrix} -1\\ 3\\ 1\\ 0\\ 0\end{bmatrix}x_3 + \begin{bmatrix}-1\\ 2\\ 0\\ 1\\ 1 \end{bmatrix}x_5$, but I don't know how to do next to get the right answer $C(A)\cap C(B) = span{(1,1,−1)}$:( $\endgroup$ – Bowen Peng Apr 17 '19 at 14:14
  • $\begingroup$ Thanks for putting your solution here. Now, what are the possible $(a,b)^T$ in $(x,y,z,a,b)^T$ in this solution? $\endgroup$ – amsmath Apr 17 '19 at 14:16
  • $\begingroup$ @amsmath Sadly, it seems like I just couldn't make it through. Maybe I just couldn't understand the hint from you or the answer well. $\endgroup$ – Bowen Peng Apr 17 '19 at 14:25
  • $\begingroup$ Dude, the solution space of the extended system is $\{[-c-d,3c+2d,c,d,d]^T : c,d\in\mathbb R\}$. This is just another way of writing your solution -- with $x_3$ and $x_5$ replaced by $c$ and $d$, respectively. Now, just project onto the last two entries. What will that be? $\endgroup$ – amsmath Apr 17 '19 at 14:28
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So $U$ is spanned by $\{(5,3,-1),(2,1,0),(-1,0,1)\}$ and $B$ is spanned by $\{(4,-2,1),(-3,3,-2)\}$.

Any vector in $U$ is of the form $$a(5,3,-1)+ b(2,1,0)+ c(-1,0,1)= (5a+2b-c,3a+b,-a+c), \quad \textrm{for some } a,b,c \in \mathbb{R}$$

and any vector in $V$ is of the form

$$ x(4,-2,1)+ y(-3,3,-2)= (4x-3y,-2x+3y,x-2y), \textrm{for some } x, y \in \mathbb{R}.$$

Any vector in the intersection of $U$ and $V$ can be written in both forms so we must have $(5a+2b-c,3a+b,-a+c)= (4x-3y,-2x+3y,x-2y)$ or $5a+2b-c= 4x-3y$, $3a+b= -2x+3y$, and $-a+c= x- 2y$. Now,

  1. Add the first and last equations to get $4a+ 3b= 5x- 5y$.
  2. Multiply the second equation by 3 to get $9a+ 3= -6x+9y$.
  3. Subtract: $5a= (-6x+9y)-(5x-5y)= -11x+ 14y$ so $a= -(11/5)x+ (14/5)y$. Then, $$b= -2x+ 3y- 3a= -2x+ 3y+ (33/5)x- (42/5)y= (23/5)x- (27/5)y$$ and $$ c= x-2y+a= x-2y-(11/5)x+(14/5)y= -(1/5)x+ (4/5)y.$$
  4. Put those for $a, b$, and $c$ in $a(5,3,-1)+ b(2,1,0)+ c(-1,0,1)$.

  5. Multiply that last equation by 2 and add it to the second equation to get $a+ b+ 2c= -y$ or $y= -a-b-2c$. Then $x= -a+c+2y= -a+c-2a-2b-4c= -3a-2b-3c$.

Any vector in the intersection of $U$ and $V$ is of the form $$(-3a-2b-3c)(4,-2,1)+ (-a-b-2c)(-3,3,-2)= (-12a-8b-12c+3a+3b+6c,6a+4b+6c-3a-3b-6c,$$ (…)

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  • $\begingroup$ Very, very bad. $\endgroup$ – amsmath Apr 17 '19 at 14:14
  • $\begingroup$ Please consider using MathJax. I cannot even decide whether this is right or wrong as it is unreadable. $\endgroup$ – M. Winter Apr 23 '19 at 18:26

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