0
$\begingroup$

Let $U = span(u_1,u_2)$ and $V = span(v_1, v_2)$ be subspaces of $\mathbb{R}^4$ with $u_1 = (1,1,1,2), u_2 = (2,1,0,3)$ and $v_1 = (1,-1,1,0), v_2 = (-1,2,1,1)$ Determine a basis of $U \cap V$. Let a,b,c,d $\in \mathbb{R}$

Let $x \in U \cap V$ then $x = a*(1,1,1,2) + b*(2,1,0,3) = c*(1,-1,1,0) + d*(1,2,1,1)$

This gives us $(a + 2b, a+b, a, 2a+3b) = (c-d, -c + 2d, c+d,d)$ How do I go from here?

1 Answers 1

1

Hint: Rearrange to put everything on the left and write as $Ax = 0$. Then you get

$$ \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \in \mathrm{ker} \begin{bmatrix} 1 & 2 & -1 & -1 \\ 1 & 1 & 1 & -2 \\ 1 & 0 & -1 & -1 \\ 2 & 3 & 0 & -1 \end{bmatrix}. $$

  • 0
    and now try to reach row echelon form? – 2017-01-19
  • 0
    Yes, do that for the augmented matrix using the methods outlined in this answer: http://math.stackexchange.com/questions/88301/finding-the-basis-of-a-null-space – 2017-01-19
  • 0
    One more question, how did you get the last two columns? The given vectors are different? – 2017-01-19
  • 0
    Ah, you're correct, there was a mistake in the last column first row - it should have been $-1$. – 2017-01-20
  • 0
    And they are negatives of the vectors being considered – 2017-01-20
  • 0
    @AndrewWhelan They are, but you could just as well use $v_1$ and $v_2$ directly, since both $\{v_1,v_2\}$ and $\{-v_1,-v_2\}$ span the same space. – 2017-02-07