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I this question from an old exam:

Determine all vectors located in Cola) and Col (B). Explain why all the vectors contained in both Col (A) and Col (B) form a subspace iR3 and calculate its dimension.

A below:

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B below:

enter image description here

The thing is i understand the calculations but what i don't understand is that they take the cross-product of the vectors in Matrix A and same thing in Matrix B. And after that they take Cross-product of the both vectors and get a another 90 degrees vector and say this vector is both Col(A) and Col(B) how come that the Cross-Product can give a vectors in both spaces i don't understand i am very confused normally you use the Cross-Product to calculate a Equation for a plane?

Solution below:

enter image description here

enter image description here

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Why can cross-product solve this problem i don't understand i am very confused ??

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  • $\begingroup$ The cross product does give you a vector in both spaces, but you are correct that that vector will not necessarily span $R(A) \cap R(B)$. For example, if $A = B = \begin{bmatrix} e_1 & e_2 \end{bmatrix}$, then you get $e_3 \times e_3 = 0$. $\endgroup$
    – Mason
    Aug 12 at 18:38
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Two 2-dimensional subspaces of $\mathbb{R}^3$ are either identical or have a 1-dimensional intersection. Clearly these two subspaces aren't identical, so we just need to find one vector in there intersection to find them all.

$v\times w, v, w$ forms a basis of $\mathbb{R}^3$ whenever $v, w$ are linearly independent. Furthermore, any vector orthogonal to $v\times w$ lies in the span of $v, w.$

So, $(v\times w) \times (v'\times w')$ will lie in the intersection of the subspace spanned by $v, w$ and the subspace spanned by $v', w'.$ Since the cross product is not 0 in this example, it must be spanning since the intersection in your case is 1-dimensional.

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  • $\begingroup$ So the Cross-Product gives a vector that is a bridge between two vectors? $\endgroup$
    – john
    Aug 12 at 18:47
  • $\begingroup$ @john What do you mean by bridge? A cross product gives a vector perendicular to both vectors. $\endgroup$
    – user147556
    Aug 12 at 18:47
  • $\begingroup$ I mean Vector that goes through both vectors? I understand what you mean by Perendicular! $\endgroup$
    – john
    Aug 12 at 18:49
  • $\begingroup$ @john It doesn't really go through either vector in any meaningful sense. It is perpendicular to both. Try graphing $(1, 0, 0), (0, 1, 0),$ and their cross product $(0, 0, 1)$ to see what the general geometric configuration (up to scaling and rotation) looks like. $\endgroup$
    – user147556
    Aug 12 at 18:57
  • $\begingroup$ Is the intersection in the subspaces then? $\endgroup$
    – john
    Aug 12 at 19:00

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