Basis of the intersection of two spaces

In summary: This is true for any two spaces, and is summarized by the mathematical statement that spaces "intersect."
  • #1
Zero2Infinity
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Homework Statement


Consider two vector spaces ##A=span\{(1,1,0),(0,2,0)\}## and ##B=\{(x,y,z)\in\mathbb{R}^3 s.t. x-y=0\}##. Find a basis of ##A\cap B##.

I get the solution but I also inferred it without all the calculations. Is my reasoning correct

Homework Equations



linear dependence definition.

The Attempt at a Solution



First of all I look for a basis of ##B##, which is ##(1,1,0),(0,0,1)##.
Now, the complete method for finding the basis of ##A\cap B## is the following one. I set
\begin{equation}
a_{1}(1,1,0)+a_{2}(0,2,0)=b_{1}(1,1,0)+b_{2}(0,0,1)
\end{equation}
or, equivalently (the coefficients are unknown so the sign is not that important as I can name ##-b_{i}=b_{i}##),
\begin{equation}
a_{1}(1,1,0)+a_{2}(0,2,0)+b_{1}(1,1,0)+b_{2}(0,0,1)=(0,0,0)
\end{equation}
By solving the linear system I get:
\begin{equation}
\begin{cases} a_{1}+0+b_{1}+0=0\\
a_{1}+2a_{2}+b_{1}+0=0 \\ 0+0+0+b_{2}=0 \end{cases}
\Leftrightarrow \begin{cases} a_{1}=-b_{1}\\
a_{1}+2a_{2}+b_{1}=0 \\ b_{2}=0 \end{cases}
\end{equation}
##a_{1}=-b_{1},a_{2}=b_{2}=0##. Thus, the generic vector ##\textbf{v}\in A\cap B## can be written as the linear combination of the basis of ##A## with ##a_{1}## and ##a_{2}## as coefficents.
\begin{equation}
\textbf{v}=-b_{1}(1,1,0)+0\cdot(0,2,0)=-b_{1}(1,1,0)
\end{equation}
The conclusion is that the searched basis is ##(1,1,0)##.

Now, I know this is the correct procedure but is it correct to say that, since I know a basis of ##A## and a basis of ##B##, the basis of ##A\cap B## are the common vectors to both basis, if there is at least one?
This way I could have immediately said that ##\mathcal{B}_A:(1,1,0),(0,2,0), \mathcal{B}_B:(1,1,0),(0,0,1)\Rightarrow \mathcal{B}_{A\cap B}:(1,1,0)##.
 
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  • #2
I think the answer is yes...but you have to be careful. If you are able to write a basis vector for one space in terms of basis vectors from another, then clearly, those vectors will be in the intersection.
However, your space A is simply the xy-plane. It could have been written with other basis vectors that would not be so simple as [1,1,0] to pick out. Consider [1,0,0] and [0,1,0].
If you have two 2-D spaces, as above, your intersection is either all of the 2D space (if A = B) , spanned by a single vector, or empty (parallel planes).
 
  • #3
You could get the dimension of ##A \cap B## from considering the dimensions of the subspaces. If the second basis vector for ##B## is not in ##A##, then the intersection must be the 1D subspace spanned by the common basis vector.

Geometrically, two planes through the origin, if not equal, intersect in a line through the origin.
 

FAQ: Basis of the intersection of two spaces

What is the basis of the intersection of two spaces?

The basis of the intersection of two spaces is the set of vectors that are common to both spaces. These vectors are linearly independent and span the intersection space.

How is the basis of the intersection of two spaces calculated?

The basis of the intersection of two spaces can be calculated by finding the basis of each individual space and then finding the vectors that are common to both bases. These common vectors will form the basis of the intersection space.

Can the basis of the intersection of two spaces be empty?

Yes, it is possible for the basis of the intersection of two spaces to be empty. This occurs when the two spaces do not have any vectors in common, meaning their intersection space is the zero vector.

What is the dimension of the intersection space?

The dimension of the intersection space is the number of vectors in the basis of the intersection. This is equal to the number of common vectors in the basis of each individual space.

Why is the basis of the intersection of two spaces important?

The basis of the intersection of two spaces is important because it allows us to understand the structure and relationships between different vector spaces. It also plays a crucial role in solving problems in linear algebra and other areas of mathematics.

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