Hi! I just want to ask you, what is the principle of finding section between two vector subspaces. Let's say: U={(a,b,0) | a, b \(\in R\)} and W={(a,b,c) | a+b+c=0, a,b,c \(\in R\)} U, W are vector subspaces from \(R^3\) P.S this is not homework question, just example, for my better understanding. Thanks.
OK well in that case, you've got to find the subspace that simultaneously satisfies the requirements for membership in both of the subspaces. *U requires a zero 3rd coordinate. *W requires the sum of the coordinates to be zero. Taken together, we have: \(U\cap W=\{(a,b,0)|a+b=0\}\) We can do even better. Since \(a+b=0\) it follows that \(b=-a\), and so: \(U\cap W=span\{(1,-1,0)\}\)
And what will be the process of solving this task: Find \(U\cap W\)\(\subseteq\)\(R^3\) U={(a,b,c) | a=b, a,b,c \(\in R }\) W={(0,b,c) | b,c \(\in R }\)
U is such that (a,b,c) is constainted to (a,a,c) W is such that (a,b,c) is constrainted to (0,b,c) \(U \cap W\) has BOTH constaints so b=a from U and a=0 from W gives (a,b,c) constrainted to (0,0,c). 'Simple' definitions like those in your questions can be easily combined, you just apply then one at a time to (a,b,c,....,d). Each constraint will typically allow you to remove one variable from (a,b,....,d). Can you see why?
In general you can write subspaces as solution sets of systems of linear equations and then solve the systems simultaneously as one system to find the intersection.
Consider 2 dimensions, so you're just working in the (x,y) plane. Without any constraints, you have a 2 dimensional space, the entire plane. If I give you a linear equation for x and y, say Ax+By=C for A,B,C real numbers you now have a constraint. You started with 2 dimensions. You have 1 constraint, so the subspace produced by that constraint is 1 dimensional. Infact, you are familiar with such a subspace because you know how to plot lines of the form \(y = \frac{C-Ax}{B}\), you've been doing it in school for years. Suppose I now give you another constraint, ax+by=c. This alone just produces another 1d subspace, another line. Now you have a 2d space with 2 linear constraints. 2-2=0 so the subspace produced by applying BOTH constraints (ie the intersection of the two 1d subspaces) will be 0 dimensional. You know how to work out that subspace too. Given two lines, \(y = \frac{C-Ax}{B}\) and \(y = \frac{c-ax}{b}\) you just eliminate a variable and work out the values of x and y, which will mean the subspace is just a single point. In general, if you are in N dimensions and you have M linear constraints (which are linearly independent) the subspace they form will be N-M dimensions. For instance, consider 3 dimensional space. The linear constraint x=0 produces a 2d plane (ie the entire (y,z) plane). Similarly y=0 gives the entire (x,z) plane. Their intersection gives a 1 dimensional subspace, the z axis. Intersecting this with the z=0 plane then gives you a point, (0,0,0). Try drawing that to see how they build up and intersect. Obviously z=0, y=0 and x=0 are simple cases, 3 general linear equations would be of the form : a) \(A_{1}x+A_{2}y+A_{3}z = A_{4}\) b) \(B_{1}x+B_{2}y+B_{3}z = B_{4}\) c) \(C_{1}x+C_{2}y+C_{3}z = C_{4}\) I can elimate x by working out say \(B_{1}\)a - \(A_{1}\)b. Or I could eliminate y or z. It just comes down to solving those simultaneous equations, just as finding the intersection of the two lines I just gave as an example, \(y = \frac{C-Ax}{B}\) and \(y = \frac{c-ax}{b}\) comes down to eliminating y or x, so \( \frac{C-Ax}{B} = \frac{c-ax}{b}\) Solve for x etc.
Ok, I have one more question. What if: \(U=\{(0,y,z)|2y+3z=0\}\) of \(\mathbb{R}^3\) \(W=\{(x,y,z)|x+y+z=0\}\) of \(\mathbb{R}^3\) How we will find \(U\cap W\)? I mean, is it possible?
