How to find this is vector space or not in linear algebra?
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Let's look at the following example:
W={(a,b,c,d)∈R4∣a+3b−2c=0}.
The vector space W consists of all solutions (x,y,z,w) to the equation
x+3y−2z=0.
How do we write all solutions? Well, first of all, w can be anything and it doesn't affect any other variable. Then, if we let y and z be anything we want, then that will force x and give a solution. So we have three degrees of freedom: a free choice of w, a free choice of z, and a free choice of y. Then x will be forced. This suggests dimension 3.
How does the choice of w affect x, y, and z? In absolutely no way. Since choosing w does not affect x, y, or z, this gives the vector (0,0,0,1): the choice of w (the 1) does not affect the others.
How does the choice of z affect x, y, and w? It doesn't affect y and w. But if z=1, then x needs to be 2: that is, we need to get two xs for every z. This gives the vector (2,0,1,0).
Finally, now does the choice of y affect x, z, and w? It doesn't affect z and w (they are free), but for every y, we need to have −3xs. That gives the vector (−3,1,0,0).
So a basis for my W consists of (−3,1,0,0), (2,0,1,0), and (0,0,0,1). You can verify that all of them lie in W, and that every vector in W can be written as a linear combination of these three in a unique way.
W={(a,b,c,d)∈R4∣a+3b−2c=0}.
The vector space W consists of all solutions (x,y,z,w) to the equation
x+3y−2z=0.
How do we write all solutions? Well, first of all, w can be anything and it doesn't affect any other variable. Then, if we let y and z be anything we want, then that will force x and give a solution. So we have three degrees of freedom: a free choice of w, a free choice of z, and a free choice of y. Then x will be forced. This suggests dimension 3.
How does the choice of w affect x, y, and z? In absolutely no way. Since choosing w does not affect x, y, or z, this gives the vector (0,0,0,1): the choice of w (the 1) does not affect the others.
How does the choice of z affect x, y, and w? It doesn't affect y and w. But if z=1, then x needs to be 2: that is, we need to get two xs for every z. This gives the vector (2,0,1,0).
Finally, now does the choice of y affect x, z, and w? It doesn't affect z and w (they are free), but for every y, we need to have −3xs. That gives the vector (−3,1,0,0).
So a basis for my W consists of (−3,1,0,0), (2,0,1,0), and (0,0,0,1). You can verify that all of them lie in W, and that every vector in W can be written as a linear combination of these three in a unique way.
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