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spring_rolls
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I have a few questions here, my main problem is not understanding the notations used, hence not understanding the questions.
1. Do the vectors [tex]j_{1}[/tex]= (1,0,-1,2) and [tex]j_{2}[/tex]= (0,1,1,2) form a basis for the space W = {(a,b,c,d) l a - b + c = 0, -2a - 2b + d =0} ?
2. Find a basis for W = { (a,b,c,d) : a - b + 2d = 0 , 3a + c + 3d = 0 }
3. Find the dimension of the following subspaces of [tex]\Re^{3}[/tex]
a) span { (1,0,1),(0,1,1),(2,0,0) }
b) span { (1,0,1),(2,2,4),(2,1,7),(-1,-1-2) }
4. Suppose that a subspace W [tex]\subset[/tex] [tex]\Re^{3}[/tex] has a basis { (1,2,3),(1,0,1) }.
a)Is { (-1,-2,-3), (3,2,5) } a basis for W? Why?
How do you do this? Is there a working? or can you solved this by inspection.
1. i substitute each vector in a,b,c,d equation and find the equality, so happen when i substitute in all returns 0, which implies that it does form a basis. Is there a proper working for this, cause i can do this by inspection.
2. I was wondering forming a matrix and row reduced the matrix and get a linear dependence equation and plug in some value for c & d to find a & b.
1 -1 0 2 0 ~ 1 0 1/3 1 0
3 0 1 3 0 0 1 1/3 -1 0
So: a = -1/3c - d & b= d - 1/3c
Let c =1 and d =1
a= -1/3 (1) - 1 = -4/3 & b = 1 - 1/3 (1) = 2/3
so basis (-1/3,2/3,1,1)
3 & 4. No idea what to do.
Homework Statement
1. Do the vectors [tex]j_{1}[/tex]= (1,0,-1,2) and [tex]j_{2}[/tex]= (0,1,1,2) form a basis for the space W = {(a,b,c,d) l a - b + c = 0, -2a - 2b + d =0} ?
2. Find a basis for W = { (a,b,c,d) : a - b + 2d = 0 , 3a + c + 3d = 0 }
3. Find the dimension of the following subspaces of [tex]\Re^{3}[/tex]
a) span { (1,0,1),(0,1,1),(2,0,0) }
b) span { (1,0,1),(2,2,4),(2,1,7),(-1,-1-2) }
4. Suppose that a subspace W [tex]\subset[/tex] [tex]\Re^{3}[/tex] has a basis { (1,2,3),(1,0,1) }.
a)Is { (-1,-2,-3), (3,2,5) } a basis for W? Why?
Homework Equations
How do you do this? Is there a working? or can you solved this by inspection.
The Attempt at a Solution
1. i substitute each vector in a,b,c,d equation and find the equality, so happen when i substitute in all returns 0, which implies that it does form a basis. Is there a proper working for this, cause i can do this by inspection.
2. I was wondering forming a matrix and row reduced the matrix and get a linear dependence equation and plug in some value for c & d to find a & b.
1 -1 0 2 0 ~ 1 0 1/3 1 0
3 0 1 3 0 0 1 1/3 -1 0
So: a = -1/3c - d & b= d - 1/3c
Let c =1 and d =1
a= -1/3 (1) - 1 = -4/3 & b = 1 - 1/3 (1) = 2/3
so basis (-1/3,2/3,1,1)
3 & 4. No idea what to do.