quick linear algebra....

[Semidevil]

ok, so let S be the set of vectors in R3 having the form

2z - 3y
y
z

I need to prove that S is a subspace of R3....

so this means that I need to show this preserves vector addition and multiplication.

my problem is that I dont know what this Vector is. is it a matrix that looks like

0x -3y + 2z
0x +1y +0z
0x +0y +z


?

see, I dot even know how to treat the vector. To check for vector adition means that I need to check U + V. so what is U and what is V?

lost....

 

[Ryan]

Ummm, I guess 7.

 

[eakers]

you need to take 2 vectors in S

so (3y+2z, y, z) + (6y+4z, 2y, 2z)
and show they add.

your teacher might want you to use variables instead of numbers though.

for multiplication perhaps you are to use the transpose of one?

i dunno.

its been a while.

 

[Semidevil]

Originally posted by: eakers
you need to take 2 vectors in S

so (3y+2z, y, z) + (6y+4z, 2y, 2z)
and show they add.

your teacher might want you to use variables instead of numbers though.

for multiplication perhaps you are to use the transpose of one?

i dunno.

its been a while.

where did the 6y + 4z, 2y, 2z come from?

 

[eakers]

Originally posted by: Semidevil

Originally posted by: eakers
you need to take 2 vectors in S

so (3y+2z, y, z) + (6y+4z, 2y, 2z)
and show they add.

your teacher might want you to use variables instead of numbers though.

for multiplication perhaps you are to use the transpose of one?

i dunno.

its been a while.

where did the 6y + 4z, 2y, 2z come from?

i just multiplied by 2 (as it has the form 2y+2z...)

another thing you could do which is probabley more right is say "let a, b be 2 vectors in S" and then say let a be alpha((3y+2z, y, z)) and b be beta((3y+2z, y, z))

then show that a+ b is in S and that ab is in S.

 

[RaynorWolfcastle]

ok, I'll show you how to do it.
You need to show that your system is closed under addition and linear multiplication. I'll use the notation y[2] means "y subscript 2"

addition:
Assume you have 2 vectors of V[1] = {2z[1]-3y[1], y[1],z[1]} and V[2] = {2z[2]-3y[2], y[2],z[2]}. Both these vectors belong to the space you have defined.
You find that V[1] + V[2] = {2(z[1]+z[2]) - 3(y[1]+y[2]), (y[1]+y[2]), (z[1] +z[2])} which is in fact a vector that is part of your subspace (that has parameters (y[1]+y[2]) and (z[1] + z[2]) as expected.

You have shown that S is closed under addition

Linear multiplication:
Use as parameters z' = 3z and y' = 3y
If you find the vector corresponding to these parameters, you can easily show that v' = 3 {2z-3y, y,z} and so your system is closed under linear multiplication
You have shown that S is closed under linear multiplication

Finally, you show that your subspace S contains the zero vector by letting z = 0 and y = 0 and you have shown that S is in fact a subspace

 

[Semidevil]

wow......thanx you. that helps a lot.