1. ## Subspace of R3

I know that for a set u of vectors to be called a subspace in R^n, it must satisify the conditions:

1- 0 E u
2- x, y E u --> x+y E u
3- x E u --> ax E u (a E R)

But I still cant manage to determine which sets are a subspace for R^n..

an example of a question is:

Which of the following sets are subspaces of R^3?

A. {(-3x + 5y, 8x +4y, -2x-4y)| x, y arbitrary numbers}
B. {(x,y,x)| x<y<z}
C. {(6,y,z)| y, z arbitrary numbers}
D. {(x,0,0)| x arbitrary number}
E. {(x,y,z)| x+y+z = 3}
F. {(x,y,x)| 6x+9y-7z=0}

a
c
d
f

but apparently it's wrong...so I don't know

Another question is:

Which of the following subsets of R^(3x3) are subspaces of R^(3x3) ?

A. The invertible matrices
B. The matrices whose entries are all integers
C. The symmetric matrices
D. The matrices whose entries are all greater than or equal to
E. The matrices with all zeros in the third row
F. The diagonal matrices

Wouldn't it be all because they're all subsets of R^(3x3) ??

Could someone kindly explain to me how to find the answers, and what the answer would be to that question?

This question has to do with linear dependency:
Which of the following sets of vectors are linearly independent?

A. { ( -2, 8, 0 ), ( 7, -9, 0 ), ( 6, -5, 0 ) }
B. { ( 3, -4, -7 ), ( 2, -3, -8 ), ( 1, -1, 1 ) }
C. { ( -9, 6 ) }
D. { ( -7, 2 ), ( -3, -8 ) , ( 4, 5 ) }
E. { ( -6, -1, 5 ), ( -1, 9, 4) }
F. { ( 3, -4, -7 ), ( 2, -3, -8 ), ( 1, 1, -1 ), ( 3, 11, -1 ) }
G. { ( -8, -3, -1, 5 ), ( -1, 5, -3, 2) }
H. { ( -3, 0, 2 ), ( 6, 5, -4 ), ( -1, -8, 9 ) }

the answer I get are B, G and H..but it still shows up as wrong..

HopefulMii

2. Originally Posted by HopefulMii
A = {(-3x + 5y, 8x +4y, -2x-4y)| x, y arbitrary numbers}
I do this first one.

Let $\bold{x} = (-3x_1 + 5x_2, 8x_1 + 4x_2, -2x_1 - 4x_2)$ and $\bold{y} = (-3y_1 + 5y_2, 8y_1 + 4y_2, - 2y_1 - 4y_2)$.

This means, $\bold{x} + \bold{y} = ( - 3(x_1 + y_1) + 5(x_2+y_2), 8(x_1+y_1) + 4(x_2+y_2), - 2(x_1+y_1) - 4(x_2+y_2) ) \in A$

And, $k\bold{x} = (-3(kx_1) + 5(kx_2), 8(kx_1) + 4(kx_2), - 2(kx_1) - 4(kx_2))\in A$.

Therefore $A$ is a vector space.

3. okay..

but how do I prove that for the ones that are just like:

{(x,y,z)}| 6x + 9y -7z=0} ?

What I think I'm supposed to do is isolate z

so z = (6x+9y)/7

and then I have the vector (x,y, (6x+9y)/7)

but then from there I don't know what to do...

4. Originally Posted by HopefulMii
okay..

but how do I prove that for the ones that are just like:

{(x,y,z)}| 6x + 9y -7z=0} ?
If $6x_1 + 9x_2 - 7x_3 = 0$ and $6y_1 + 9y_2 - 7y_3 = 0$.
This means (their sum) $6(x_1+y_2) + 9 (x_2 + y_2) - 7 (x_3 + y_3) = 0$.