Are My Basis Calculations for R3 and R4 Subspaces Correct?

[Carmen12]

Homework Statement

Find the basis for the subspaces of R³ and R⁴ below.

Homework Equations

A) All vectors of the form (a,b,c), where a=0
B) All vectors of the form (a+c, a-b, b+c, -a+b)
C) All vectors of the form (a,b,c), where a-b+5c=0

The Attempt at a Solution

I honestly had no idea what I was doing. I just scratched what I could make of it down. I'm going to do more research online but I'm terrified of the time limit coming up so I thought I'd post my work here meanwhile. I hope you all don't mind!
Here's what I got (don't laugh please!):

A) {(0,1,1),(1,1,0)}
B) {(0,-1,0,1),(1,0,-1,0)}
C) {(5,0,1),(1,0,5)}

-Carmen

 

[lanedance]

Homework Statement

Find the basis for the subspaces of R³ and R⁴ below.

Homework Equations

A) All vectors of the form (a,b,c), where a=0

A) {(0,1,1),(1,1,0)}

-Carmen

start with A)
(a,b,c), where a=0 means

(0,b,c) for any b or c...

your first vector is in the set, but your 2nd has a = 1 so is not

it may help to picture it geometrically in which case it is the yz plane

 

[jakncoke]

A) Basis spans your subspace. So your answer indicates, for example that,

<br /> \left[ \begin{array}{cccc}0 &amp; 1 \\ 1 &amp; 1 \\ 1 &amp; 0 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 2 \\ 5 \end{array} \right]<br />

should have a solution, but under ur basis, this eqn has no solution. so ur basis cannot span this subspace

Now if we use

<br /> \left[ \begin{array}{cccc}0 &amp; 0 \\ 1 &amp; 0 \\ 0 &amp; 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ a \\ b \end{array} \right]<br />

this guy has a solution for all a, b. Now you have to show this is independent. Do this by showing<br /> \left[ \begin{array}{cccc}0 &amp; 0 \\ 1 &amp; 0 \\ 0 &amp; 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 0 \\ 0 \end{array} \right]<br />

has only the trivial soln. which it does.