Are My Basis Calculations for R3 and R4 Subspaces Correct?

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SUMMARY

The discussion focuses on finding the basis for specific subspaces in R3 and R4. For subspace A, the correct basis is {(0,1,0), (0,0,1)}, representing the yz-plane, as the initial submission included an incorrect vector. Subspace B's basis is {(0,-1,0,1), (1,0,-1,0)}, while subspace C's basis is {(5,0,1), (1,0,5)}. The participants emphasize the importance of ensuring that the chosen basis spans the subspace and is linearly independent.

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  • Understanding of vector spaces and subspaces in linear algebra
  • Familiarity with the concepts of basis and dimension
  • Knowledge of linear independence and spanning sets
  • Ability to perform operations with vectors in R3 and R4
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Carmen12
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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 :eek:
 
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Carmen12 said:

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 :eek:

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

[tex] \left[ \begin{array}{cccc}0 & 1 \\ 1 & 1 \\ 1 & 0 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 2 \\ 5 \end{array} \right][/tex]

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

Now if we use

[tex] \left[ \begin{array}{cccc}0 & 0 \\ 1 & 0 \\ 0 & 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ a \\ b \end{array} \right][/tex]

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

has only the trivial soln. which it does.
 
Last edited:

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