# Prove the Matrix has a subspaceWatch

#1
I Need help with this question please

Thank you
Last edited by NoahMal; 4 weeks ago
0
#2
Here is what I have so far but im not sure if its correct
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4 weeks ago
#3
(Original post by NoahMal)
Here is what I have so far but im not sure if its correct
It seems rather confused. What is W, when V is the label of the subspace? Also you need to show two things in V add to give something in V? The matrices you consider aren't in V to begin with, they're general 2x2 matrices.
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#4
(Original post by RichE)
It seems rather confused. What is W, when V is the label of the subspace? Also you need to show two things in V add to give something in V? The matrices you consider aren't in V to begin with, they're general 2x2 matrices.
Last edited by NoahMal; 4 weeks ago
0
4 weeks ago
#5
(Original post by NoahMal)
Definitely better. You're still calling it W and is your (2,2) entry an "a" or a poorly looking "d"? Should be a "d".
1
4 weeks ago
#6
So, you'd need to show that v is closed under matrix addition and scalar multiplication.
It's been a while since I've done this, but why have you used w throughout your working instead of v ?

You'd want to say for two matrices in v, the addition is also in v, that is, is in v, which can be shown if you use the condition a+b+c=0 and e+f+g = 0, then you just have (a+e)+(b+f)+(c+g)=0. i.e adding the equations to each other. The second matrix shouldn't have the trailing diagonal as f, as you're making another assumption that these two elements should be equal which isn't a condition given in the question.

Also show why it is closed under scalar multiplication (even if it's obvious).

Then you're good.
1
#7
(Original post by RichE)
Definitely better. You're still calling it W and is your (2,2) entry an "a" or a poorly looking "d"? Should be a "d".
It's A poorly written d

(Original post by NotNotBatman)
.
Thank you
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