/sci/ - Science & Math » Thread #11789976

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Anonymous Fri 12 Jun 16:06:20 2020 No.11789976 View Reply Original Report

Quoted By: >>11790030 >>11790431 >>11790669 >>11790769

How would I prove these are linearly independent.
I already put them as vectors in R9 and got the sum of all three is [0 0 0 0 0 0 0 0 0]

Anonymous

Anonymous Fri 12 Jun 2020 16:21:06 No.11790030 Report

Quoted By:

>>11789976
literally by the definition: show that the only way to write
a E1 + b E2 + c E3 = 0
is by having a = b = c = 0

Anonymous

Anonymous Fri 12 Jun 2020 17:57:53 No.11790431 Report

Quoted By:

>>11789976
anon if you see zeros in different entries across a set of vectors they're almost always linearly independent. You can by inspection tell that these are linearly independent

Anonymous

Anonymous Fri 12 Jun 2020 19:12:13 No.11790654 Report

Quoted By:

You have to show that the equation

$c_1 E_1 + c_2 E_2 + c_3 E_3 = 0$

has only the trivial solution $c_1 = c_2 = c_3 = 0$

this is the definition of an independent set. Hint: look at where the zeroes are.

Anonymous

Anonymous Fri 12 Jun 2020 19:17:58 No.11790669 Report

Quoted By:

>>11789976
>I already put them as vectors in R9 and got the sum of all three is [0 0 0 0 0 0 0 0 0]
How? The first entry alone sums to 2.

Note the isomorphism and proceed from there, this isn't a difficult exercise.

Anonymous

Anonymous Fri 12 Jun 2020 19:51:41 No.11790769 Report

Quoted By: >>11790840

>>11789976
If you row reduce the matrices and end up with the identity, they are linearly independent (it shows that the rank is 3, null space is 0, and they form a basis for $\mathbb{R}^{3}$ ). Not sure if that is the (rather simple) solution that is intended though.

Anonymous

Anonymous Fri 12 Jun 2020 20:22:41 No.11790840 Report

Quoted By:

>>11790769
You are retarded

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