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# Linear independence - is this ok? watch

1. To check for linear (in)dependence, we need to see if, for every real number x:

If we take

If we take

Then, as we know that

Now, seeing as we know that

However, this is where I get stuck.

If these vectors a linearly independent, then

But, if they are linearly dependent, then there exists a non-trivial solution (ie at least one of )

Since we know that

We have to show that or

So, we are now left with the equation:

For
or

Then, is either trivial, or non-trivial.

But, seeing as this is for all real values of x, then:

Hence,

Hence, they are linearly independent.

Is that right?
2. They are indeed linearly independent (and this should be obvious from the fact that the first function is periodic, the second and third are increasing) so you could never find 3 constants that give 0 for all values of x. I'm wondering if a shorter method could be achieved by contradiction using the above fact.

So assume that there exists non-zero constants such that

for all , it would suffice to show that these constants do not give:

for all

Which could perhaps be done by looking at because however the other two functions have increased and so

3. (Original post by Noble.)
They are indeed linearly independent (and this should be obvious from the fact that the first function is periodic, the second and third are increasing) so you could never find 3 constants that give 0 for all values of x. I'm wondering if a shorter method could be achieved by contradiction using the above fact.

So assume that there exists non-zero constants such that

for all , it would suffice to show that these constants do not give:

for all

Which could perhaps be done by looking at because however the other two functions have increased and so

I see what you've done, and that does make sense.

However, I don't think that's something I would have thought of in an exam, which is worrying.
Is my method wrong?
4. (Original post by ekudamram)
I see what you've done, and that does make sense.

However, I don't think that's something I would have thought of in an exam, which is worrying.
Is my method wrong?
No it isn't wrong. Once you've deduced that both you're left with and clearly this can only be true for all if which gives linear independence.

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Updated: April 13, 2013
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