Linearly independent vs dependent functions

  • #1
member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For part(b),
1717397036962.png

My solution is,
##(c_1, c_2, c_3, c_4) = (c_1, 5c_1, -c_1, -3c_1)##

They have taken the case that c_1 = -1, which gives their expression for a linear dependent function as they have shown. However, I'm confused that the functions are linearly dependent for any value of ##c_1 \in \mathbb{R}##. Since if we take ##c_1 = 0##, then ##c_1 = c_2 = c_3 = c_4 = 0##. I agree thought they the functions are linear dependent for ##c_1 \in \mathbb{R}\{0\} ## (Set of real numbers without zero)

Does someone please know why they did not include that case?

Thanks!
 
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  • #2
ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For part(b),
View attachment 346379
My solution is,
##(c_1, c_2, c_3, c_4) = (c_1, 5c_1, -c_1, -3c_1)##

They have taken the case that c_1 = -1, which gives their expression for a linear dependent function as they have shown. However, I'm confused that the functions are linearly dependent for any value of ##c_1 \in \mathbb{R}##. Since if we take ##c_1 = 0##, then ##c_1 = c_2 = c_3 = c_4 = 0##. I agree thought they the functions are linear dependent for ##c_1 \in \mathbb{R}\{0\} ## (Set of real numbers without zero)

Does someone please know why they did not include that case?

Thanks!
Heya! From the definition of linear dependence, the functions
$$f_1,f_2,...,f_k$$ are linearly dependent if there exist scalars $$a_1,a_2,..,a_k,$$ not all zero, such that
$$a_1f_1+\cdots f_ka_k=0.$$
 
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  • #3
ChiralSuperfields said:
They have taken the case that c_1 = -1, which gives their expression for a linear dependent function as they have shown. However, I'm confused that the functions are linearly dependent for any value of ##c_1 \in \mathbb{R}##. Since if we take ##c_1 = 0##, then ##c_1 = c_2 = c_3 = c_4 = 0##. I agree thought they the functions are linear dependent for ##c_1 \in \mathbb{R}\{0\} ## (Set of real numbers without zero)
That is not at all what they are saying. Being linearly dependent means that there is a non-zero solution to
$$
\sum_{i = 1}^4 c_i f_i(t) = 0
$$
and their solution has all ##c_i \neq 0##. It makes no sense to say that the functions are linearly independent for some particular values of ##c_i##. Being linearly independent just means that there exists a non-zero solution for which the above equation holds for all ##t \in \mathbb R## (not ##c_i \in \mathbb R##), which is what they are saying.

Edit: Note: The correct ##\LaTeX## for the set you are referring to (the reals apart from zero) is ##\mathbb R \setminus \{0\}##.
 
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  • #4
If ##c1f_1+c_2f_2+ c_3 f_3+ c_4f4=0##, then ##-( c_1f_1+ c2f_2+ c_3f_3+c_4 f_4)=0##. Notice each of your terms is,the negative of those in the solution from the book.
 
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