Respuesta :
Answer:
See the proof below.
Step-by-step explanation:
What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where [tex] n\geq 2[/tex]. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"
Proof
Since we have a if and only if w need to proof the statement on the two possible ways.
If X is linearly dependent, then a vector is a linear combination
We suppose the set [tex] X= (x_1, x_2,....,x_n)[/tex] is linearly dependent, so then by definition we have scalars [tex] c_1,c_2,....,c_n[/tex] in C such that:
[tex] c_1 x_1 +c_2 x_2 +.....+c_n x_n =0[/tex]
And not all the scalars [tex] c_1,c_2,....,c_n[/tex] are equal to 0.
Since at least one constant is non zero we can assume for example that [tex] c_1 \neq 0[/tex], and we have this:
[tex] c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n [/tex]
We can divide by c1 since we assume that [tex] c_1 \neq 0[/tex] and we have this:
[tex] v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n [/tex]
And as we can see the vector [tex] v_1[/tex] can be written a a linear combination of the remaining vectors [tex] v_2,v_3,...,v_n[/tex]. We select v1 but we can select any vector and we get the same result.
If a vector is a linear combination, then X is linearly dependent
We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select [tex] v_1[/tex] and we have this:
[tex] v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n[/tex]
For scalars defined [tex] c_2,c_3,...,c_n[/tex] in C. So then we have this:
[tex] v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0[/tex]
So then we can conclude that the set X is linearly dependent.
And that complet the proof for this case.
