Question

Prove that if {x1x2.......xk}isany

Answer 1

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 $n\geq 2$. 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 $X= (x_1, x_2,....,x_n)$ is linearly dependent, so then by definition we have scalars $c_1,c_2,....,c_n$ in C such that:

$c_1 x_1 +c_2 x_2 +.....+c_n x_n =0$

And not all the scalars $c_1,c_2,....,c_n$ are equal to 0.

Since at least one constant is non zero we can assume for example that $c_1 \neq 0$, and we have this:

$c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n$

We can divide by c1 since we assume that $c_1 \neq 0$ and we have this:

$v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n$

And as we can see the vector $v_1$ can be written a a linear combination of the remaining vectors $v_2,v_3,...,v_n$. 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 $v_1$ and we have this:

$v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n$

For scalars defined $c_2,c_3,...,c_n$ in C. So then we have this:

$v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0$

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.