Prove that vectorsv1, v2, ... ,vn of the same dimension are linearly dependent if and only if one of them can be written as a linear combination of the others.
I'm not sure how exactly I would correctly go about proving this. Thanks!
Prove that vectorsv1, v2, ... ,vn of the same dimension are linearly dependent if and only if one of them can be written as a linear combination of the others.
I'm not sure how exactly I would correctly go about proving this. Thanks!
Suppose that $\displaystyle v_k = \alpha _1 v_1 + \cdots \alpha _{k - 1} v_{k - 1} + \alpha _{k + 1} v_{k + 1} \cdots + \alpha _n v_n $.
But that means $\displaystyle 0 = \alpha _1 v_1 + \cdots \alpha _{k - 1} v_{k - 1} + v_k + \alpha _{k + 1}{\color{red}-1} v_{k + 1} \cdots + \alpha _n v_n$.
Do you know what it means for the set $\displaystyle \left\{ {v_1 ,v_2 , \cdots v_{n - 1} ,v_n } \right\}$ to be linearly independent?
If you do then you see the above proves that the set is not linearly independent.
Moreover, you can reverse the argument.
Well, this can be proven directly:
Theorem: $\displaystyle \bold{v_1}, \bold{v_2}, ..., \bold{v_n}$ of the same dimension are linearly dependent iff one of them can be written as a linear combination of the others.
Proof:
Let the set $\displaystyle \{\bold{v_1}, \bold{v_2}, ..., \bold{v_n}\}$ be linearly dependent. This means that there exists scalars $\displaystyle s_1, s_2, ..., s_n$, not all zero, such that:
$\displaystyle s_1\bold{v_1} + s_2\bold{v_2} + ... + s_n\bold{v_n} = 0$
So, assume that $\displaystyle s_1 \neq 0$. We can rewrite this as:
$\displaystyle \bold{v_1} = \left(-\frac{s_2}{s_1}\right)\bold{v_2} + \left(-\frac{s_3}{s_1}\right)\bold{v_3} + ... + \left(-\frac{s_n}{s_1}\right)\bold{v_n}$
Therefore, we get $\displaystyle \bold{v_1}$ as a linear combination of the others.
Now conversely, we assume that $\displaystyle \bold{v_1}$ is a linear combination of $\displaystyle \bold{v_2}, \bold{v_3}, ... , \bold{v_n}$ which means that there exists scalars $\displaystyle c_2, c_3, ..., c_n$ such that:
$\displaystyle \bold{v_1} = (-c_2)\bold{v_2} + (-c_3)\bold{v_3} + ... + (-c_n)\bold{v_n}$
We can rewrite this to be:
$\displaystyle 1\bold{v_1} + c_2\bold{v_2} + c_3\bold{v_3} + ... + c_n\bold{v_n} = 0$
Therefore, the set $\displaystyle \{\bold{v_1}, \bold{v_2}, ... , \bold{v_n}\}$ is linearly dependent.