Linear Algebra: Linear Independence

[DerivativeofJ]

Homework Statement

Let S be a basis for an n-dimensional vector space V. Show that if v1,v2,...,vr form a linearly independent set of the vectors in V, then the coordinate vectors (v1)s, (v2)s,...,(vr)s form a linearly independent set in the Rn, and conversely.

Homework Equations

The Attempt at a Solution

I tried working this problem but i got stuck almost at the end. i know that to show that the coordinate vectors form a linearly independent set that the following equation

k1((v1)s)+ k2((v2)s) +...+ kr((v)s)=0 has to have only the trivial solution. Could i please get some help. I wrote v1, v2,..vn as a linear combination of the set S which i defined as S={w1,w2,...,wn}. Help please.

 

[Dick]

If v1 is a vector, then what is (v1)s supposed to mean?

 

[DerivativeofJ]

It is notation. It is called the coordinate vector of v1 relative to S.

for example v1 can be written as a linear combination of the basis S

v1= c1(w1)+ c2(w2)+...+ cn(wn)

thus

(v1)s= {c1,c2,...,cn}

 

[Fredrik]

You want to prove that the two implications
$$\sum_i k_i v_i=0\ \Rightarrow\ \forall i~~k_i=0$$ and
$$\sum_i k_i (v_i)_S=0\ \Rightarrow\ \forall i~~k_i=0.$$ are either both true or both false. You can do this by proving that $\sum_i k_i (v_i)_S=\big(\sum_i k_i v_i)_S$. This is a matrix equation, so it holds if and only if the jth components of the left-hand side and the right-hand side are equal for all j.

The following observation is useful. For all vectors x, we have
$$
\begin{align}
x &=\sum_j x_j w_j\\
x_S &=\sum_j (x_S)_j e_j=\sum_j x_j e_j,
\end{align}
$$ where the $e_j$ are the standard basis vectors for $\mathbb R^n$. The important detail here is that $(x_S)_j=x_j$, by definition of the "S" notation.

 

[DerivativeofJ]

Thank You!