- #1

mathmari

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Let $V$ be a vector space. Let $b_1, \ldots , b_n\in V$ and let $\displaystyle{b_k':=\sum_{i=1}^kb_i}$ for $k=1, \ldots , n$.

I want to show that $\{b_1, \ldots , b_n\}$ is a basis of $V$ iff $\{b_1', \ldots , b_n'\}$ is a basis of $V$. I have done the following:

Let $B:=\{b_1, \ldots , b_n\}$ be a basis of $V$.

That means that the set is linearly independent and $|B|=\dim (V)$, right?

We want to show that the set $B':=\{b_1', \ldots , b_n'\}$ is also linearly independent and $|B'|=\dim (V)$, don't we?

Since the elements of $B$ is linearly independent, we have that $c_1b_1+c_2b_2+\ldots c_nb_n=0 \Rightarrow c_1=c_2=\ldots =c_n=0$.

We have that \begin{align*}\gamma_1b_1'+\gamma_2b_2'+\ldots \gamma_nb_n'=0 &\Rightarrow \gamma_1\sum_{i=1}^1b_i+\gamma_2\sum_{i=1}^2b_i+\ldots \gamma_n\sum_{i=1}^nb_i=0 \\ & \Rightarrow b_1\sum_{i=1}^n\gamma_i+b_2\sum_{i=2}^n\gamma_i+\ldots +b_n\sum_{i=n}^n\gamma_i=0 \\ & \Rightarrow \sum_{i=n}^1\gamma_i=\sum_{i=n}^2\gamma_i=\ldots =\sum_{i=n}^n\gamma_i=0\\ & \Rightarrow \gamma_1=\gamma_2=\ldots =\gamma_n=0\end{align*}

Since the set $B'$ has $n$ elements as the set $B$ it follows that $|B'|=\dim (V)$.

Therefore $B'$ is a basis of $V$.

Is at this direction everything correct? (Wondering) For the other direction:

Let $B'=\{b_1', \ldots , b_n'\}$ be a basis of $V$.

That means that the set is linearly independent and $|B'|=\dim (V)$.

We want to show that the set $B:=\{b_1, \ldots , b_n\}$ is also linearly independent and $|B|=\dim (V)$.

Since the elements of $B'$ is linearly independent, we have that $\gamma_1b_1'+\gamma_2b_2'+\ldots \gamma_nb_n'=0 \Rightarrow \gamma_1=\gamma_2=\ldots =\gamma_n=0$.

How do we get from $c_1b_1+c_2b_2+\ldots c_nb_n=0 $ the linear combination with the elements $b_i'$ ? (Wondering)