MHB Sum of basis elements form a basis

[mathmari]

Hey!

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)

 

[I like Serena]

Is at this direction everything correct? (Wondering)

Hey mathmari!

Yep. It's all correct so far. (Bow)

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 $b_i'$ ?

How about substituting the definitions of $b_i'$ and continue in a similar fashion as the other direction? (Wondering)

 

[mathmari]

How about substituting the definitions of $b_i'$ and continue in a similar fashion as the other direction? (Wondering)

We have the following:
\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 \ \text{ since } \gamma_1=\gamma_2=\ldots =\gamma_n=0\end{align*}
From that it follows that the elements of $B$ are linearly independent.

Is that correct? (Wondering)

Since the set $B$ has $n$ elements as the set $B'$ it follows that $|B|=\dim (V)$ and so $B$ is also a basis. We use here the fact that if a set is linearly independent and has the same number of elements than a basis which means that it spans the vector space then that set is also a basis, right? (Wondering)

 

[I like Serena]

Yep. All correct. (Nod)

 

[mathmari]

Yep. All correct. (Nod)

Great! Thanks a lot! (Mmm)