# Sum of basis elements form a basis

• MHB
• mathmari
In summary, the conversation discusses the equivalence between two sets being bases of a vector space. The main points are that a set is a basis if it is linearly independent and spans the vector space, and if two sets have the same number of elements, then they are both bases. The conversation also includes a step-by-step explanation of the proof for the equivalence.
mathmari
Gold Member
MHB
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)

mathmari said:
Is at this direction everything correct? (Wondering)

Hey mathmari!

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

mathmari said:
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)

Klaas van Aarsen said:
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)

Yep. All correct. (Nod)

Klaas van Aarsen said:
Yep. All correct. (Nod)

Great! Thanks a lot! (Mmm)

## 1. What does it mean for the sum of basis elements to form a basis?

When we say that the sum of basis elements forms a basis, it means that any vector in a vector space can be written as a linear combination of those basis elements. In other words, these basis elements span the entire vector space.

## 2. How do you prove that the sum of basis elements forms a basis?

In order to prove that the sum of basis elements forms a basis, we need to show that the basis elements are linearly independent and span the entire vector space. This can be done by using the definition of linear independence and showing that any vector in the vector space can be written as a linear combination of the basis elements.

## 3. Can the sum of basis elements form a basis for any vector space?

Yes, the sum of basis elements can form a basis for any finite-dimensional vector space. This is because any finite-dimensional vector space has a finite number of basis elements, making it possible for their sum to span the entire space.

## 4. Are there any cases where the sum of basis elements does not form a basis?

Infinite-dimensional vector spaces are an example where the sum of basis elements may not form a basis. This is because the sum of infinitely many basis elements may not span the entire vector space.

## 5. What is the significance of the sum of basis elements forming a basis?

The fact that the sum of basis elements forms a basis is important because it allows us to easily represent any vector in a vector space as a linear combination of the basis elements. This makes it easier to perform calculations and solve problems within the vector space.

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