MHB What are the basis subsets of a 5-element vector space with additional vectors?

[mathmari]

Hey!

Let $V$ be a vector space with with a 5-element basis $B=\{b_1, \ldots , b_5\}$ and let $v_1:=b_1+b_2$, $v_2:=b_2+b_4$ and $\displaystyle{v_3:=\sum_{i=1}^5(-1)^ib_i}$.

I want to determine all subsets of $B\cup \{v_1, v_2, v_3\}$ that form a basis of $V$.
Are the desired subsets the following ones?

- $B$
- $\{v_1, b_2, b_3, b_4, b_5\}$
- $\{v_1, b_1, b_3, b_4, b_5\}$
- $\{v_2, b_1, b_2, b_3, b_5\}$
- $\{v_2, b_1, b_3, b_4, b_5\}$
- $\{v_3, b_1, b_2, b_3, b_4\}$
- $\{v_3, b_1, b_2, b_3, b_5\}$
- $\{v_3, b_1, b_2, b_4, b_5\}$
- $\{v_3, b_1, b_3, b_4, b_5\}$
- $\{v_3, b_2, b_3, b_4, b_5\}$
- $\{v_1, v_2, b_2, b_3, b_5\}$ Are there also more sets? (Wondering)

 

[I like Serena]

Hey mathmari!

Yep. Those are all a basis of $V$.
And yes, there are more sets.
Currently you have all the ones with exactly one of $\{v_1,v_2,v_3\}$, but only those.
For starters, shouldn't $\{b_1, b_2, b_3, b_4, b_5\}$ be in the list?
And how about $\{v_1, v_2, b_1, b_3, b_5\}$? (Wondering)

 

[mathmari]

Currently you have all the ones with exactly one of $\{v_1,v_2,v_3\}$, but only those.
For starters, shouldn't $\{b_1, b_2, b_3, b_4, b_5\}$ be in the list?

That is the set $B$, so I have added that one. (Thinking)

And how about $\{v_1, v_2, b_1, b_3, b_5\}$? (Wondering)

The elements $b_1$ and $b_2$ can be replaced by $v_1$, the elements $b_2$ and $b_4$ can be replaced by $v_2$ and the all the elements $b_i$ can be replaced by $v_3$, right? (Wondering)

So, we get the following sets:

- $B=\{b_1, b_2, b_3, b_4, b_5\}$
- $\{v_1, b_2, b_3, b_4, b_5\}$
- $\{b_1, v_1, b_3, b_4, b_5\}$
- $\{b_1, v_2, b_3, b_4, b_5\}$
- $\{b_1, b_2, b_3, v_2, b_5\}$
- $\{v_3, b_2, b_3, b_4, b_5\}$
- $\{b_1, v_3, b_3, b_4, b_5\}$
- $\{b_1, b_2, v_3, b_4, b_5\}$
- $\{b_1, b_2, b_3, v_3, b_5\}$
- $\{b_1, b_2, b_3, b_4, v_3\}$
- $\{v_1, v_2, b_3, b_4, b_5\}$
- $\{v_1, b_2, b_3, v_2, b_5\}$
- $\{b_1, v_1, b_3, v_2, b_5\}$
- $\{v_1, v_3, b_3, b_4, b_5\}$
- $\{v_1, b_2, v_3, b_4, b_5\}$
- $\{v_1, b_2, b_3, v_3, b_5\}$
- $\{v_1, b_2, b_3, b_4, v_3\}$
- $\{v_3, v_1, b_3, b_4, b_5\}$
- $\{b_1, v_1, v_3, b_4, b_5\}$
- $\{b_1, v_1, b_3, v_3, b_5\}$
- $\{b_1, v_1, b_3, b_4, v_3\}$
- $\{v_3, v_2, b_3, b_4, b_5\}$
- $\{b_1, v_2, v_3, b_4, b_5\}$
- $\{b_1, v_2, b_3, v_3, b_5\}$
- $\{b_1, v_2, b_3, b_4, v_3\}$
- $\{v_3, b_2, b_3, v_2, b_5\}$
- $\{b_1, v_3, b_3, v_2, b_5\}$
- $\{b_1, b_2, v_3, v_2, b_5\}$
- $\{b_1, b_2, b_3, v_2, v_3\}$

I think I have now all the popssible sets, or not? (Wondering)

 

[I like Serena]

You have 29 sets, but I think there are 33. (Thinking)

I have:
\begin{align*}
\{b_1, b_2, b_3, b_4, b_5\} \\
\{v_1, b_1, b_3, b_4, b_5\} \\
\{v_1, b_2, b_3, b_4, b_5\} \\
\{v_1, v_2, b_1, b_3, b_5\} \\
\{v_1, v_2, b_2, b_3, b_5\} \\
\{v_1, v_2, b_3, b_4, b_5\} \\
\{v_1, v_2, v_3, b_1, b_3\} \\
\{v_1, v_2, v_3, b_1, b_5\} \\
\{v_1, v_2, v_3, b_2, b_3\} \\
\{v_1, v_2, v_3, b_2, b_5\} \\
\{v_1, v_2, v_3, b_3, b_4\} \\
\{v_1, v_2, v_3, b_3, b_5\} \\
\{v_1, v_2, v_3, b_4, b_5\} \\
\{v_1, v_3, b_1, b_3, b_4\} \\
\{v_1, v_3, b_1, b_3, b_5\} \\
\{v_1, v_3, b_1, b_4, b_5\} \\
\{v_1, v_3, b_2, b_3, b_4\} \\
\{v_1, v_3, b_2, b_3, b_5\} \\
\{v_1, v_3, b_2, b_4, b_5\} \\
\{v_1, v_3, b_3, b_4, b_5\} \\
\{v_2, b_1, b_2, b_3, b_5\} \\
\{v_2, b_1, b_3, b_4, b_5\} \\
\{v_2, v_3, b_1, b_2, b_3\} \\
\{v_2, v_3, b_1, b_2, b_5\} \\
\{v_2, v_3, b_1, b_3, b_4\} \\
\{v_2, v_3, b_1, b_4, b_5\} \\
\{v_2, v_3, b_2, b_3, b_5\} \\
\{v_2, v_3, b_3, b_4, b_5\} \\
\{v_3, b_1, b_2, b_3, b_4\} \\
\{v_3, b_1, b_2, b_3, b_5\} \\
\{v_3, b_1, b_2, b_4, b_5\} \\
\{v_3, b_1, b_3, b_4, b_5\} \\
\{v_3, b_2, b_3, b_4, b_5\} \\
\end{align*}

 

[mathmari]

Ahh ok! But the idea is the following, or not?

We consider the basis $B=\{b_1, b_2, b_3, b_4, b_5\}$. The vector $v_1$ replace each of the vectors $b_1$ and $b_2$, the vector $v_2$ can replace each of the vectors $b_2$ and $b_4$, the vector $v_3$ can replace each of the 5 vectors $b_1, b_2, b_3, b_4, b_5$, and the set that we get is again a basis since the elements remain linearly independent.

(Wondering)

 

[I like Serena]

Ahh ok! But the idea is the following, or not?

We consider the basis $B=\{b_1, b_2, b_3, b_4, b_5\}$. The vector $v_1$ replace each of the vectors $b_1$ and $b_2$, the vector $v_2$ can replace each of the vectors $b_2$ and $b_4$, the vector $v_3$ can replace each of the 5 vectors $b_1, b_2, b_3, b_4, b_5$, and the set that we get is again a basis since the elements remain linearly independent.

That sounds about right.
I also think that it's easy to make mistakes. (Worried)

 

[mathmari]

I also think that it's easy to make mistakes. (Worried)

What do you mean? (Wondering)