MHB Statements with linearly independent vectors

[mathmari]

Hey!

Let $\mathbb{K}$ a field and let $V$ a $\mathbb{K}$-vector space. Let $1\leq m, n\in \mathbb{N}$ and $n=\dim_{\mathbb{K}}V$. Let $v_1, \ldots , v_m\in V$ be linearly independent.

1. Let $\lambda_1, \ldots , \lambda_m, \mu_1, \ldots , \mu_m\in \mathbb{K}$ such that $\displaystyle{\sum_{i=1}^m\lambda_iv_i=\sum_{i=1}^m\mu_iv_i}$. Then show that $\lambda_i=\mu_i$ for all $1\leq i\leq m$.
2. Let $w\in V$. Then show that $w\notin \text{span}(v_1, \ldots , v_m) \iff v_1, \ldots , v_m, w$ linearly independent.
3. Let $2\leq q\in \mathbb{N}$ with $|\mathbb{K}|=q$. Then show that $ \text{span}(v_1, \ldots , v_m) $ has $q^m$ elements and determine the number of vectors $w\in V$ such that $v_1, \ldots, v_m, w$ are linearly independent.
4. Let $2\leq q\in \mathbb{N}$ with $|\mathbb{K}|=q$. Then show that $$\prod_{i=0}^{n-1}(q^n-q^i)=(q^n-1)(q^n-q)\cdots (q^n-q^{n-2})(q^n-q^{n-1})$$ the number of (ordered) bases of $V$.

For question 1, I have done the following:
\begin{align*}\sum_{i=1}^m\lambda_iv_i=\sum_{i=1}^m\mu_iv_i &\Rightarrow \sum_{i=1}^m\lambda_iv_i-\sum_{i=1}^m\mu_iv_i=0\\ & \Rightarrow \sum_{i=1}^m\left (\lambda_i-\mu_i\right )v_i=0 \\ & \ \overset{v_1, \ldots , v_m\text{ linear unabhängig}}{\Longrightarrow } \ \lambda_i-\mu_i=0, \ \forall i\in \{1, \ldots , m\} \\ & \Rightarrow \lambda_i=\mu_i, \ \forall i\in \{1, \ldots , m\}\end{align*}

For question 2, do we have to reformulate the equivalence into:
\begin{equation*}w\in \text{span}(v_1, \ldots , v_m) \iff v_1, \ldots , v_m, w\text{ linearly dependent }\end{equation*}
or can we show the equivalance in the given form?
But what would then $w\notin \text{span}(v_1, \ldots , v_m)$ mean? That $\displaystyle{w\neq\sum_{i=1}^m\alpha_iv_i }$ ? Could you give me a hint for questions 3 & 4 ?

 

[HOI]

What does "|K|" mean?

 

[mathmari]

What does "|K|" mean?

It's the dimension.

 

[GJA]

Hi mathmari,

Very interesting set of questions.

Question 2
Yes, nicely done. We can reformulate the statement this way and it is a good idea to do so.

Question 3
For the number of elements in the span, note that for each $v_{i}$, there are $q$ choices from $\mathbb{K}$ for $v_{i}$'s coefficient. Hence, by the fundamental counting principle, there are $q^{m}$ total elements in their span.

For the second part, I will start off with an example that highlights all the important features of the general proof. Take a look and see if you can extend these ideas. If anything is unclear, certainly feel free to let me know.

Suppose $n=5$ and $m=2$. Let $v_{3},v_{4},$ and $v_{5}$ be vectors such that $\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}$ is a basis for $V$. Consider a vector of the form $$w=\sum_{i=1}^{5}a_{i}v_{i}.$$ The key here is to determine what condition must be placed on $a_{3}, a_{4}$ and $a_{5}$ to guarantee that $w$ will be linearly independent from $v_{1}$ and $v_{2}$. Can you think of anything that would do the trick? I will leave it at this for now but am certainly happy to help more if need be.

Question 4
This question relies heavily on having a proof/formula from Question 3, so I will refrain on the details of this until then. However, using the formula we discover in Question 3, we simply use it repeatedly to deduce the desired result.

 

[mathmari]

Question 3
For the number of elements in the span, note that for each $v_{i}$, there are $q$ choices from $\mathbb{K}$ for $v_{i}$'s coefficient. Hence, by the fundamental counting principle, there are $q^{m}$ total elements in their span.

For the second part, I will start off with an example that highlights all the important features of the general proof. Take a look and see if you can extend these ideas. If anything is unclear, certainly feel free to let me know.

Suppose $n=5$ and $m=2$. Let $v_{3},v_{4},$ and $v_{5}$ be vectors such that $\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}$ is a basis for $V$. Consider a vector of the form $$w=\sum_{i=1}^{5}a_{i}v_{i}.$$ The key here is to determine what condition must be placed on $a_{3}, a_{4}$ and $a_{5}$ to guarantee that $w$ will be linearly independent from $v_{1}$ and $v_{2}$. Can you think of anything that would do the trick? I will leave it at this for now but am certainly happy to help more if need be.

Do we use here the question 2, and $w$ must not be in the span of the vectors? :unsure:

 

[GJA]

Exactly right, great job connecting the ideas. Can you determine a condition/requirement on $a_{3}, a_{4}$ and $a_{5}$ that would ensure $w$ is not in the span of $v_{1}$ and $v_{2}$?

 

[mathmari]

Exactly right, great job connecting the ideas. Can you determine a condition/requirement on $a_{3}, a_{4}$ and $a_{5}$ that would ensure $w$ is not in the span of $v_{1}$ and $v_{2}$?

The coefficients $a_3,a_4,a_5$ should not be all equal to zero, right?

 

[GJA]

Yeah, that's it! So now we know that $a_{1}$ and $a_{2}$ can be any elements from $\mathbb{K}$, and that $a_{3}, a_{4},$ and $a_{5}$ can't all be zero. Can you use these pieces of information to count the total number of elements $w$ from $V$ such that $v_{1}, v_{2}, w$ are linearly dependent?

 

[mathmari]

Do we do for question 3 the following?

We consider the vectors $w\in V$ such that $v_1, \ldots , v_m, w$ are linearly independent. For the question 2 it holds that $w\notin \text{Lin}(v_1, \ldots , v_m)$.
The dimension of $V$ is $n$ and we have that $|\mathbb{K}|=q$.
It holds that \begin{equation*}q^n=\left (\#\text{ linearly independent vectors to }v_1, \ldots , v_m \right )+\left (\#\text{ linearly dependent vectors to }v_1, \ldots , v_m \right )\end{equation*}
The number of dependent vectors to $v_1, \ldots , v_m$ is the number of vectors in the span of $v_1, \ldots , v_m$. Since the span contains $q^m$ elements, tehre are $q^m$ elements $w$ such that $\displaystyle{w=\sum_{i=1}^m\alpha_iv_i}$.
Therefore the number of vectors that are linear independent to $v_1, \ldots , v_m$ is equal to $q^n-q^m$.

:unsure:For question 4:

Do we use question 3 for different values of $m$ ?

Or do we do the following:

There are $q^n$ elements in the vector space and so for the first choice for the basis there are $q^n-1$ ways ($0$ is excluded).
This vector generates a subspace of dimension $q$ and so for the second one there are $q^n-q$ ways.
If we continue we get that the number of ordered bases of $V$ is equal to \begin{equation*}(q^n-1)(q^{n-1}-q)\cdots (q^n-q^{n-2})(q^n-q^{n-1})=\prod_{i=0}^{n-1}(q^n-q^i)\end{equation*}

:unsure:

 

[GJA]

Yes, this is all correct. Nicely done!

Edit: A minor point of correction. Your post says " This vector generates a subspace of dimension $q$..." This is not correct. A single vector generates a 1-dimensional subspace. There are $q$ total elements in this 1-dimensional subspace, but the space itself is still 1-dimensional. This is just like how the $x$-axis in $\mathbb{R}^{2}$ contains infinitely many vectors but, as a whole, is 1-dimensional.

The slightly corrected version of your idea would be to select a single non-zero vector, say $b_{1}$, to be the first basis element for $V$. Because we must exclude zero, there are $q^{n}-1$ ways to do this. The next vector we choose must be linearly independent of $b_{1}$. Since the span of $b_{1}$ is 1-dimensional, from Question 3 we know there are $q^{n}-q$ choices for $b_{2}.$ By induction, if the first $m-1$ basis elements are selected, Question 3 tells us there are $q^{n}-q^{m-1}$ ways to select the next basis element $b_{m}$ so that $b_{m}$ is linearly independent of $b_{1}, b_{2},\ldots, b_{m-1}$. Equivalently, $b_{m}$ is not part of the $m-1$ dimensional subspace spanned by $b_{1}, b_{2},\ldots, b_{m-1}$. As you've correctly stated, multiplying these values together gives the total number of ordered bases for $V$ over the field $\mathbb{K}$.

 

[HOI]

It's the dimension.

K is a field, not a vector space. What do you mean by "dimension" of a field?