Exercise on Vector Space Bases 4
We aim to determine whether every basis of the vector space $V = \mathbb{R}^n$ has exactly $n$ elements.
$$ V = \mathbb{R}^n $$
By definition, a vector space has dimension $n$ if any basis $ \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \} $ consists of exactly $n$ linearly independent vectors:
$$ B = \{ \vec{v}_1 , \vec{v}_2 , \dots , \vec{v}_n \} $$
To verify this, we consider a well-known candidate: the set of standard unit vectors.
$$ B = \left\{ \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}, \dots, \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} \right\} $$
Step 1: Proving that the set spans ℝn
The set $\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \}$ spans $\mathbb{R}^n$ if any vector $\vec{v} = (a_1, a_2, \dots, a_n)$ can be written as a linear combination of these vectors:
$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 + \dots + k_n \vec{v}_n = \vec{v} $$
Substituting the basis vectors gives:
$$ k_1 \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} + k_2 \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix} + \dots + k_n \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} $$
This is equivalent to the system:
$$ \begin{cases} k_1 = a_1 \\ k_2 = a_2 \\ \vdots \\ k_n = a_n \end{cases} $$
Each equation has a unique solution: simply take $k_i = a_i$ for all $i$.
Thus, the set $\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \}$ spans $\mathbb{R}^n$.
Step 2: Proving linear independence
A set of vectors is linearly independent if the only solution to the equation:
$$ k_1 \vec{v}_1 + k_2 \vec{v}_2 + \dots + k_n \vec{v}_n = \vec{0} $$
is the trivial one: $k_1 = k_2 = \dots = k_n = 0$.
Substituting the vectors:
$$ k_1 \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} + k_2 \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix} + \dots + k_n \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} $$
This results in the system:
$$ \begin{cases} k_1 = 0 \\ k_2 = 0 \\ \vdots \\ k_n = 0 \end{cases} $$
which clearly admits only the trivial solution.
Therefore, the vectors $\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \}$ are linearly independent.
Step 3: Conclusion
Since the set $\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \}$ both spans $\mathbb{R}^n$ and is linearly independent, we conclude that:
It forms a basis for $\mathbb{R}^n$.
Note: The vectors $\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_n \}$ - where $\vec{v}_1 = (1, 0, \dots, 0)$, $\vec{v}_2 = (0, 1, \dots, 0)$, ..., $\vec{v}_n = (0, 0, \dots, 1)$ - make up the standard basis of $\mathbb{R}^n$.
According to the basis dimension theorem, all bases of a vector space have the same cardinality - that is, the same number of vectors.
Since the standard basis contains $n$ vectors, it follows that every basis of $\mathbb{R}^n$ consists of exactly $n$ linearly independent vectors.
And so on.
