0
$\begingroup$

Let $V$ be an $n$-dimensional vector space. Then any linearly independent set of vectors $\{v_1, v_2, \ldots, v_n\}$ is a basis for $V$

Proof:

Let $w \in V$. Then $\{v_1, v_2,\ldots, v_n, w \}$ is linearly dependent. Then exists at least one $c_i \ne 0$ s.t. $\displaystyle{c_0w +\sum_{i = 1}^nc_iv_i} = \vec 0.$ If $c_0 = 0,$ then there's at least one $c_i \ne 0$ in the sum $\displaystyle{\sum_{i = 1}^nc_iv_i} = \vec 0.$ Contradiction as $\{v_1, \ldots, v_n\}$ are linearly independent. Hence $c_0 \ne 0$ meaning we can solve for $w: \ w= \displaystyle{\sum_{i = 1}^n\left(-\frac{c_i}{c_0}\right)v_i}.$ Thus $v_i$ span $V$.

We know $\{v_1, v_2,\ldots, v_n, w \}$ is linearly dependent because $\dim(V) = n$ and $v_i \in V$ meaning $v_i$ span $V$. The proof above shows that $\text{span}(v_1,\ldots,v_n) = V.$ Are we proving something we already know from the definition of basis? Is there subtlety here that I am missing?

$\endgroup$
0
$\begingroup$

The definition of basis is a set of vectors that is linearly indepedent and spans the space. This theorem states that in a finite-dimensional vector space, as long as your set has $n$-many vectors and it's linearly indepedent, then it also spans the space. Therefore it is a basis.

Also, we know $B=\{v_1, v_2,\ldots, v_n, w \}$ is linearly depedent because the length of any linearly indepedent list must be less than or equal to the length of a spanning set. Since $\dim V=n$, the spanning sets of $V$ have length $n$, thus since $|B| \not \leq n$, it must be linearly depedent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.