13
$\begingroup$

I don't understand why this statement is wrong:

$V$ is a vector space, and $W$ is a subspace of $V$. $K$ is a basis of $V$. We can manage to find a subset of $K$ that will be a basis of $W$.

Sorry if my English is bad... and if you can show me an example of something that contradicts it, it'd be great.

$\endgroup$
4
  • 7
    $\begingroup$ a basis has finitely many subsets, but a vector space usually has infinitely many subspaces $\endgroup$
    – mercio
    Jun 30 '13 at 9:38
  • $\begingroup$ this argument is wonderful! $\endgroup$
    – Ant
    Jun 30 '13 at 12:36
  • $\begingroup$ There is however a way to find a basis for W from K. Namely, there is a subset L of K such that projection of V onto the subspace spanned by L restricts to an isomorphism on W, and then the preimage in W of the set L is a basis for W. There is an effective algorithm for this, called row reduced echelon form. $\endgroup$
    – roy smith
    Jun 10 '20 at 2:30
  • $\begingroup$ (In previous comment, you must start from a spanning set for W.) $\endgroup$
    – roy smith
    Jun 10 '20 at 2:51
18
$\begingroup$

Take for example $V = \mathbb{R}^2$, $W = \{(x,y)\in\mathbb{R}^2 : x = y\}$, and $K = \{(1,0),(0,1)\}$.

$W$ is a subspace of $V$, but there is no subset of $K$ that gives a basis for $W$.

$\endgroup$
0
10
$\begingroup$

A simple counter-example is $V=\mathbb R^2$, $W$ = the $y$-axis, and the basis of $V = \{ (1,0),(1,1)\}$.

Clearly neither of the two linearly independent vectors in this basis lie along the $y$-axis.

$\endgroup$

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.