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.

  • 7
    $\begingroup$ a basis has finitely many subsets, but a vector space usually has infinitely many subspaces $\endgroup$
    – mercio
    Commented Jun 30, 2013 at 9:38
  • 1
    $\begingroup$ this argument is wonderful! $\endgroup$
    – Ant
    Commented Jun 30, 2013 at 12:36
  • 1
    $\begingroup$ 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
    Commented Jun 10, 2020 at 2:30
  • $\begingroup$ (In previous comment, you must start from a spanning set for W.) $\endgroup$
    – roy smith
    Commented Jun 10, 2020 at 2:51

2 Answers 2


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 $W$.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .