Question: Prove that any system of generators of $\mathbb R^{n}$ contains a subset that is a basis of $\mathbb R^{n}$.

This is a question in one of my exam papers and it looks a lot like another question that I have seen. (Prove that any minimal set of generators of a linear system is a basis)

My Attempt

Let {$ \vec{e_{1}},\vec{e_{2}},...,\vec{e_{n}}$} be our generating system for $\mathbb R^{n}$ so that not all vectors are 0.

If this set is linearly independent then it will form a basis. If the set it not linearly independent then at least one vector can be written as a linear combination of the other.

Let's say $\vec{e_{n}}$ is a linear combination of the other vectors in the set, we can remove it

$span${$\vec{e_{1}},\vec{e_{2}},...,\vec{e_{n-1}}$} $= span${$\vec{e_{1}},\vec{e_{2}},...,\vec{e_{n}}$}

If this is linearly independent it's a basis. We can continue to remove vectors until we get a linearly independent set and that subset will be a basis of $\mathbb R^{n}$

Is this okay? Is it an acceptable proof for both of those questions and is there an easier way to prove this?

  • $\begingroup$ What do you know about the dimension of a space? Your proof starts off assuming the generating set has the same cardinality as the dimension of the space, which is a big assumption. $\endgroup$ – pjs36 Nov 15 '16 at 15:46
  • $\begingroup$ $dim (\mathbb R^{n}) = n $ $\endgroup$ – Patrick Moloney Nov 15 '16 at 15:52
  • $\begingroup$ Also since $\mathbb R^{n}$ is a finite-dimensional vector space, every basis of $\mathbb R^{n}$ has the same number of vectors. $\endgroup$ – Patrick Moloney Nov 15 '16 at 15:55

Your Answer

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

Browse other questions tagged or ask your own question.