0
$\begingroup$

enter image description here I'm trying to solve parts b and c for this textbook practice problem and I can't seem to understand how to construct the basis for either of them. Would part b be a subspace because they are multiplying by vector v to get the zero vector?

$\endgroup$
  • $\begingroup$ Part b is indeed a subspace, because it is the null space of the linear map $v \mapsto v\cdot a$. This should also give you a hint regarding its dimension. $\endgroup$ – Bungo Oct 18 '17 at 17:11
  • $\begingroup$ Therefore the zero vector has dimension zero. Does that mean that the basis would just be an empty set for part B? $\endgroup$ – CluelessCoder Oct 18 '17 at 17:21
  • $\begingroup$ The dimension of the null space of the linear map $v \mapsto v \cdot a$ is $3$, not $0$. This is because the dimension of the image is $1$ (as a subspace of $\mathbb R$ it could only be $0$ or $1$; it's not $0$ because, for example, $a\cdot a \neq 0$). Then rank-nullity tells us that the null space must have dimension $4 - 1 = 3$. So, your basis should have three vectors. $\endgroup$ – Bungo Oct 18 '17 at 17:24
  • $\begingroup$ Please take the time to enter the key parts of your question instead of linking to/uploading an image so that the gist of your question shows up in summaries. You can find a quick reference on how to use MathJax to format your equations and mathematical expressions here. If you aren’t going to take the time to help others understand what your question is, why do you expect them to take their own time to help you? $\endgroup$ – amd Oct 18 '17 at 18:13

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.