2
$\begingroup$

So, I understand in general that if you have three linearly independent vectors in $\mathbb{R}^4$, they span a space in $\mathbb{R}^4$, not all of $\mathbb{R}^3$, because vectors with four entries live in $\mathbb{R}^4$ not $\mathbb{R}^3$. However, what if you have three linearly independent vectors like:

$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix} $

Do these span $\mathbb{R}^3$? It seems like they would occupy the same space as $\mathbb{R}^3$ no matter how many zeroes you add.

$\endgroup$
  • 5
    $\begingroup$ Only a subset of a space can span that space. So these don't span $\mathbb R^3$, instead the span the vector space $\{ (x,y,z,0)\}$, where $x,y,z \in \mathbb R$. As it turns out, this subspace of $\mathbb R^4$ is isomorphic to $\mathbb R^3$, but it is incorrect to call it $\mathbb R^3$ itself, although this is often done in textbooks, with the meaning to be implicitly understood. $\endgroup$ – астон вілла олоф мэллбэрг May 2 '17 at 0:50
  • 3
    $\begingroup$ They span a three dimensional subspace of $\Bbb R^4$ which is isomorphic to $\Bbb R^3$, however the space that they span contains no vectors from $\Bbb R^3$, those are all vectors which live in $\Bbb R^4$. Note that $\Bbb R^4\cap \Bbb R^3=\emptyset$ $\endgroup$ – JMoravitz May 2 '17 at 0:50

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.