0
$\begingroup$

Am I correct in my understanding that given a basis $\left\{e_1,e_2,e_3\right\}$ for a vector space $V$, the corresponding basis elements $\left\{w_1,w_2,w_3\right\}$ of the dual space are orthogonal to the original basis? I.e., the plane $w_1$ is orthogonal to $e_1$, $w_2$ orthogonal to $e_2$, etc? I'm still trying to wrap my mind around the dual space. Thanks!

$\endgroup$
  • $\begingroup$ $w_1$, $w_2$, and $w_3$ are linear functionals on $V$ defined in a particular way. Their null spaces can be identified with planes, but they are not planes. At any rate, how are you defining orthogonal? $\endgroup$ – Umberto P. Sep 17 '15 at 16:59
  • $\begingroup$ Hmm I guess I'm not conceptualizing what the dual space is so maybe I'll review other questions first and rephrase. I was defining orthogonal as e1 dot w1 = 0 $\endgroup$ – Brutus Sep 17 '15 at 17:06
0
$\begingroup$

Maybe you have in mind the duality pairing between elements (functionals) in the dual space $V^*$ and elements in the original space $V$. For $w\in V^*$ and $e\in V$ it is denoted by $\langle w,e\rangle_{V^*\times V}$ or if there is no danger of confusion which is the space and its dual, just $\langle w,e\rangle$. This notion looks like the notion for inner product $(.,.)$. If the original space $V$ is a Hilbert space, then using Riesz'reprezentation theorem, you can indeed write the action of the functional $w$ on the element $e$ using inner product: $\langle w,e\rangle=(Rw,e)=(z,e)$ where $R: V^*\to V$ is the Riesz izomorphism and $Rw$ is the image of $w$ in the space $V$. In this case the property of the basis that $\langle w_i,e_j\rangle=\delta_{ij}$ can be transformd to orthogonality between the images of the $w_i,\,\,i=1,2,3,.. $ under thr Riesz' izomorphism $R$ and the basis of $V$ $e_j, \,\,j=1,2,3,..$, i.e $\langle w_i,e_j\rangle =(Rw_i,e_j)=\delta_{ij}$

$\endgroup$
0
$\begingroup$

The dual space $V^*$ of vector space $V$ is the space of linear functionals on $V$, so it is a different space as $V$, so you can not define an inner product between tvo different spaces and you can not define the notion of orthogonality between elements of these different spaces.. And note that a basis in $V*$ is such that ( using your notation): $w_i(e_j)=\delta_{ij}$ so: $$ w_1(e_1)=1 \qquad w_2(e_2)=1 \qquad w_3(e_3)=1 $$

$\endgroup$
  • $\begingroup$ Okay cool. What happens if we define an inner product between the two spaces? $\endgroup$ – Brutus Sep 17 '15 at 17:14
  • $\begingroup$ I am not sure you can define such a thing. $\endgroup$ – Tyler Hilton Sep 17 '15 at 17:15
  • $\begingroup$ I don't see how we can define such an ''inner product'' :) $\endgroup$ – Emilio Novati Sep 17 '15 at 17:15
  • $\begingroup$ Okay thanks! I'll keep reading $\endgroup$ – Brutus Sep 17 '15 at 17:16

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.