# Dual Vector Space of linear functionals

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!

• $w_1$, $w_2$, 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? – Umberto P. Sep 17 '15 at 16:59
• 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 – Brutus Sep 17 '15 at 17:06

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 $\langle w_i,e_j\rangle =(Rw_i,e_j)=\delta_{ij}$

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: $$w_1(e_1)=1 \qquad w_2(e_2)=1 \qquad w_3(e_3)=1$$

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