# Unique Linear Functionals

• Sep 8th 2011, 01:20 PM
jnava
Unique Linear Functionals
I am working on the following...

Let 𝐵 = { ${\bf v}_1,{\bf v}_2,...,{\bf v}_n$} be a basis for vector space 𝔙 .
Show that there is a unique linear functional $v_\imath ^{\ast}$ on 𝔙, such that $v_\imath ^{\ast} ({\bf v}_\jmath)$ = $\delta_{\imath \jmath}$
Also, show that the set of 𝑛 distinct linear functionals on 𝔙 obtained form 𝐵, are linearly independent.

I have no idea on what to do, could someone help me out?

Thanks
• Sep 8th 2011, 02:47 PM
Drexel28
Re: Unique Linear Functionals
Quote:

Originally Posted by jnava
I am working on the following...

Let 𝐵 = { ${\bf v}_1,{\bf v}_2,...,{\bf v}_n$} be a basis for vector space 𝔙 .
Show that there is a unique linear functional $v_\imath ^{\ast}$ on 𝔙, such that $v_\imath ^{\ast} ({\bf v}_\jmath)$ = $\delta_{\imath \jmath}$
Also, show that the set of 𝑛 distinct linear functionals on 𝔙 obtained form 𝐵, are linearly independent.

I have no idea on what to do, could someone help me out?

Thanks

The basic idea is this. Suppose that $V$ is a $F$-space. You can completely specify a linear transformation from $V$ to any other $F$-space $W$ by demanding that $v_i\mapsto w_i$ where $\{v_1,\cdots,v_n\}$ is a basis for $V$ and the $w_i$ are just any vectors in $W$ (not necessarily different). How? Well, say you have made a choice about what the $v_i$ go to, you still haven't defined a transformation on $V$ itself. That said, if the map which takes $v_i\to w_i$ is to be a linear transformation you must take each $v=\alpha_1v_1+\cdots+\alpha_n v_n$ to $\alpha_1w_1+\cdots+\alpha_nw_n$. Thus, if $T:V\to W$ is linear and satisfies $T(v_i)=w_i$ then the function must be defined by the rule $T(v)=\alpha_1w_1+\cdots+\alpha_nw_n$ where $\alpha_1v_1+\cdots+\alpha_nv_n$ is the UNIQUE representation of $v$ as a linear combination of the basis $\{v_1,\cdots,v_n\}$. Conversely, you can check that the map defined that way does, in fact, satisfy the condition of being a linear transformation $V\to W$ with $v_i\mapsto w_i$. So, in your case you have that $W=F$ and $w_j=\delta_{i,j}\in F$

For the second part, what would happen if $\displaystyle \sum_{i=1}^{n}v^\ast_i=0(v)$ (where I put $0(v)$ to emphasize that it's the zero function)? What happens if you plug in $v_j$ for $j=1,\cdots,n$?
• Sep 8th 2011, 03:04 PM
jnava
Re: Unique Linear Functionals
Quote:

Originally Posted by Drexel28
The basic idea is this. Suppose that $V$ is a $F$-space. You can completely specify a linear transformation from $V$ to any other $F$-space $W$ by demanding that $v_i\mapsto w_i$ where $\{v_1,\cdots,v_n\}$ is a basis for $V$ and the $w_i$ are just any vectors in $W$ (not necessarily different). How? Well, say you have made a choice about what the $v_i$ go to, you still haven't defined a transformation on $V$ itself. That said, if the map which takes $v_i\to w_i$ is to be a linear transformation you must take each $v=\alpha_1v_1+\cdots+\alpha_n v_n$ to $\alpha_1w_1+\cdots+\alpha_nw_n$. Thus, if $T:V\to W$ is linear and satisfies $T(v_i)=w_i$ then the function must be defined by the rule $T(v)=\alpha_1w_1+\cdots+\alpha_nw_n$ where $\alpha_1v_1+\cdots+\alpha_nv_n$ is the UNIQUE representation of $v$ as a linear combination of the basis $\{v_1,\cdots,v_n\}$. Conversely, you can check that the map defined that way does, in fact, satisfy the condition of being a linear transformation $V\to W$ with $v_i\mapsto w_i$. So, in your case you have that $W=F$ and $w_j=\delta_{i,j}\in F$

For the second part, what would happen if $\displaystyle \sum_{i=1}^{n}v^\ast_i=0(v)$ (where I put $0(v)$ to emphasize that it's the zero function)? What happens if you plug in $v_j$ for $j=1,\cdots,n$?

when you plug back in $v_j$ you will get zero, meaning orthogonality, meaning linear independence?
• Sep 8th 2011, 04:21 PM
Drexel28
Re: Unique Linear Functionals
Quote:

Originally Posted by jnava
when you plug back in $v_j$ you will get zero, meaning orthogonality, meaning linear independence?

Uh, not sure what you mean. You get that $\alpha_j=0$, right?
• Sep 8th 2011, 04:38 PM
jnava
Re: Unique Linear Functionals
Quote:

Originally Posted by Drexel28
Uh, not sure what you mean. You get that $\alpha_j=0$, right?

Yes all $\alpha_j=0$ due to 0(v) correct? Since it will be zero, the the functionals are orthogonal implying linear dependence? This kind of math hurts my head way too much lol
• Sep 8th 2011, 05:54 PM
Drexel28
Re: Unique Linear Functionals
Quote:

Originally Posted by jnava
Yes all $\alpha_j=0$ due to 0(v) correct? Since it will be zero, the the functionals are orthogonal implying linear dependence? This kind of math hurts my head way too much lol

Haha, you have proven that all the $\alpha$'s are zero, which shows that they are linearly independent.
• Sep 8th 2011, 07:33 PM
jnava
Re: Unique Linear Functionals
Quote:

Originally Posted by Drexel28
Haha, you have proven that all the $\alpha$'s are zero, which shows that they are linearly independent.

Thank you for the help!