Dimension of range($\Gamma$) where $\Gamma$ is the linear map from $V'\to\mathbb{F}$ defined $\Gamma(\varphi)=(\varphi(v_1),...,\varphi(v_m))$?

Question

Suppose $V$ is finite-dimensional and $v_1,...,v_m\in V$. Define a linear map $\Gamma: V'\to \mathbb{F}^m$ by

$$\Gamma(\varphi)=(\varphi(v_1),...,\varphi(v_m))\tag{1}$$

where $V'$ is the dual map of $V$. That is $V'=L(V,\mathbb{F})$, the linear space of all linear maps from $V$ to $\mathbb{F}$.

My question is what is the dimension of $\mathrm{range}(\Gamma)$?

Here is my attempt at answering this question.

Assume $\varphi\in \text{null }\Gamma$.

Then $\Gamma\varphi=(0,0,...,0)$, and so $\varphi v_i=0, i=1,...,m$.

Now, $v_1, ..., v_m$ may be linearly dependent or independent. In either case we can reduce this set of vectors to a linearly independent set, say, $v_1, ..., v_n$.

The $m-n$ vectors $v_{n+1},...,v_{m}$ can be written as linear combinations of the linearly independent set $v_1,...,v_n$.

Thus,

$$\Gamma\varphi=\left (\varphi v_1,...,\varphi v_n,\varphi_{n+1}, \sum_{i=1}^{n}a_{n+1,i}\varphi v_i, \sum_{i=1}^{n}a_{m,i}\varphi v_i\right )$$

and the nullspace of $\Gamma$ is formed by linear functionals $\varphi$ such that

$$\varphi v_i=0,\ \ i=1,...,n$$

That is, $n\leq\text{ dim null }\varphi \leq \text{dim } V$.

There are two possible cases.

In the first case, dim $V=n$, in which case, dim null $\varphi=n$ and so dim range $\varphi=0$.

$\varphi$ is the zero element of $V'$, and we conclude that dim range $\Gamma=$ dim $V'=n$.

Intuitively, when defining a $\varphi\in V'$, we can specify $\varphi v_i$ for each $i=1,...,n$ independently.

In the second case, we have dim $V>n$.

$$\text{dim range } \varphi=\text{ dim } V - \text{ dim null } \varphi$$

then we have linear functionals $\varphi$ that have range with dimension between 0 and $\text{dim }V-n$ (while having null space with dimension between dim $V$ and $n$).

Extend $v_1, ..., v_n$ to a basis of $V$

$$v_1, ...,v_n, w_1, ...,w_m$$

The nullspace of $\Gamma$ is

$$\text{null }\Gamma=\{ \varphi\in V':\varphi v_i=0, i=1,...,n \}$$

Define for $j=1$ to $m$ the m linear functionals

$$\varphi_j(w_k)=\begin{cases} 1, \text{ if } k=j, \\ 0, \text{ if } k\neq j\end{cases}$$

$$\varphi_j(v_i)=0, \text{ for } i=1,...,n$$

Then $\varphi_j\in\text{ null }\Gamma$.

If we show that this set of vectors is linearly independent and spans null $\Gamma$, then we can conclude that the set is a basis for null $\Gamma$ which thus has dimension $m$.

From this we have

$$\text{dim range } \Gamma=\text{ dim }V'-\text{ dim null }\Gamma$$

$$=(n+m)-m=n$$

To show this, assume

$$\sum_{i=1}^m a_i\varphi_i=0$$

In particular, this means that for the vectors $w_j, j=1,...,m$ we have

$$\sum_{i=1}^m a_i\varphi_i w_j = a_j, j=1,...,n$$

and so $a_i=0$ for $i=1,...,m$.

That is, the functions $\varphi_i, i=1,...,m$ are linearly independent.

Now, let $\varphi\in\text{ null }\Gamma$. Let $v\in V$. Then,

$$\varphi v=\varphi\left ( \sum_{i=1}^n a_iv_i+ \sum_{i=1}^m b_i w_i \right )$$

$$=\varphi\left (\sum_{i=1}^m b_iw_i\right )=\sum_{i=1}^mb_i\varphi w_i$$

$$=\sum_{i=1}^m (\varphi_i v) \varphi w_i$$

$$=\left (\sum_{i=1}^m\varphi w_i \varphi_i\right )v$$

which I think (but I am not sure) proves that $\varphi$ is a linear combination of $\varphi_i$, $i=1,...,m$.

If this is true, then since the $\varphi_i$ are linearly independent they form a basis of $\text{ null }\Gamma$.

Answer 1

Your ideas are fine and can be reworded more shortly: let $$W:=\operatorname{span}(v_1,\dots,v_m)$$ and denote by $\pi:V\to V/W$ the canonical projection. The map $$(V/W)^*\to\operatorname{null}\Gamma,\quad\psi\mapsto\psi\circ\pi$$ is an isomorphism, hence $$\begin{align}\dim(\operatorname{null}\Gamma)&=\dim((V/W)^*)\\&=\dim(V/W)\\&=\dim(V)-\dim(W),\end{align}$$ and therefore $$\dim(\operatorname{range}\Gamma)=\dim(W).$$

Answer 2

I also tried to provide a proof and hope the professors at seat can pose some suggestions.

Supposing $\dim V =n$,since $\Gamma$ is a isomorphism,then $Im \Gamma\simeq V'/Ker \Gamma$,wich implies that $$ rank(\Gamma)=\dim (Im\Gamma) =\dim (V'/Ker \Gamma)=\dim V'-\dim Ker \Gamma $$ Hence,we transform the problem to consider the dimension of $Ker \Gamma$.Note that $$ \begin{aligned} \varphi\in Ker\Gamma&\Longleftrightarrow \varphi(v_1)=...=\varphi(v_m)=0\\ &\Longleftrightarrow v_1,...,v_m\in Ker\varphi\\ &\Longleftrightarrow \left \langle v_1,...,v_m \right \rangle \subseteq Ker\varphi\\ \end{aligned} $$ We take a maximum linearly independent system W of $\left \{ v_1,...,v_m \right \} $.Without loss of generality, let $W=\left \{ v_1,...,v_r \right \}$.Now,we expand it to a base of $V$, $ v_1,...,v_r,v_{r+1},...,v_n $,and the set $\left \{ f_j \right \} $ of linear map is a basis of $V'$,where $f_j(v_i)=\delta_{ji}$.

On the one hand,we have $\left \langle f_{r+1},...,f_n \right \rangle \subseteq Ker\Gamma$.On the other hand,since we can write $\varphi\in Ker\Gamma$ as $\varphi=\sum_{i=1}^{n}k_if_i $,then we can obtain $k_1=...=k_r=0 $ by taking $\varphi(v_i)=0,i=1,...,r$.It follows that we can writes $\varphi$ as $\varphi=\sum_{i=r+1}^{n}k_if_i $. Hence we get $$ Ker\Gamma \subseteq \left \langle f_{r+1},...,f_n \right \rangle $$ Above all,we have $Ker\Gamma = \left \langle f_{r+1},...,f_n \right \rangle$. In the end, we have $ \dim (Ker\Gamma)=n-r$,which implies that $$ rank(\Gamma)=\dim V-\dim Ker \Gamma=n-(n-r)=r=\dim\left \langle v_1,...,v_m \right \rangle $$