# Linear dependency in dual space

Let $V$ be a finite $\mathbb{R}$-Vectorspace of dimension $n \in \mathbb{N}.$ Prove $$\varphi_1, ..., \varphi_n \in V^* \,\text{linearly independent} \Leftrightarrow \bigcap_{i=0}^{n} \text{ker}\,\varphi_i = 0$$

I already proved the implication '$\Rightarrow$' by contraposition:

$$\text{Let} \bigcap_{i=0}^{n} \text{ker}\,\varphi_i \neq 0.$$

Then there is a subvectorspace $U \subseteq V, U \neq 0$ with $\varphi_1, ..., \varphi_n \in U^0$. In particular $\text{span}(\varphi_1, ..., \varphi_n) \subseteq U^0$. We know:

$$m := \text{dim }U^0 = \underbrace{\text{dim }V}_{n} - \underbrace{\text{dim }U}_{\geq 1} < n$$

$\Rightarrow \varphi_1, ..., \varphi_n$ are $n$ vectors in the subvectorspace $U^0$ of dimension $m < n$ and thus linearly dependent. $\hspace{625px}\square$

Sadly I struggle to prove the other implication to complete the proof. Every hint or tip is welcome.

R$^2$ is 2 dimensional vector space over R.Projection onto first coordinate and projection onto second coordinate are linearly independent elements of dual space but intersection of their karnal is trivial.

• Sorry, I mixed up an important = and $\neq$. I edited my question. – user462931 Sep 12 '17 at 12:17

Suppose $\phi_1 = \sum_{i\geq 2} \mu_i \phi_i$. Then it follows that $\bigcap_{i\geq 2} ker(\phi_i) \subseteq ker(\phi_1)$. What does this say about the dimension of $\bigcap_{i\geq 2} ker(\phi_i) \cap ker(\phi_1)$?

Choose a basis of $V$ to be $\{e_{1},e_{2},...e_{n}\}$, Let $\{e_{1}^{*},e_{2}^{*},...,e_{n}^{*}\}$ be the dual basis in $V^{*}$.

$\forall 1\le{k}\le{n}$ we can write $$\varphi_{k}=\sum_{j=1}^{n}a_{jk}e_{j}^{*}$$ where $a_{jk}=\varphi_{k}(e_{j})$. Let matrix $A=\{a_{jk}\}_{1\le{j,k}\le{n}}$

Consider $$\sum_{k=1}^{n}c_{k}\varphi_{k}$$ and denote $C=(c_{1},c_{2},...,c_{n})^{T}$ and $X=(x_{1},x_{2},...,x_{n})^{T}$

$\forall \alpha\in{V}$ let $\alpha=\sum_{j=1}^{n}x_{j}e_{j}$ $$\sum_{k=1}^{n}c_{k}\varphi_{k}(\alpha)=\sum_{k=1}^{n}c_{k}\varphi_{k}(\sum_{j=1}^{n}x_{j}e_{j})=\sum_{k=1}^{n}\sum_{j=1}^{n}c_{k}x_{j}\varphi_{k}(e_{j})=\sum_{k=1}^{n}\sum_{j=1}^{n}c_{k}x_{j}a_{jk}=X^{T}AC$$

"$\Longrightarrow$" If $\{\varphi_{k}\}$ is linearly independent, then $\{\varphi_{1},\varphi_{2},...,\varphi_{n}\}$ is a basis of $V^{*}$ Since $A$ is the transformation matrix between $\{e_{1},e_{2},...e_{n}\}$ and $\{\varphi_{1},\varphi_{2},...,\varphi_{n}\}$, $A$ is invertible.

$\forall\alpha\in⋂_{i=0}^{n}ker\varphi_{i}$, (we write $\alpha=\sum_{j=1}^{n}x_{j}e_{j}$)

then $\forall 1\le{i}\le{n}$ $\varphi_{i}(\alpha)=0$ $$\forall C\in\mathbb{R}^{n} \sum_{k=1}^{n}c_{k}\varphi_{k}(\alpha)=0$$ Equivalently $$\forall C\in\mathbb{R}^{n} X^{T}AC=0$$ which leads to $X^{T}A=0$. And because $A$ is invertible, $X=0$ which means $\alpha=0$. So $⋂_{i=0}^{n}ker\varphi_{i}=0$

"$\Longleftarrow$" $⋂_{i=0}^{n}ker\varphi_{i}=0$

$\forall{0}\neq{X}\in\mathbb{R}^{n}$ $\alpha=\sum_{j=1}^{n}x_{j}e_{j}\neq0$. Because $⋂_{i=0}^{n}ker\varphi_{i}=0$ we have $$(\varphi_{1}(\alpha),\varphi_{2}(\alpha),...,\varphi_{n}(\alpha))\neq(0,0,...,0)........(1)$$

While $\varphi_{i}(\alpha)=X^{T}A_{i}$ where $A_{i}$ denotes the i-th column of $A$, we rewrite (1) to be $$X^{T}A\neq0;\forall{0}\neq{X}\in\mathbb{R}^{n}$$ Thus $$A^{T}X=0\iff{X=0}$$ Whic means $A$ is invertible. And thus $\{\varphi_{1},\varphi_{2},...,\varphi_{n}\}$ is a basis of $V^{*}$ and is independent.