True or false? $ \dim_{\mathbb{K}}(U) - \dim_{\mathbb{K}}(\bigcap_{\varphi\in U}$Kernel$ (\varphi)) = 6$.

Question

How can I check if this statement is true or false?

Does there exist a subspace $ U$ of $ (\mathbb{K}^{15})^*$, so that $ \dim_{\mathbb{K}}(U) - \dim_{\mathbb{K}}(\bigcap_{\varphi\in U}$Kernel$ (\varphi)) = 6$.

Answer 1

I believe this is false. Hopefully I have not blundered, here's my reasoning:

Let $V=k^{15}$. Any $U$ subspace of $V^*$ is isomorphic to a the image of a linear operator $T:V \to V$. Think about the map $T(\vec{v})=\begin{bmatrix}\varphi_1(\vec{v})\\ \varphi_2(\vec{v})\\ \vdots \\ \varphi_{15}(\vec{v})\end{bmatrix}$.

Now the rank of $T$ is equal to $\dim \left< \varphi_i \mid i=1,\ldots, 15 \right>=\dim U$

Also the nullity of $T$ is $\dim\ker T=\dim (\cap \ker \varphi_i)$

If the statement you said is true then

$\textrm{Rank}(T)-\dim\ker(T)=6$

add $2\dim\ker T$ to both sides

$\textrm{Rank}(T)+\dim\ker(T)=6+2\dim\ker T$

Then the left is equal to 15 by the Rank-Nulity theorem so you have

$6+2\dim\ker T=15$ or $\dim\ker T=\frac{9}{2}$ which is not possible