Show that $\ker(L_1)\cap \ker(L_2)\cap\ker(L_3)\neq \{0\}$

Question

Let $L_1:\mathbb{R}^{31}\to \mathbb{R}^{14},L_2:\mathbb{R}^{31}\to \mathbb{R}^{10},L_3:\mathbb{R}^{31}\to \mathbb{R}^{4}$ be three linear transformations. Show that $\ker(L_1)\cap \ker(L_2)\cap\ker(L_3)\neq \{0\}$

Suppose $\ker(L_1)\cap \ker(L_2)\cap\ker(L_3)= \{0\}$

then let $x\in \ker(L_1)\cap \ker(L_2)\cap\ker(L_3) $

ie. $x\in \ker(L_1),x \in \ker (L_2), x\in \ker (L_3)$

is i am in right way i missing something from here

Answer 1

Your argument isn't currently making sense -- the $x$ you chose is $0$ since you just finished assuming that $0$ was the only thing in the intersection, so you won't learn anything interesting by thinking about $x.$ Here's one way you could make the argument, although there are other equivalent ways to formulate the same idea.

Consider the map $L:=L_1 \oplus L_2 \oplus L_3: \mathbb{R}^{31} \to \mathbb{R}^{14} \oplus \mathbb{R}^{10} \oplus \mathbb{R}^4 \cong \mathbb{R}^{28}.$ It is easy to see that $\ker(L) = \ker(L_1) \cap \ker(L_2) \cap \ker(L_3).$ But $\dim \text{im}(L) \leq 28,$ so by Rank-Nullity, $\dim \ker(L) \geq 31-28 = 3,$ and in particular $\ker(L)$ is non-trivial.