Showing that $\mathbb{C}^5/\ker(P)\simeq \mathbb{C}^2$

Question

Consider the vector space $\mathbb{C}^5$ and the map $P:\mathbb{C}^5\mapsto\mathbb{C}^5,\,P(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0,0)$ According to my lecture notes, this implies: $\mathbb{C}^5/\ker(P)\simeq \mathbb{C}^2$. But what exactly does this mean? That $\mathbb{C}^5\ni x\mapsto \mathbb{C}^5/\ker(P)$ is isomorphic to $\mathbb{C}^2$?

Answer 1

For a vector space $V$ and a subspace $U$ the quotient $V/U$ is defined as the set of equivalence classes of $v\sim w\iff v-w\in U$ equipped with the inherited operations. You can show that these are well-defined and $V/U$ is in fact a vector space itself. Elements of $V/U$ are commonly written as $[v]:=v+U$.

In your case $V=\Bbb C^5$ and $U=\ker P$ where $P$ is the projection $$P\colon\Bbb C^5\to\Bbb C^5,\,(z_1,z_2,z_3,z_4,z_5)\mapsto(z_1,z_2,0,0,0)$$ The kernel of $P$ is readily identified with $\Bbb C^3$. The isomorphism theorem (or also called rank-nullity theorem in case of vector spaces) yields that ${\rm im}\,P=\Bbb C^2$ and the desired isomorphism $\Bbb C^5/\ker P\cong\Bbb C^2$.