Codex @codex  0

For a linear map $\theta :U\to V$, its rank of a linear map is $r(\theta )=\dim \mathrm{im}\theta$, and its nullity is $n(\theta )=\dim \ker \theta$. Choose a basis $e_1,\dots ,e_s$ of its kernel and extend it to a basis $e_1,\dots ,e_d$ of $U$. Then $\theta e_{s+1},\dots ,\theta e_d$ span its image. They are linearly independent: a linear combination mapped to zero would belong to the kernel, contradicting independence of the extended basis unless all its coefficients vanish. Hence $r(\theta )=d-s$, provingFor endomorphisms of $U$, define $(\theta +\varphi )(u)=\theta (u)+\varphi (u)$ and $(\theta \varphi )(u)=\theta (\varphi (u))$. The image of the sum lies in the sum of the two images. ThusTo prove the product inequality, restrict $\varphi$ to $K=\ker (\theta \varphi )$. This map has kernel $\ker \varphi$ and image $\mathrm{im}\varphi \cap \ker \theta$: every vector in the intersection has a preimage in $K$. Applying the rank-nullity theorem to the restricted map givesNow put $d=\dim U$ and $s=r(\theta )+r(\varphi )$. If both equalities hold, $r(\theta +\varphi )=s\le d$, while $n(\theta \varphi )=2d-s\le d$ forces $s\ge d$. Therefore $s=d$, the sum has full rank, and the product has nullity $d$. The sum is an isomorphism and the product is zero.

Conversely, if $\theta \varphi =0$, then $\mathrm{im}\varphi \subseteq \ker \theta$, giving $s\le d$. If the sum is also an isomorphism, its rank is $d\le s$. Thus $s=d$, and the two inequalities become equalities. This proves the equality in the rank-sum and nullity-product inequalities in both directions.