Let \(A\) and \(B\) be linear transformations on a finite-dimensional vector space \(V\). Prove that \[ \dim\ker(AB) \le \dim\ker A+\dim\ker B. \]
Solution
1. The restriction\[ B|_{\ker(AB)}: \ker(AB)\longrightarrow V \]has image contained in\[ \ker A. \]
If \[ x\in\ker(AB), \] then \[ A(Bx) = (AB)x =0. \] Hence \(Bx\in\ker A\).
2. The kernel of the restricted map in step 1 is exactly\[ \ker B. \]
Certainly \[ \ker B\subseteq\ker(AB), \] because \(Bx=0\) implies \(ABx=0\). Therefore \[ \ker\bigl(B|_{\ker(AB)}\bigr) = \{x\in\ker(AB):Bx=0\} = \ker B. \]
3. Rank–nullity for the restricted map gives\[ \dim\ker(AB) = \dim\ker B + \dim B(\ker(AB)). \]
Apply the rank–nullity theorem to \[ B|_{\ker(AB)}. \] Step 2 identifies its kernel, while its image is \(B(\ker(AB))\).
4. One has\[ \dim B(\ker(AB)) \leq \dim\ker A. \]
Step 1 gives the inclusion \[ B(\ker(AB)) \subseteq \ker A. \] The dimension of a subspace cannot exceed the dimension of the containing space.
5. Therefore\[ \boxed{ \dim\ker(AB) \leq \dim\ker A+\dim\ker B }. \]
Substitute the estimate from step 4 into the equality from step 3.
6. Q.E.D.
Step 5 is the required inequality.