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I know how to prove this inequality using bases for the row space, however, my professor presented a proof of \begin{align*} \operatorname*{rank}(A+B) \leq \operatorname*{rank}A + \operatorname*{rank}B \end{align*} using rank-nullity theorem. He fell sick so he just sent the notes, which I don't understand, can someone please help me. \begin{align*} n - \operatorname*{null}(A+B) &\leq 2n - \operatorname*{null}A -\operatorname*{null}B \\ \operatorname*{null}(A+B) &\geq -n + \operatorname*{null}A + \operatorname*{null}B \\ \operatorname*{null}A + \operatorname*{null}B &\leq n + \operatorname*{null}(A+B) \end{align*} Then it suffices to prove this inequality. \begin{align*} N(A) \cap N(B) \subseteq N(A+B) \end{align*} I understand why this is true, but I don't understand the next line \begin{align*} \operatorname*{null}A + \operatorname*{null}B \setminus (N(A) \cap N(B)) \leq n \end{align*} Not only does this not make sense from the sense of 'where does $n$ come from', it also, if I understand correctly, makes no sense from the mathematical standpoint. He probably meant $\dim(N(A) \cap N(B))$

Hence, \begin{align*} \operatorname*{null}A + \operatorname*{null}B \leq n + \dim(N(A) \cap N(B)) \leq n + {\rm null}(A+B) \end{align*} This last statement makes sense from the fact above that $N(A) \cap N(B) \subseteq N(A+B)$

If anyone knows how to prove this using rank-nullity theorem or understands what happened here, please let me know.

Rodrigo de Azevedo
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Maxim
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3 Answers3

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Assuming $A,B$ are linear maps defined on a vector space $E$ of dimension $n$ (this answers your 'where does $n$ come from', which you should have asked earlier in your post), your professor first used the rank-nullity theorem, to reduce the task to proving $$\operatorname*{null}A + \operatorname*{null}B\le n + \operatorname*{null}(A+B).$$

Then, he implicitely applied Grassmann's formula: for any two subspaces $F,G$ of $E,$ $$\dim(F)+\dim(G)-\dim(F\cap G)=\dim(F+G)\le n$$ to $F=N(A)$ and $G=N(B).$ This is why he (nearly) wrote $$\operatorname{null}(A)+\operatorname{null}(B)-\dim(N(A)\cap N(B))\le n.$$ So you rightly guessed his "$\setminus(N(A)\cap N(B))$" was a typo and meant "$-\dim(N(A)\cap N(B))$".

Since the rest was clear, you are done.

Anne Bauval
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For any two subspaces $A$ and $B$, $$\dim(A)+\dim(B)-\dim(A\cap B)=\dim(A+B)$$ Indeed, let $A∩B$ have basis $\{u_i\}$, A have basis $\{u_i\}∪\{a_i\}$, and B have basis $\{u_i\}∪\{b_i\}$. Suppose there is a nontrivial linear combination of these basis vectors with a nonzero coefficient for some $a_i$. Then there exists a vector $x$ in $B$ that can be expressed as a linear combination of $a_i$'s. However, $x$ is also contained in $A∩B$ and can thus be expressed as a linear combination of $u_i$'s, which contradicts the fact that $\{u_i\}∪\{a_i\}$ is a basis of $A$. Therefore,$ \{u_i\}∪\{a_i\}∪\{b_i\}$ is a basis of $A+B$ and the formula is proven.

Rodrigo de Azevedo
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LuckyJollyMoments
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Presumably $A,B$ are defined on a linear space $V$ of dimension $n$. There seems to be a typo en the solution given to the OP or a transliteration. In any case here is a short proof of

\begin{align*} \operatorname*{null}A + \operatorname*{null}B \leq n + \dim(N(A) \cap N(B)) \leq n + {\rm null}(A+B) \end{align*}

where $\operatorname{null}(A):=\dim(N(A))$, $N(A):=\{x\in V: Ax=0\}$.

Let $A'$ be the restriction of $A$ to $N(B)$. Then $N(A')=N(A)\cap N(B)$ and so $$\dim(N(B))=\dim(N(A)\cap N(B))+\operatorname{rank}(A')$$

Hence \begin{align} \dim(N(A))+\big(\dim(N(B))-\dim(N(A)\cap N(B))\big)&=\dim(N(A))+\operatorname{rank}(A')\\ &\leq \dim(N(A))+\operatorname{rank}(A)=n \end{align}

Mittens
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