I know that for two $n\times n$- matrices $A$ and $B$ we have $$\operatorname{rank}(A+B)=\dim\operatorname{Im}(A+B)\leq \operatorname{rank}(A) + \operatorname{rank}(B) - \dim \operatorname{Im}(A)\cap \operatorname{Im}(B).$$

But is there another relation between $\dim\operatorname{Im}(A+B)$ and $\dim \operatorname{Im}(A)\cap\operatorname{Im}(B)$? I'm looking for something that connects these two but that is rather an equality than an inequality.

  • $\begingroup$ By $A$ and $B$, you mean the representative matrices of two different linear transformations $T_A$ and $S_B$? $\endgroup$ – Itay4 Mar 12 '17 at 13:17
  • $\begingroup$ @Itay4 yes, that's what I meant. $\endgroup$ – user160919 Mar 12 '17 at 13:29

Consider a vector space $V$ of dimension $n$ (over a field of characteristic $\ne2$), with a basis $\{v_1,\dots,v_n\}$. For $1\le k\le n$, consider the maps $$ f_k\colon V\to V, \qquad f_k(v_i)=\begin{cases} v_i & \text{if $1\le i\le k$} \\[6px] 0 & \text{if $k<i\le n$} \end{cases} $$ and $$ g\colon V\to V, \qquad g(v_i)=\begin{cases} 0 & \text{if $i=1$} \\[6px] v_i & \text{if $1<i\le n$} \end{cases} $$ extended by linearity.

Observe that the rank of $f_k+g$ is $n$ for every $k$. Moreover \begin{gather} \dim(\operatorname{Im}f_1\cap\operatorname{Im}g)=0 \\ \dim(\operatorname{Im}f_2\cap\operatorname{Im}g)=1 \\ \dim(\operatorname{Im}f_3\cap\operatorname{Im}g)=2 \\ \vdots \end{gather}

With different choices of $g$ you can get the rank of the sum smaller than $n$.

This shows that you cannot establish a general relation between the rank of the sum and the dimension of the intersection of the images.


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