0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.