0
$\begingroup$

Is it the following statement correct?

Given three matrices $A, B, C \in \Bbb R^{N\times N}$, if $\def\rank{\operatorname{rank}}\rank(A)=N>\rank(B)$ (so that the columns of $A$ span the whole space $\Bbb R^{N}$ while those of $B$ just span a subspace of $\Bbb R^{N}$), and $\rank(A)=N=\rank(B+C)$, then the columns of $C$ span the orthogonal complement of the subspace spanned by those of $B$.

Edit: what if, in addition to $\rank(A)=N=\rank(B+C)$, we also impose A=B+C? Is it true in this case that the columns of $C$ span the orthogonal complement of the subspace spanned by those of $B$?

Thanks

$\endgroup$
2
$\begingroup$

It is false. Take, for instance, $N=2$, $A=\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$, $B=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$, and $C=\left[\begin{smallmatrix}0&1\\0&1\end{smallmatrix}\right]$.

$\endgroup$
3
  • $\begingroup$ Ok Josè thanks a lot. If we add the condition A=B+C, in addition to rank(A)=rank(B+C) does the statement hold? @JoséCarlosSantos $\endgroup$
    – JMallin
    Sep 18 '19 at 18:35
  • $\begingroup$ No; I've edited my answer. Now, $A=B+C$. $\endgroup$ Sep 18 '19 at 19:07
  • $\begingroup$ Ok thanks a lot $\endgroup$
    – JMallin
    Sep 18 '19 at 19:16

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.