Rank-one update of singular matrix

Question

My problem is the following:

Given a column vector $\mathbf{e''}\in\mathbb{R}^{3}$ with $||\mathbf{e''}||=1$ and a $3\times 3$ matrix $B$ such that $rank\{B\}=3$, I would like to find a vector $\mathbf{v}\in\mathbb{R}^3$ garantying that $rank\{A(\mathbf{v})\}=3$, where $A(\mathbf{v})=(\mathbf{e''}\mathbf{e''}^T-I_{3\times 3})B+\mathbf{e''}\mathbf{v}^T$.

One approach would be to try to maximize $|det\{A(\mathbf{v})\}|$ with respect to $\mathbf{v}$, but I think this would involve non-linear optimization. As my problem seems to relate to rank-one updates, I would rather find a closed-form formula if it exists. I found this thread, which is quite close but only provides a solution to rank-one updates of non-singular matrices. This is not the case here as $rank\{(\mathbf{e''}\mathbf{e''}^T-I_{3\times 3})B\}\leq rank\{\mathbf{e''}\mathbf{e''}^T-I_{3\times 3}\}=2$.

Can someone think of a way to solve this ?

Thanks

copper.hat · Accepted Answer · 2013-11-29T09:55:43.250

0

Write $A(v) = (w w^T -I)B+w v^T = ((w w^T -I)+w v^T B^{-1}) B = (w w^T -I + w (B^{-T}v)^T)B$.

Then if we choose $B^{-T}v = -w$, or $v = - B^T w$, we get $A(-B^Tw) = -B$.

edited Nov 29 '13 at 09:55

answered Nov 29 '13 at 09:01

copper.hat

172,524

Nice idea. However, there's a typo: $A(-B^Tw)=-B$, not $-I$, hence B would need to be invertible for A to be of rank 3. That should work for me though, thanks! – BConic Nov 29 '13 at 09:23
@AldurDisciple: Good catch! Thanks. – copper.hat Nov 29 '13 at 09:56

Rank-one update of singular matrix

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