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Could you guys give me some hints on this homework?

Find inverse of $\mathbf{I} + \mathbf{ab}^\intercal$. Hint: try to form $c\mathbf{I} + d\mathbf{ab}^\intercal$ and solve for $c,d$. What happens when $\mathbf{b}^\intercal \mathbf{a} = -1$?

Attempt:

Not quite sure what the hint is trying to say. Perhaps Inverse is of the form $c\mathbf{I}+d\mathbf{ab}^\intercal$.

$\begin{align*} (\mathbf{I}+\mathbf{ab}^\intercal)(c\mathbf{I}+d\mathbf{ab}^\intercal) &= \mathbf{I}\\ c\mathbf{I}+(c+d)\mathbf{ab}^\intercal + d(\mathbf{ab}^\intercal)(\mathbf{ab}^\intercal) &= \mathbf{I} \end{align*}$

if $\mathbf{b}^\intercal\mathbf{a} = -1$ then

$\begin{align*} c\mathbf{I} + c\mathbf{ab}^\intercal &= \mathbf{I} \end{align*}$

This doesn't look like I am going down the right path.

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First notice that $b^Ta$ is a scalar so $$ab^Tab^T=(b^Ta)ab^T$$ and then we find

$$cI+(c+d(1+b^Ta))ab^T=I$$ hence in the case $1+b^Ta\ne0$ we choose $c=1$ and $d=\frac{-1}{1+b^Ta}$ to find the desired inverse.

In the case $b^Ta=-1$ then $$(I+ab^T)a=a-a=0$$ so $a$ is an eigenvector associated to the eigenvalue $0$ hence $I+ab^T$ isn't invertible.

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  • $\begingroup$ ah thanks, I guess I was on the right track $\endgroup$ – bdeonovic Sep 6 '14 at 16:45
  • $\begingroup$ Yes you were! You're welcome. $\endgroup$ – user63181 Sep 6 '14 at 16:48

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