Let $v \neq 0$ be any vector in $\Bbb R^n$ and let, $U_v = \{x \in \Bbb R^n : \langle v,x \rangle \neq 1\}$. Then:
(i) Show that the matrix $I_n – v^t x$ is invertible $\forall x \in U_v$.
(ii) Compute the derivative of the map $f:U_v \to GL_n(\Bbb k)$ ($\Bbb k = \Bbb R \text{ or } \Bbb C$) defined by $$f(x)={(I-v^tx)}^{-1} .$$
My attempt:
(i) I have tried to do it using determinant but couldn't do it.
(ii) Observed $f=f_2 \circ f_1$, where $$f_1 : U_v \to GL_n(\Bbb R)$$ $$x \mapsto (I_n – v^t x)$$ and $$ f_2 : GL_n(\Bbb R) \to GL_n(\Bbb R)$$ $$ A \mapsto A^{-1}$$
Then derived, ${Df_1}_{(x)} (h)= -v^t h$ and ${Df_2}_{(A)}(H)= -A^{-1}HA^{-1} $.
Applied Chain Rule to obtain, $$Df_x (h)= {(I_n – v^tx)}^{-1} (v^t h){(I_n – v^tx)}^{-1}$$
I think I have done part (ii) only for $\Bbb k = \Bbb R$.
How to prove part (i) and if there is any mistake in my attempt in part $(ii)$, then please point it out. Thanks in advance for help!
Best Answer
(i) One option is to try and construct explicitly an inverse for $I_n - v^t x$. This is a modification of the identity matrix by a rank-$1$ matrix, so it's natural to guess that an inverse would have the same property: We thus write the candidate inverse as $I_n + w^t y$ for some vectors $w, y$ (all vectors here are row vectors). By definition, we have $$I_n = (I_n - v^t x)(I_n + w^t y) .$$ Multiplying both sides on the right by $y^T$ shows that $v \parallel w$, and instead multiplying both sides on the left by $w$ shows that $x \parallel y$, so the inverse must have the form $$(I_n - v^t x)^{-1} = I_n + \lambda v^t x$$ for some $\lambda$. Using again the definition of matrix inverse and solving gives $\lambda$. (Note that the expression is undefined when $v x^t = 1$, which hints at why that condition was necessary.)
Remark This gives a special case of the Sherman-Morrison Formula, itself a special case of the Woodbury Matrix Identity.
Alternatively, one can compute the determinant of $I_n - v^t x$ using the Matrix Determinant Lemma, which gives the determinant of a rank-$1$ update of a matrix. An easy proof for the case here (where the starting matrix is $I_n$) is given here.
(ii) Your solution looks good to me.