maybe this is an idiot question, however I could not figure out how to solve it. Let $X =M_n(\mathbb{R})$ be the space of $n \times n$ matrix over the reals, then there exists two open neighborhoods of the identity, $U$ and $V$, such that the function $\phi: V \longrightarrow U$, $\phi(A) = \sqrt{A}$ is well defined and is differentiable at the identity $I$.Furthermore, what's $d\phi(I)(T)$ ? I was thinking in inverting the matrix $A = I - B$ by the usual $\sum_i B^{i}$ and then, somehow, find the unique square root.

Thanks in advance.

  • 1
    $\begingroup$ Maybe you can try inverse function theorem? The derivative of the squaring operator should be easier to compute. $\endgroup$ – ronno Dec 6 '13 at 2:12
  • $\begingroup$ @ronno Oh, you're right! I'm so dumb. I just have to use $f(A) = A^2$, I thought this question was harder. $\endgroup$ – user40276 Dec 6 '13 at 2:19

This looks like a straightforward application of inverse function theorem. Consider $$\psi \colon M_n(\mathbb{R}) \to M_n(\mathbb{R}); A \mapsto A^2$$

Then $\psi$ is continuously differentiable everywhere with derivative $d\psi(A)(T) = AT+TA$. At $A = I$, this is $T \mapsto 2T$, which is invertible.

Now conclude by the inverse function theorem (statement taken from wikipedia):

If the total derivative of a continuously differentiable function $F$ defined from an open set $U$ of $\mathbb{R}^n$ into $\mathbb{R}^n$ is invertible at a point $p$ (i.e., the Jacobian determinant of $F$ at $p$ is non-zero), then $F$ is an invertible function near $p$. That is, an inverse function to $F$ exists in some neighborhood of $F(p)$. Moreover, the inverse function $F^{-1}$ is also continuously differentiable.

Further, $$J_{F^{-1}}(F(p)) = [J_F(p)]^{-1}$$ where $[\cdot]^{-1}$ denotes matrix inverse and $J_G(q)$ is the Jacobian matrix of the function $G$ at the point $q$.

  • $\begingroup$ Can you elaborate on this please? I am stuck at a similar question and found this post. Would the derivative of the square root operator be $(2T)^{-1}$ at a neighborhood of $I$? Where did $AT + TA$ come from? $\endgroup$ – user191919 Jan 31 '14 at 21:24
  • $\begingroup$ @user191919 The derivative at $A$ is $AT + TA$, which at $A = I$ simplifies to $2T$. But the inverse is of the linear map $T \mapsto 2T$, that is, $T \mapsto \frac12T$. That is, the derivative of this (local) square root at identity is $\frac12 T$. $\endgroup$ – ronno Feb 2 '14 at 3:51
  • $\begingroup$ But how do I compute the derivative? Why isnt the inverse ${1 \over 2}{T^{-1}}$? $\endgroup$ – user191919 Feb 2 '14 at 10:43
  • $\begingroup$ @user191919 Do you know how is the derivative of a bilinear form? Matrix multiplication is bilinear. Furthermore $(2T)^{-1} = (1/2) T^{-1}$. $\endgroup$ – user40276 Feb 2 '14 at 13:24
  • $\begingroup$ Nope... I tried applying the usual limits definition for a matrix norm, i.e.,$$lim_{\mathbb{||}T\mathbb{||}}{(A+T)^2-A \over \mathbb{||}T\mathbb{||}} = {AT + TA - T^2 \over \mathbb{||}T\mathbb{||}}$$. I can somewhat intuitively see that the derivative should be AT + TA by making analogy with the univariate real case, but I can't rigorously see why this would actually be the case. ps: I clearly have a gap of knowledge here, so if you prefer to indicate any reference that would be great already. thanks. $\endgroup$ – user191919 Feb 2 '14 at 13:57

Those answers don't look quite right. Let $A$ and $B=A^{1/2}$ be $n$ by $n$ matrix functions on $\mathbb{R}$ with derivatives $A_1, B_1.$ Since $A=B^2, A_1/2=BB_1+B_1B$. One has to solve this for $B_1$.


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.