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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.

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    $\begingroup$ 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
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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 $q$.

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  • $\begingroup$ 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
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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 $B_1$.

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