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I have an equation like $$A = (B + mI)^{-1}D$$ $I$ = Identity matrix
$m$ = Scalar such that $m > 0$

Based on the answer in this

$$\frac{dA}{dm} = \frac{d}{dm} (B + mI)^{-1} D$$ $$\frac{dA}{dm} = -(B + mI)^{-1}\frac{d}{dm}(B + mI)(B + mI)^{-1} D$$

Is this differentiation correct and can it be further simplified(middle term)?
Also, does the negative sign in the differentiation term specifies that as $m$ increases $A$ decreases and vice versa?

Coolboy
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  • Your differentiation is correct, and the result can be simplified to $$\frac{\partial A}{\partial m}=-(B+mI)^{-1}A$$ Since $A$ is a matrix, the concept of increasing or decreasing doesn't really apply. – greg Mar 02 '22 at 18:04

1 Answers1

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Let $A(m) = (B+mI)^{-1}D$, since $\det$ and the entries of the adjugate are polynomials in $m$, it is clear that if $B+mI$ is invertible at $m$ then $A$ is a smooth function of $m$, so all that remains is to determine the formula.

We have $(B+mI) A(m) = D$, differentiating gives $I A(m) + (B+mI) A'(m) = 0$, or $A'(m) = - (B+mI)^{-1} A(m) = - (B+mI)^{-2} D$.

Since $A(m)$ is a matrix, it does not make sense to talk about $A$ increasing or decreasing.

Note:

I am not sure if this helps answer the increasing or decreasing question.

If you take $D=\operatorname{diag}(-1,1)$ and $B=-D$. Then $A(m) = \operatorname{diag}({1 \over 1+m}, {1 \over 1-m})$ and so $\|A(m)\|_2 = {1 \over 1-|m|}$ for $|m|<1$. That is, $m=0$ is a local minimum of $\|A\|_2$.

copper.hat
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  • This question is an extension of the original question of trying to understand the effect of change in m on A. That was the reason for my second question of "m increases A decreases". See this. Also, can you please write the answer without skipping any steps? I am not much familiar with the differentiation of matrices. – Coolboy Feb 27 '22 at 07:28
  • For finding the effect of m on A, is differentiation the right technique? – Coolboy Feb 27 '22 at 07:32
  • I am not sure what you mean by effect. To some extent the point is that the derivative gives an estimate $A(m+h) \approx A(m) + h A'(m)$. – copper.hat Feb 27 '22 at 19:15
  • You can see the effect on $A(m){ij}$ of perturbing $m$ by looking at ${\partial A(m){ij} \over \partial m}$. But it doesn't really make sense to talk about $A$ increasing or decreasing, it is an array of numbers. – copper.hat Feb 27 '22 at 19:17
  • My question relates to whats the interpretation of negative sign? This is just an abstract question which is part of a bigger question in which I am trying to prove if $m_1 > m_2$ then $||A_1|| < ||A_2||$ – Coolboy Feb 27 '22 at 22:54
  • I added a note which may clarify. – copper.hat Feb 27 '22 at 23:18
  • I am not sure I completely understand your note. But let me try to clarify what I want. I have a function $f(x,y,m) = x^2 + y^2 - m^2$. I want to prove that as $m$ increase $f(x,y,m)$ decreases. Now, $\frac{\partial f(x,y,m)}{\partial m} = -2m$. Can I say that as $m$ increases $f(x,y,m)$ decreases because I get negative derivative? – Coolboy Feb 27 '22 at 23:55
  • This seems to be straying from the question, but the sign of the derivative depends on the value of $m$. If $m>0$ then locally $f$ will decrease. My example shows that the norm of $A$ may have a local minimum, so no matter what you do to $m$ (locally) you cannot decrease the norm, so there is no (local) monotonicity. – copper.hat Feb 27 '22 at 23:57
  • Yes, I think that I forgot to mention this but $m > 0$. I'll add this constraint in the question as well. – Coolboy Feb 28 '22 at 00:00