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Consider the symmetric, positive definite matrix $\mathbf{A}$. I'd like to find a general form for

$$(\mathbf{I} + \mathbf{A})^{-1}$$

that only involves $\mathbf{A}^{-1}$, i.e., no other inverse appears in the solution (as, for instance, in the Woodbury matrix identity).

I've tried to derive the inverse by hand but I could only obtain a result up to he $4 \times 4$ case as follows.

  • $2 \times 2$:

$$(\mathbf{I} + \mathbf{A})^{-1} = \frac{\mathrm{det}(\mathbf{A}) \mathbf{A}^{-1} + \mathbf{I}}{\mathrm{det}(\mathbf{A}) + \mathrm{tr}(\mathbf{A}) + 1}$$

  • $3 \times 3$:

$$(\mathbf{I} + \mathbf{A})^{-1} = \frac{\mathrm{det}(\mathbf{A}) \mathbf{A}^{-1} - \mathbf{A} + \big( \mathrm{tr}(\mathbf{A}) + 1 \big) \mathbf{I}}{\mathrm{det}(\mathbf{A}) + \mathrm{det}(\mathbf{A}) \mathrm{tr}(\mathbf{A}^{-1}) + \mathrm{tr}(\mathbf{A}) + 1}$$

  • $4 \times 4$:

$$(\mathbf{I} + \mathbf{A})^{-1} = \frac{\mathrm{det}(\mathbf{A}) \mathbf{A}^{-1} + \mathbf{A}^2 - \mathrm{tr}(\mathbf{A}) \mathbf{A} - \mathbf{A} + \big( \tfrac{1}{2}(\mathrm{tr}(\mathbf{A})^2-\mathrm{tr}(\mathbf{A}^2)) + \mathrm{tr}(\mathbf{A}) + 1 \big) \mathbf{I}}{\mathrm{det}(\mathbf{A}) + \tfrac{1}{2}(\mathrm{tr}(\mathbf{A})^2-\mathrm{tr}(\mathbf{A}^2)) + \mathrm{det}(\mathbf{A}) \mathrm{tr}(\mathbf{A}^{-1}) + \mathrm{tr}(\mathbf{A}) + 1}$$

Can we find a general expression for higher dimensions that builds on the ones above? Even obtaining the $5 \times 5$ case is prohibitive, and so far I haven't been able to spot the pattern.

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  • $\begingroup$ I think the next case ($4\times 4$) is the crucial one. If you replace $A$ by $tA$ to keep track of the terms the middle term, the $t^2$ one of the denominator, is the trace of the matrix of $2\times 2$ cofactors, $\mathrm{tr}(A^{(2)})$ and I can't see how to get this in the form you want. $\endgroup$ Commented Mar 7, 2018 at 16:40
  • $\begingroup$ By the way I think that if you can do the case of $A$ diagonal you can do the general case by a change of basis. $\endgroup$ Commented Mar 7, 2018 at 16:41
  • $\begingroup$ Well, if $\mathbf{A}$ is diagonal, then the inverse is trivial. Can you please be more specific? $\endgroup$
    – TheDon
    Commented Mar 7, 2018 at 23:41
  • $\begingroup$ It is trivial when $A$ is diagonal to write the inverse in terms of the eigenvalues, but the trick is to write it in terms of $A$ etc. This is not so easy. $\endgroup$ Commented Mar 8, 2018 at 7:47
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    $\begingroup$ I still think that if one can get the $4\times 4$ case then one will be able to spot the pattern. I do not have a computer algebra package to check, but I think the answer is something like $I + (\mathrm{tr}(A)I-A) +(\mathrm{tr}({A^{(2)})} I -\mathrm{tr}(A)A +A^2)+\mathrm{det}(A)A^{-1}$. $\endgroup$ Commented Mar 8, 2018 at 10:30

2 Answers 2

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Let $p(x)=\sum_{k=0}^na_kx^k$ (with $a_n=1$) and $q(x)$ be the characteristic polynomials of $A$ and $B=I+A$ respectively. By Cayley-Hamilton theorem, $B^{-1}=g(B)$ where \begin{align} g(x) &=\frac{q(0)-q(x)}{xq(0)}\\ &= \frac{p(-1)-p(x-1)}{xp(-1)}\\ &=\frac{-\sum_{j=0}^n a_j[(x-1)^j - (-1)^j]}{xp(-1)}\\ &=\frac{-\sum_{j=1}^n a_j \sum_{k=0}^{j-1}(-1)^{j-1-k}(x-1)^k} {\sum_{j=0}^n (-1)^ja_j}\\ &=\frac{-\sum_{j=0}^{n-1} a_{j+1} \sum_{k=0}^j(-1)^{j-k}(x-1)^k}{\sum_{j=0}^n (-1)^ja_j}. \end{align} Therefore, $(I+A)^{-1}$ can be expressed in terms of $A$ by $f(A)$, where \begin{align} f(x) &=\frac{-\sum_{j=0}^{n-1} a_{j+1} \sum_{k=0}^j(-1)^{j-k}x^k} {\sum_{j=0}^n (-1)^ja_j}\\ &=\frac{-\sum_{k=0}^{n-1} x^k \sum_{j=k}^{n-1} (-1)^{j-k} a_{j+1}} {\sum_{j=0}^n (-1)^ja_j}\\ &=\frac{-\sum_{k=0}^{n-1} x^k \sum_{j=0}^{n-1-k} (-1)^j a_{j+k+1}} {\sum_{j=0}^n (-1)^ja_j}. \end{align} That is, $$ (I+A)^{-1}=\frac{-\sum_{k=0}^{n-1}\left[\sum_{j=0}^{n-1-k} (-1)^j a_{j+k+1}\right] A^k} {\sum_{j=0}^n (-1)^ja_j}.\tag{1} $$ It is well-known that each coefficient $a_i$ can be expressed in terms of the traces of the powers of $A$. More specifically, by Girard-Waring formula, if we define $s_k=\operatorname{tr}(A^k)$, then $$ a_{n-m} = \sum_{\substack{k_1+2k_2+\cdots+mk_m=m\\ k_1,k_2,\ldots,k_m\ge0}}(-1)^{k_1+k_2+\cdots+k_m}\frac{1}{k_1!k_2!\cdots k_m!}\left(\frac{s_1}1\right)^{k_1}\left(\frac{s_2}2\right)^{k_2}\cdots\left(\frac{s_m}m\right)^{k_m}.\tag{2} $$ In particular, when $n=5$, we have $a_5=1,\ a_0=-\det(A)$ and \begin{align} a_1 &=-\frac{s_4}4 + \frac{s_2^2}8 + \frac{s_1s_3}3 - \frac{s_1^2s_2}4 + \frac{s_1^4}{24},\\ a_2&= -\frac{s_3}3 + \frac{s_1s_2}2 - \frac{s_1^3}6,\\ a_3&= \frac{s_1^2-s_2}2,\\ a_4&= -s_1, \end{align} but surely, if $A^{-1}$ is also known, we would express $a_1$ as $\det(A)\operatorname{tr}(A^{-1})$ instead. Plug these into $(1)$, you can express $(I+A)^{-1}$ as a weighted sum of the powers of $A$ when $n=5$.

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  • $\begingroup$ Thank you for your answer, this is closer to what I was looking for. Is there a way to simplify the expression of the coefficients $a$ if $\mathbf{A}^{-1}$ is known? For what I see, we have the following (for any $n$): $a_{n} = 1$, $a_{n-1} = - \mathrm{tr} (\mathbf{A})$, $a_{n-2} = \tfrac{1}{2} \big( (\mathrm{tr} (\mathbf{A}))^{2} - \mathrm{tr} (\mathbf{A}^{2}) \big)$, $\ldots$, $a_{1} = \mathrm{det} (\mathbf{A}) \mathrm{tr} (\mathbf{A}^{-1})$, $a_{0} = (-1)^{n} \mathrm{det} (\mathbf{A})$. $\endgroup$
    – TheDon
    Commented Mar 13, 2018 at 16:22
  • $\begingroup$ @TheDon Not sure, but I don't think so. Note that $A^{-1}=\frac{-1}{a_0}\sum_{k=1}^na_kA^{k-1}$. If you subtract a multiple of $A^{-1}$ from $(1)$, there are still at least $n-1$ terms remain. So, you'll have a sum of $n$ terms anyway. $\endgroup$
    – user1551
    Commented Mar 13, 2018 at 17:15
  • $\begingroup$ Do you have any answer to this?: math.stackexchange.com/questions/3184171/… $\endgroup$ Commented Apr 11, 2019 at 19:51
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Let $A\in M_n(\mathbb{C})$, $spectrum(A)=(\lambda_i)$ and $B=I+A$ (we assume that $-1\notin spectrum(A)$); note that $spectrum(B)=(1+\lambda_i)$.

Proposition. $(I+A)^{-1}$ is a polynomial of degree $n-1$ in $A$, the coefficients of which, are rational functions of the $(trace(A^k))_{k\leq n}$.

Proof (for $n=5$). Let $(\sigma_k)$ be the elementary symmetric functions of the $(\lambda_i)$ and $(\tau_k)$ be the elementary symmetric functions of the $(1+\lambda_i)$. According to Hamilton-Cayley:

$B^{-1}=\dfrac{1}{\tau_5}(B^4-\tau_1B^3+\tau_2B^2-\tau_3B+\tau_4I_5)$.

Each $B^k$ is a polynomial of degree $k$ in $A$.

On the other hand, each $\tau_k$ is a symmetric function of the $(\lambda_i)$, then a polynomial in $\sigma_1,\cdots,\sigma_k$. Since each $\sigma_k$ is a polynomial in $trace(A),\cdots,trace(A^k)$ (use the Newton's formulae), $\tau_k$ is a polynomial in $trace(A),\cdots,trace(A^k)$.

Finally, $(I+A)^{-1}$ is a polynomial in $A$, the coefficients of which, are rational functions of the $(trace(A^k))_{k\leq 5}$.

Remark. Of course, we may replace the characteristic polynomial of $A$ with its minimal polynomial.

EDIT. Answer to @ancientmathematician ; cf. below a procedure (in Maple) that works for every dimension $m$ (I use the Bell function).

***Decomposition of $(I+A)^{-1}$; the result is in "res"; $a$ denotes $A\in M_m$ and $t[i]=t_i$ denotes $Trace(A^i)$.

restart;

with(LinearAlgebra);

m:=5:

TAU:=Vector(m): for n from 1 to m do U:=Matrix(n,n): for i to n-1 do U[i+1, i] := -1 end do; S := Vector(n, symbol = s); X := Vector(n, symbol = x); for i to n do x[i] := -factorial(i-1)*s[i] end do;

for i to n-1 do for k to i do U[i+1-k, i] := x[k]*binomial(n-i-1+k, k-1) end do end do;

for i to n do U[i, n] := x[n-i+1] end do;

TAU[n] := (-1)^n*Determinant(U)/factorial(n);

T:=Vector(n,symbol=t):

for k from 1 to n do

rr:= (1+y)^(k): ss:=m:

for i from 1 to k do ss:=ss+coeff(rr,y,i)*t[i]: od: s[k]:=ss: od:

TAU[n]:=expand(TAU[n]): x:='x':s:='s':t:='t': od: ;

***$(I+A)^{-1}$ as a polynomial in $A$

res := (1+a)^(m-1);

for i from m-2 by -1 to 0 do res := res+(-1)^(i-m+1)TAU[m-i-1](1+a)^i end do;

res := (-1)^(m-1)*collect(expand(res), a)/TAU[m];

***VERIFICATION

A := RandomMatrix(m,m);

B := 1/(IdentityMatrix(m)+A);

a := A; for i to m do t[i] := Trace(A^i) end do;

simplify(res-B);

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    $\begingroup$ I agree with all this. But I am left with a nagging doubt about whether the final answer simplifies to something more recognisable. After all we've got the expansion of powers of $(I+A)$; the replacement of the $\tau_j$ by expressions in the $\sigma_j$ [which can also be expressed as $\chi_{A+I}(x)=\chi_{A}(x-1)$]; and the replacement of the elementary symmetric functions by the powers sum ones. $\endgroup$ Commented Mar 10, 2018 at 12:13
  • $\begingroup$ Thanks for your answer. Can you please clarify if there is a closed-form expression for the coefficients $\tau_{k}$? $\endgroup$
    – TheDon
    Commented Mar 12, 2018 at 10:10
  • $\begingroup$ Of course, no. I wrote the above procedure to show to @ancientmathematician (now he disappeared...) that the method works. In particular, for $n=10$ the time of calculation is $0"5$ but the result fills 2.5 Maple sheets. $\endgroup$
    – user91684
    Commented Mar 12, 2018 at 23:04
  • $\begingroup$ @loupblanc, I have no doubt the method works, I was about to post an account of it, but you beat me to it! But I am still not convinced there is not a way to express the answer in terms of the theory of symmetric functions. $\endgroup$ Commented Mar 13, 2018 at 7:52
  • $\begingroup$ Do you have any answer to this?: math.stackexchange.com/questions/3184171/… $\endgroup$ Commented Apr 11, 2019 at 19:52

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