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In the book is the following formula:

Let $c$ and $d$ be $n \times 1$ nonzero columns such that $1+d^Tc \neq 0$, then $$(I + cd^T)^{-1} = I - \frac{cd^T}{1 + d^Tc}.$$

If $1+d^Tc \neq 0$, then $\big(I + cd^T\big)\big(I - \frac{cd^T}{1 + d^Tc}\big) = I$ so the equation is correct. But I am trying to derive the formula. I tried the solve the following equation $(I + cd^T)= (I + X)$, but without success.

How can I derive such formula? Thank you.

littleO
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Kapur
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  • $\big(I + cd^T\big)\big(I - \frac{cd^T}{1 + d^Tc}\big) = I$ is in fact a derivation. – user619894 Dec 11 '19 at 08:30
  • Ok differently. If someone ask me what is inverse of $\big(I + cd^T\big)$. How can I find the answer, if guessing is forbidden? – Kapur Dec 11 '19 at 08:47
  • ok, maybe try expanding formally – user619894 Dec 11 '19 at 12:20
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    To expand on what I think @user721481 is saying: Use the formal von Neumann series $(I + A)^{-1} = \sum_{n = 0}^\infty (-1)^n A^n$ where $A = cd^T$ in your case. Note that $(cd^T)^n = (d^Tc)^{n-1} cd^T$ (because $d^Tc$ is a scalar). Use that to factor out $-cd^T$ from all but the first term and you are left with $I$ minus $cd^T$ times the same series in parenthesis, except now for $A = d^Tc$ instead. This gives the formula. – Eike Schulte Dec 11 '19 at 12:45
  • You might ask about convergence but actually we don’t really care: We just use this to get an idea about what the result might be. Once we have it, it is easy to check that it is correct. – Eike Schulte Dec 11 '19 at 12:48
  • @EikeSchulte Yep it looks like a geometric series, indeed. – mathreadler Dec 11 '19 at 12:52
  • (Oh “Neumann series”, not “von Neumann series”. Different person.) – Eike Schulte Dec 11 '19 at 12:55
  • Yep Carl Neumann, not John von Neumann – mathreadler Dec 11 '19 at 15:20
  • Here is a derivation: https://math.stackexchange.com/questions/2202735/inverse-of-matrix-of-ones-ni/. Also https://math.stackexchange.com/questions/2300320/in-a-proof-of-the-sherman-morrison-formula-why-does-iwvt-1-i-fracwvt?noredirect=1&lq=1 – StubbornAtom Dec 11 '19 at 16:44

2 Answers2

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The series approach in the comments works but the following might be simpler: We start with your ansatz $$(I + cd^T)^{-1} = (I + X)$$ and compute $$ I = (I + cd^T)(I + X) = I + cd^T + X + cd^TX $$ but also $$ I = (I + X)(I + cd^T) = I + cd^T + X + X cd^T. $$ Comparing these, we find $$ Xcd^T = cd^T X. $$ So $X$ and $cd^T$ commute. Certainly all polynomials in $cd^T$ commute with $cd^T$ but let’s hope that we can make our life simpler and try it with the easiest kind of polynomial that might work first, i.e. with $X = a cd^T$ where $a$ is some scalar. Going back to the equation above (subtracting $I$ on both sides), we have $$ 0 = cd^T + acd^T + acd^Tcd^T.$$ Now we have to look closely and see that $d^Tc$ in the last term is a scalar as well and hence can be moved to the front (a very useful trick!): $$ 0 = cd^T + a cd^T + a(d^Tc) cd^T$$ Factoring out $cd^T$ on the right-hand side, we get a product of a matrix and a scalar. We know that the matrix is nonzero, so for the product to be zero the scalar has to be: $$0 = 1 + a + a(d^Tc)$$ Solving for $a$ results in $$a = \frac {-1} {1 + d^Tc}$$ if $1 + d^Tc \neq 0$.

Eike Schulte
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  • Nice! That's exactly what I was hinting for. – mathreadler Dec 11 '19 at 15:38
  • Thank you. I understand everythink except the sentence "Ceraintly all polynamials in $cd^T$ commute with $cd^T$...". What does mean that "polynomials in $cd^T$"? – Kapur Dec 11 '19 at 18:04
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    Whenever you have a matrix $A$ (with entries from some field $F$) and a polynomial $p(T) = a_n T^n + \cdots + a_0$ with coefficients $a_i \in F$ you can compute $p(A) = a_n A^n + \cdots + a_0 I$. This is also a matrix and by “a polynomial in $cd^T$” I mean a matrix of the form $p(cd^T)$ where $p$ is some polynomial. Matrices of the type $p(A)$ are the ones that “obviously” commute with $A$; so if you want to have two commuting matrices it is a good idea to first try if you can take one to be a polynomial in the other. – Eike Schulte Dec 12 '19 at 07:42
  • @Kapur By the way, if this answered your question it would be nice if you could mark it as answered by clicking the checkmark next to it. – Eike Schulte Dec 14 '19 at 11:30
  • @EikeSchulte As long as I do not completely understand your post I cannot accept your answer. I do not have time to read your answer now. But do not worry, I will acept your answer if I understand your additional explanation (see my history of posts, I do not have a problem with accepted answers for my questions). – Kapur Dec 14 '19 at 15:52
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There are formulas for that, but maybe you are not allowed to use them. This is a so called "rank 1 perculation". Also known as Sherman-Morrison formula.

$$({\bf A+uv}^T)^{-1} = {\bf A}^{-1} - \frac{{\bf A}^{-1}{\bf uv}^T{\bf A}^{-1}}{1+{\bf v}^T{\bf A}^{-1}{\bf u}}$$

Now set $\bf A = I$ and you can probably see what happens.

mathreadler
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  • Thank you, I want to use the equation above to derive Sherman-Morrison formula. So, I cannot use Sherman-Morisson formula. – Kapur Dec 11 '19 at 09:44
  • Ah okay, then I suppose it does not help so much. I will see if I can think of anything else. – mathreadler Dec 11 '19 at 09:49