If $B$ is nilpotent and $B^{k} = 0$ (and B is square), how should I go around proving that $I + B $ is invertible? I tried searching for a formula - $I = (I + B^{k}) = (I + B)(???)$
But I didn't get anywhere :(
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4 
- 1$\begingroup$ Expand $1/(I+B)$ using the formula for a geometric series. $\endgroup$Andrew Dudzik– Andrew Dudzik2015-11-23 17:08:41 +00:00Commented Nov 23, 2015 at 17:08
- 6$\begingroup$ $$I+B^k = (I+B)(I-B + B^2 - B^3 + \dots +(-1)^{k-1}B^{k-1})$$ $\endgroup$Crostul– Crostul2015-11-23 17:09:03 +00:00Commented Nov 23, 2015 at 17:09
- 4$\begingroup$ See math.stackexchange.com/questions/352383/… and math.stackexchange.com/questions/140348/… $\endgroup$Martin Sleziak– Martin Sleziak2015-11-23 17:57:20 +00:00Commented Nov 23, 2015 at 17:57
- $\begingroup$ @Crostul: $I+B^k$, or $I-B^k$? $\endgroup$Brian Tung– Brian Tung2015-11-23 21:26:37 +00:00Commented Nov 23, 2015 at 21:26
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4 Answers
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5 Heuristically, "expand" \begin{align*} \frac{I}{I+B}&=I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}+(-1)^kB^k+\cdots\\ &=I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}. \end{align*} To have a rigorous solution, verify directly that $$ (I+B)(I-B+B^2+\cdots+(-1)^{k-1}B^{k-1})=I. $$
- $\begingroup$ This is not true for $(1+x^{4})$ which is not equal to $(1+x)(1-x+x^2-x^3)$ $\endgroup$Lisa– Lisa2015-11-23 18:05:29 +00:00Commented Nov 23, 2015 at 18:05
- 5$\begingroup$ @Lisa This answer doesn't claim that it is. But $(1+x)(1-x+x^2-x^3) = 1-x^4$. $\endgroup$Andrew Dudzik– Andrew Dudzik2015-11-23 18:22:29 +00:00Commented Nov 23, 2015 at 18:22
- $\begingroup$ So what does $(1+x^4)$ equal to? $\endgroup$Lisa– Lisa2015-11-23 19:59:54 +00:00Commented Nov 23, 2015 at 19:59
- $\begingroup$ @Lisa: $1+x^4 = (1-\sqrt{2}x+x^2)(1+\sqrt{2}+x^2)$. But what is being used here is the idea that if $B^k = 0$, then obviously $I-B^k = I$. But $I-B^k$ can be factored as $(I+B)(I-B+B^2+\cdots+(-1)^{k-1}B^{k-1})$. Therefore $(I+B)(I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}) = I$, and thus $(I+B)^{-1}$ exists and is equal to $I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}$. $\endgroup$Brian Tung– Brian Tung2015-11-23 21:24:34 +00:00Commented Nov 23, 2015 at 21:24
- 1$\begingroup$ @KimJongUn: I think it may be better to use $I$ in the numerator of your first fraction, perhaps? $\endgroup$Brian Tung– Brian Tung2015-11-23 21:25:27 +00:00Commented Nov 23, 2015 at 21:25
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If $B$ is nilpotent then its only eigenvalue is $0$. You can see this by observing that if $(\lambda,x)$ is any eigenpair of $B$, then $0 = B^k x = \lambda^k x \implies \lambda^k = 0 \implies \lambda = 0$. Thus, $\lambda = 1$ is the only eigenvalue of $I + B$ and so $I + B$ is invertible.
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Essentially if $B$ is idempotent of order $n$ you can find a basis where
$$B=\left(\begin{array}{ccccc} 0 & b_{12} & 0 & \cdots & 0\\ 0 & 0 & b_{23} & & 0\\ 0 & 0 & 0 & & 0\\ \vdots & & & \ddots & b_{n-1n}\\ 0 & 0 & 0 & \cdots & 0 \end{array}\right)$$
In this form you can easly see what's happening when you calculate $B^k$. For example squaring $B$ we get: $$B^2=\left(\begin{array}{ccccc} 0 &0 & b_{12}b_{23} & 0 & \cdots & 0\\ 0 &0 & 0 & b_{23}b_{34} & & 0\\ 0 &0 & 0 & 0 & \ddots & \vdots \\ 0 &0 & 0 & 0 & \ddots & b_{n-2n-1}b_{n-1n}\\ \vdots&\vdots & & & \ddots &0 \\ 0 &0 & 0 & 0 & \cdots & 0 \end{array}\right)$$ Everytime you multiply for $B$ you shift the non zero columns and rows by one. You can fill out the details yourself, (for example treating the general case of $B$ idempotent of order k, doesn't change much) this was just to give you an useful picture of what's happening.... Clearly in this base $$I+B=\left(\begin{array}{ccccc} 1 & b_{12} & 0 & \cdots & 0\\ 0 & 1 & b_{23} & & 0\\ 0 & 0 & 1 & & 0\\ \vdots & & & \ddots & b_{n-1n}\\ 0 & 0 & 0 & \cdots & 1 \end{array}\right)$$ and $det(I+B)=1$ and so it's clearly invertible.
In the general case you can always write $I+B$ as un upper triangular matrix with $1$ on the diagonal and so with $det(I+B)=1$
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3 Factor $I-B^k$ and $I-B$ is one factor, but it also equals $I$ since $B^k=0$.
- $\begingroup$ +1. A little terse, perhaps, but it doesn't deserve a downvote. Maybe you should expand "it" as $I-B^k$? $\endgroup$Brian Tung– Brian Tung2015-11-23 21:26:13 +00:00Commented Nov 23, 2015 at 21:26
- $\begingroup$ @BrianTung Thank you Brian, I guess the complain of the downvoters is that I factored out $I-B$ and not $I+B$ but I meant it more as a hint than a complete answer. Appreciate the upvote :-) $\endgroup$Gregory Grant– Gregory Grant2015-11-23 21:28:05 +00:00Commented Nov 23, 2015 at 21:28
- $\begingroup$ Ahh, yes, well, I would probably change that to $I+B$, myself. Do as you will, of course. $\endgroup$Brian Tung– Brian Tung2015-11-23 21:30:16 +00:00Commented Nov 23, 2015 at 21:30