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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 :(

Martin Sleziak
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Lisa
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4 Answers4

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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 $$ (I+B)(I-B+B^2+\cdots+(-1)^{k-1}B^{k-1})=I. $$

Kim Jong Un
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  • This is not true for $(1+x^{4})$ which is not equal to $(1+x)(1-x+x^2-x^3)$ – Lisa Nov 23 '15 at 18:05
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    @Lisa This answer doesn't claim that it is. But $(1+x)(1-x+x^2-x^3) = 1-x^4$. – Andrew Dudzik Nov 23 '15 at 18:22
  • So what does $(1+x^4)$ equal to? – Lisa Nov 23 '15 at 19:59
  • @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 $I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}$. – Brian Tung Nov 23 '15 at 21:24
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    @KimJongUn: I think it may be better to use $I$ in the numerator of your first fraction, perhaps? – Brian Tung Nov 23 '15 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.

K. Miller
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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 $det(I+B)=1$

Picard
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Factor $I-B^k$ and $I-B$ is one factor, but it also equals $I$ since $B^k=0$.

Gregory Grant
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  • +1. A little terse, perhaps, but it doesn't deserve a downvote. Maybe you should expand "it" as $I-B^k$? – Brian Tung Nov 23 '15 at 21:26
  • @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 :-) – Gregory Grant Nov 23 '15 at 21:28
  • Ahh, yes, well, I would probably change that to $I+B$, myself. Do as you will, of course. – Brian Tung Nov 23 '15 at 21:30