Suppose $T^{k+1}=0$. Show that $I + T + T^2 + \dots+ T^k = (I - T)^{-1}.$

Question

$T^{k+1} = 0$, where $k$ is a positive integer. Prove that $I + T + T^2 + \dots+ T^k = (I - T)^{-1}.$

My proof: Let $T$ be a $m \times n$ square matrix with the same number of $m$ rows and $n$ columns. $K$ must be greater than zero.

$$I = I^2 * I^{-1}$$ $$T = T^2 * T^{-1}$$ $$T^{k-1} = T^k * T^{-1}$$ $$I-T^{-1} = \frac{1}{I-T}$$

$$I + T + T^2 + \dots + T^k = (I-T)^{-1}$$

$$I + T + T^k + T^2 + \dots + T^{k-1} = (I-T)^{-1}$$

$$I + T + T^k + T^2 + \dots + T^kT^{-1} = (I-T)^{-1}$$

I'm a bit stuck. It seems like one of those $P*P^{-1}$ problems.

Is what I am doing overkill to the proof?

I need directions.

Have you tried to multiply your sum ($I+T+T^2+\cdots+T^{k}$) by $I-T$? See if you can simplify what you get... — universalset, Dec 11 '13 at 17:15

Cameron Buie · Answer 1 · 2013-12-12T19:20:43.347

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There's a serious mistake in your approach: namely, that there is any such thing as $T^{-1}$. There isn't, since $T$ is nilpotent, and so has $0$ as an eigenvalue (its only one, actually). Also, I think you mean $m\times m$ or $n\times n$.

Ultimately, you're making things too complicated. Call $M=I+T+T^2+\cdots+T^k.$ Note that $$TM=T+T^2+T^3+\cdots+T^{k+1}=T+T^2+\cdots+T^k=M-I,$$ by assumption, and so we can readily see that $$(I-T)M=IM-TM=M-(M-I)=I.$$ Hence, $M=(I-T)^{-1},$ and we're done.

edited Dec 12 '13 at 19:20

answered Dec 11 '13 at 17:28

Cameron Buie

102,994

how come you do not need to distribute the equation? – user983246 Dec 11 '13 at 17:41
We did distribute. I have rewritten it to make that clearer, though. – Cameron Buie Dec 11 '13 at 17:45

Christopher Liu · Accepted Answer · 2013-12-11T17:47:38.367

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$$ \begin{align*} &I + T + T^2 + \ldots + T^k = (I - T)^{-1}\\ \iff & (I+ T + T^2 + \ldots + T^k )(I - T) = (I - T)^{-1}(I-T) = I\\ \iff & (I+ T + T^2 + \ldots + T^k ) - (T + T^2 + \ldots + T^k + T^{k+1}) = I\\ \iff & I+ (T + T^2 + \ldots + T^k ) - (T + T^2 + \ldots + T^k) - T^{k+1} = I\\ \iff & I - T^{k+1} = I\\ \iff & I - [0] = I \end{align*} $$

edited Dec 11 '13 at 17:47

answered Dec 11 '13 at 17:21

Christopher Liu

1,319

should it T^k+1 be zero instead of T^k-1? – user983246 Dec 11 '13 at 17:26
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@user983246 oops, I already edited it for him while you were writing this comment. It should indeed be $k+1$. – rschwieb Dec 11 '13 at 17:29

score 1 · Answer 3 · answered Dec 11 '13 at 17:24

The identity for real numbers....

$$1+x+x^2+...+x^k=\frac{1-x^{k+1}}{1-x}$$

holds for matrices in the form

$$1+T+T^2+...+T^{k}=\left(1-T\right)^{-1}\,\left(1-T^{k+1}\right)$$. It is easy to show by multiplying both side by the same quantities

\begin{eqnarray} (1-T)\,(1+T+T^2+...+T^{k})&=&1-T+(1-T)\,T+...+(1-T)\,T^{k}\\ &=&1-T+T-T^2+T^2-...-T^{k+1}=1-T^{k+1} \end{eqnarray}

hence:

$$ (1+T+T^2+...+T^{k})=(1-T)^{-1}\,(1-T^{k+1}) $$

In your case $T^{k+1}=0$.

score 1 · Answer 4 · answered Dec 11 '13 at 17:26

1

Hint:

$$I + T + T^2 + ...+ T^k = \frac{1-T^{k+1}}{I-T}$$

answered Dec 11 '13 at 17:26

Salech Alhasov

6,780
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I prefer $(I-T)(I + T + T^2 + ...+ T^k) = I-T^{k+1}$, since otherwise, you have also to prove $I-T$ is regular. – Jean-Claude Arbaut Dec 11 '13 at 17:31

Suppose $T^{k+1}=0$. Show that $I + T + T^2 + \dots+ T^k = (I - T)^{-1}.$

4 Answers4