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I think this proposition is right. If this is not right, could you provide a counter example? However, this is definitely right for the $\mathbb{R}^3 $ case. Here is how I proved it. Is it right and if so rigorous? Even if yes, are there more ways to prove this, I'm really curious. I put two ways, I'm not sure if either is right, or maybe one of them is rigorous and the other one is not. Could you please point out flaws in the proofs if the idea is right but it is not rigorously explained? I'm new to proofs and it's summer so I can't annoy my professors. Thanks.

$\\(I-A)^k=0 \\ det(I-A)^k=0 \Rightarrow det(I-A)=0 \\ \text{Therefore there exists a non-zero vector x, }\\ (2) \quad(I-A)x=Ix-Ax=0 \\ \text{There exists non trivial x s.t. } Ax=\lambda_{2}x \\ \text{So now } (I-A)x=x-\lambda_2x=(1-\lambda_2)x \\ \text{Multiplying both sides by (A-I) k times, we get } 0=(1-\lambda)^kx \\ \text{Since x is non-trivial, }1-\lambda=0 \Rightarrow \lambda=1 \\ \therefore \text{A has eigenvalue 1 of multiplicity k, and so A is invertible} $

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  • $\begingroup$ Not about your attempts, but you can binomial-expand $(I-A)^k$, take the $I$ to one side, and factorize $A$ out of the other side, so you instantly have an explicit inverse. For example, $(I-A)^2 = 0$ then $I^2 - 2AI+A^2 = 0$ so $I-2A+A^2 =0$ so $A^2+2A = I$ so $A(A+2) = (A+2)A = I$. $\endgroup$ Aug 2, 2021 at 1:21
  • $\begingroup$ Note that your attempt by taking the determinant is incorrect since you only show that at least one eigenvalue is 1, not that all eigenvalues are 1. $\endgroup$
    – Eric
    Aug 2, 2021 at 1:28
  • $\begingroup$ Your attempt assumes that $I-A=0$. That is not what you are given: you are given that there is a $k$ such that $(I-A)^k=0$. $\endgroup$ Aug 2, 2021 at 1:31
  • $\begingroup$ I don't assume that I-A=0. Do I assume it without noticing? I say that determinant(I-A) is true since $(I-A)^k=\prod_{i=1}^{\k} (I-A)$ and the determinant of the right is zero. Then (I-A)v=Lv, (I-A)(I-A)v=(I-A)Lv=L^2*v, you keep going until you have (I-A)^k=0=L^kv. So L must be 0. $\endgroup$
    – Goob
    Aug 2, 2021 at 2:00
  • $\begingroup$ Don't make the same comment in two different places. I've replied in your identical comment on my answer. You correctly conclude $\lambda=0$ is an eigenvalue of $I-A$, but then what you do with that is either pointless or unwarranted, unless you are asserting that $(I-A)x=0$ for all $x$. $\endgroup$ Aug 2, 2021 at 2:03

4 Answers 4

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Since $[A,I]=0$, we have

$$0=(I-A)^k=\sum_{i=0}^k\binom{k}{i}(-1)^iA^i I^{k-i}=I+\sum_{i=1}^k\binom{k}{i}(-1)^iA^i$$

Rearranging gives

$$I=-\sum_{i=1}^k\binom{k}{i}(-1)^iA^i=A\left[-\sum_{i=1}^k\binom{k}{i}(-1)^iA^{i-1}\right]$$

as desired.

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  • $\begingroup$ outside the box, nice! $\endgroup$
    – orangeskid
    Aug 2, 2021 at 1:52
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Best is to denote $B\colon = I- A$, so $A = I-B$. Now $B$ is nilpotent, ( $B^k = 0$). Check that $$( I + B + B^2 + \cdots B^{k-1})(I-B) = I$$

With your method: assume that $A v = 0$. Then $(I-A)v = v$, so by induction, $(I-A)^n v = v$ for all $n\ge 1$. Now, for $n=k$ we get $(I-A)^k v = v$. But the LHS is $0$, so $v=0$, and thus $A$ is injective, and therefore has an inverse $A^{-1}$.

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  • $\begingroup$ This is really neat thanks! This is only for left inverse. Not that it matters? $\endgroup$
    – Goob
    Aug 2, 2021 at 1:54
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    $\begingroup$ @RabbitBoy: For square matrices, if $BA=I$ then $AB=I$. $\endgroup$ Aug 2, 2021 at 2:10
  • $\begingroup$ @Rabbit Boy: won't matter, but in this case, also the permuted product gives $I$. $\endgroup$
    – orangeskid
    Aug 2, 2021 at 4:11
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Let's try this with minimal polynomial argument: Consider $p(x)=(1-x)^k$ . Then $A$ is a matrix which satisfies the polynomial $p(x)$. Now the minimal polynomial of $A$ has to be a divisor of $p(x)$( This fact can be easily proven using the divison algorithm). So the minimal polynomial of $A$ has to be of the form $m(x)=(1-x)^r$ where $r\leq k$. So the minimal polynomial of $A$ has roots 1 with multiplicity $r$. Now the roots of the minimal polynomial of $A$ are exactly the eigen values od $A$. So eigen values of $A$ is 1 and this is the only eigen value. So $0$ is not an eigen value of $A$ hence $A$ is invertible.

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About your attempt:

Your (1) is incorrect. You know that $(I-A)^k$ is the zero matrix, but you do not know that $(I-A)x=0$. There is absolutely no warrant for asserting that.

Your (2) shows that if $\lambda_2$ is an eigenvalue of $A$ and $x$ is an eigenvector corresponding to $\lambda_2$, then it is an eigenvector of $I-A$ corresponding to $1-\lambda_2$. It would be better to state it that way.

Your Method 1 is incorrect, since it assumes that $I-A=0$, which was not warranted.

Method 2 is also incorrect, because again you assume that $I-A=0$, which is not warranted.


Easy to prove facts:

  1. If $\lambda$ is an eigenvalue of $A$, then $k-\lambda$ is an eigenvalue of $kI-A$, for any scalar $k$.

  2. If $\lambda$ is an eigenvalue of $B$, then $\lambda^k$ is an eigenvalue of $B^k$.

  3. $C$ is invertible if and only if $\lambda=0$ is not an eigenvalue of $C$.

Using these facts: If $\lambda$ is an eigenvalue of $A$, then $1-\lambda$ is an eigenvalue of $I-A$ (what you prove in (2)), and $(1-\lambda)^k$ is an eigenvalue of $(I-A)^k$. Since $(I-A)^k=0$, the only eigenvalue of $(I-A)^k$ is $0$. Therefore...

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  • $\begingroup$ I don't assume I-A=0. I assume (I-A)x=0 for some non zero x. You also say I also cannot do that but why? The determinant of I-A must be zero since $(I-A)^k=\prod_{i=1}^{k} (I-A)$. The determinant of the right is zero, then so must the determinant of every one of those products. Doesn't this imply a 0 eigenvalue? $\endgroup$
    – Goob
    Aug 2, 2021 at 1:49
  • $\begingroup$ @RabbitBoy: If you are assuming that $(I-A)x=0$, then you point (1) is... pointless. Of course the only $\lambda$ for which $\lambda x=0$ is $\lambda=0$. You are essentially saying "Assuming that $x$ is an eigenvector of $0$, then it is only an eigenvector of $0$ and no other scalar". That's true for any matrix, any eigenvector. What's the point, then? And your (2) would then assume that you can pick $x$ to also be an eigenvector of $A$, and in that case you are making another unwarranted assumption. Yes, $0$ is an eigenvalue, but you are not doing valid things with that fact. $\endgroup$ Aug 2, 2021 at 2:02
  • $\begingroup$ If a matrix has a determinant zero, then the nullity of that matrix is not empty right?The determinant of (I-A) is 0. Does this not imply that there is a non zero vector x in the nullity of I-A such that (I-A)x=0? I don't understand why this is an assumption. Sorry. $\endgroup$
    – Goob
    Aug 2, 2021 at 2:25
  • $\begingroup$ @RabbitBoy: Again: you are correct that there must be an eigenvector of $0$ for $I-A$. Then, once you select an eigenvector of $0$ for $I-A$, what is it you are doing with $\lambda_1$? As far as I can tell, just showing that if $x$ is also an eigenvector corresponding to some possibly different eigenvalue $\lambda_1$, then $\lambda_1=0$. Why are you doing that? It's (i) irrelevant; and (ii) we always have for any matrix that if $x$ is an eigenvector of $\lambda$, then that's the only eigenvalue it can be an eigenvector for. (cont) $\endgroup$ Aug 2, 2021 at 2:59
  • $\begingroup$ @RabbitBoy: then you take $\lambda_2$, and assume that $x$, in addition to being an eigenvector corresponding to $0$ for $I-A$, is also an eigenvector corresponding to $\lambda_2$ for $A$. That is an assumption: that you can pick a vector which is both an eigenvector corresponding to $0$ for $I-A$, and at the same time, an eigenvector for $A$ corresponding to some $\lambda_2$. You prove that if such a thing can be done, then $\lambda_2=1$. But you are assuming it can be done in the first place. $\endgroup$ Aug 2, 2021 at 3:00

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