If I have the matrix equation:

$A^3=KA$ , k≠1

How to show that $A+I$ is invertible? I'd like to do this problem using eigenvalues, so how can I prove that $-1$ is an eigenvalue of the matrix $A$?

  • $\begingroup$ does the matrix $K$ arbitrary? $\endgroup$
    – tortue
    Aug 7, 2018 at 12:32
  • $\begingroup$ What is $K$? For $K=0$, $(A+I)$ is invertible because $(A+I)(I-A+A^2)=A-A^2+A^3+I-A-A^2=I$. $\endgroup$
    – Kusma
    Aug 7, 2018 at 12:34
  • $\begingroup$ K is arbitrary real constant, k≠1 $\endgroup$ Aug 7, 2018 at 12:38
  • $\begingroup$ You want to show that $A$ does not have an eigenvalue $-1$. Since the eigenvalues are roots of $\lambda(\lambda^2-k)=0$, can you infer it now? $\endgroup$ Aug 7, 2018 at 12:45
  • $\begingroup$ Although I know that A+I is invertible, because I can show that there's a matrix (xA^2+yA+z)(A+I)=I, that's means that -1 is an eigenvalue of it, doesn't it? $\endgroup$ Aug 7, 2018 at 12:47

2 Answers 2


The equation $A^3=kA$ tells you that $f(A)=0$ for the polynomial $f(x)=x^3-kx=x(x^2-k)$, so the minimal polynomial (a divisor of $f$) can only have the zeroes $0$ and $\pm \sqrt{k}$ (provided $k\ge 0$). The characteristic polynomial has the same zero set as the minimal polynomial, so $-1$ is not an eigenvalue of $A$. Hence $A+I$ is invertible.

  • $\begingroup$ That's it! Sorry, I misundertood some concepts, thank you"! $\endgroup$ Aug 7, 2018 at 12:52

The idea is inspired by the Neumann series $(1-x)^{-1} = \sum_{k=0}^\infty x^k$ for $|x|<1$. We want to use this idea with $x=-A$. Since $A^3=kA$, this series will contain powers of $A$ only up to order $2$. So we guess the inverse to be of the form $I+xA+yA^2$. Then we get $$ (A+I)(I + xA + yA^2) = I + xA + yA^2 + A+xA^2+yA^3 =I + (x+ky+1)A + (x+y)A^2. $$ This is equal to the identity if and only if $$x=-y, \quad x(1-k)=1.$$ Hence setting $x = \frac1{1-k}$, $y=-\frac1{1-k}$ we get $$ (A+I)^{-1} = I + \frac1{1-k}(A-A^2). $$


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