2
$\begingroup$

Is the matrice $A^2-A+I$ invertible?

$A^2=A-I /\cdot A$

$ A^3 = A^2 - A$

$A^2=A$

Thus I conclude that $\lambda=0$ is one of the eigenvalues of this matrice and it isn't invertible, my conclusion comes from the fact that the eigenvalues of the matrices $A,A^2,...$ are all connected, and if $A^3=O$ that means that those two have the same eigenvalues.

  • 0
    How does $A^2=A$ prove that $\lambda=0$ is an eigenvalue of $A^2-A+I$? – 2017-01-26
  • 0
    Sorry it only limits the eigenvalues to being 0 or 1, my conclusion was wrong, and it turns out that $A=I$ – 2017-01-26
  • 0
    If $A=I$, how is $A^3=0$? – 2017-01-26

2 Answers 2

11

$(A^2 - A + I)(A+I) = A^3 + I = I$.

To see how one knows that the matrix is invertible by just looking at it, one might recall the general fact:

If $\sum_{n=0}^{\infty}(-1)^n A^n$ is convergent, then $I+A$ is invertible and one has:

$$(I+A)^{-1} = \sum_{n=0}^{\infty}(-1)^n A^n$$

It's clearly the case here that the said series converges.

  • 0
    so it is invertible, but can you give some more insight on how would one prove that the given matrice actually is invertible by just looking at that matrice? – 2017-01-26
  • 0
    @adadaae12313412 see the edit. – 2017-01-26
2

If $\lambda$ is an eigenvalue of $A$, then $\lambda^3$ is an eigenvalue of $A^3=0$. Hence $\lambda=0$. This shows that $A^2-A+I$ has only one eigenvalue: $1$.

Therefore $0$ is not an eigenvalue of $A^2-A+I$.