Root Polynom and eigenvalue

Question

Let $A$ be an $N \times N$ matrix. We know that $A^2 - 2A + I = 0$ so we need to prove that $A = I$ or give a counterexample.

We know that $(A-I)^2 = 0$ but I don't know how we can get from the equation ( 1 eigenvalue from Algebraic multiplicity = 2 ) ?

Edit: We know that $A$ is diagonalizable

Answer 1

If $A$ is diagonalisable, so $A = S D S^{-1}$ where $D$ is diagonal, then $D^2 - 2 D + I = S^{-1} (A^2 - 2 A + I) S = 0$. The diagonal elements of $D^2 - 2 D + I$ are $\lambda^2 - 2 \lambda + 1$ corresponding to the diagonal elements $lambda$ of $D$, so $\lambda = 1$ is the only possibility. Thus $D = I$ and $A = S I S^{-1} = I$.

Answer 2

Hint:

From $(A-I)^2 = 0$ it follows that $1$ is the only eigenvalue of $A$. Since $A$ is diagonalizable we get $A = I$.