Let $A\in M_{n\times n}(\mathbb{C})$ be an invertible matrix. Show that there exist $u\in\mathbb{R}^n$ and $\lambda\in\mathbb{C}$ not null, such that $u=\frac{1}{\lambda^n}A^nu$
And I am not able to do this, say whether it is true or false and justify with a demonstration or an example, $A,B\in M_{5\times 5}(\mathbb{C})$ similar matrices and u is a vector of A then $\sqrt2u$ is a proper vector of B
On
What we need here is for $A$ to have an eigenvalue/eigenvector. Since $A^n$ are just repeated mappings of the linear transformation represented by the matrix $A$, obviously, the eigenvalue will be scaled by a power of $n$ and the eigenvector remains unchanged. That is, if $Ax=\lambda x$, then $A^nx=\lambda^n x$. This is precisely the equation in the question! Can you continue from here?
Edit: Here are some more details to help you along. As shown above we just require some $\lambda\neq0$ to exist such that $\det(A-\lambda I)=0$. $A$ is invertible, so $\det(A-0I)\neq0$ and if such a $\lambda$ exists, then it can't be $0$. So we really only need to prove that the characteristic polynomial of $A$ has a complex root. Is this obvious to you?
Hint: You just need a nonzero eigenvalue. Invertibility actually implies none of the eigenvalues are $0$.