Question on eigenvalue properties If $\lambda$ is eigenvalue of matrix $A$ and $B=S^{-1} A S$ show that $\lambda$ is also eigenvalue of matrix $B$. Also show that if matrix $A^{2015}$ is a zero matrix then its only eigenvalue is $0$.
The first problem I tackled like this. if $\lambda$ is a eigenvalue of $B$ it solves this equation.
\begin{align*}
B&=S^{-1} A S\;\;\;|\cdot x\\
Bx&=(S^{-1} A S) x\;\;\;\text{where }(S^{-1} A S)=\lambda
\end{align*}
So this means that $\lambda=S^{-1} A S$ and it should solve:
\begin{align*}
Ax&=(S^{-1} A S)x\\
A&=S^{-1}AS=B
\end{align*}
And shouldn't this hold because now while writing that I was thinking that for all matrises $X$ if you multiply it first with matrix $S$ and then with $S^{-1}$ should it be $X$ so does this hold for all matrises?:
$$B=S^{-1}AS=A$$
Isn't the second question very trivial because $A^{2015}=\mathbf{0}$
\begin{align*}
A^{2015}x&=\lambda x \\
\mathbf{0}x &= \lambda x \\
\mathbf{0} &= \lambda x\;\;\text{where } x\neq\mathbf{0} \text{ thus}\\
\lambda &= 0
\end{align*} 
 A: Similar matrices have same eigenvalues.
$$B=S^{-1}AS$$
$f_B(\lambda)=\text{det}(B-\lambda I)=\text{det}(S^{-1}AS-\lambda I)=\text{det}(S^{-1}(A-\lambda I)S)=\text{det}(A-\lambda I)=f_A(\lambda)$
Hint:
$A^{2015}=0$ implies $A$ is nilpotent.  Can you get the characteristics polynomial from this? 

 nilpotent matrices has eigenvalue $0$

A: To show that if $A$ has eigenvalue $\lambda$ then $B=S^{-1}AS$ does as well, let $x$ be an eigenvector of $A$ with the given eigenvalue, and let $y=S^{-1}x$.
$$By=(S^{-1}AS)(S^{-1}x)=S^{-1}Ax=S^{-1}(\lambda x)=\lambda S^{-1}x=\lambda y$$
Hence, $y$ is an eigenvector of $B$ with eigenvalue $\lambda$; therefore $B$ has all the same eigenvalues as $A$ (though with different eigenvectors).
To show that if $A^n=0$ for some $n$ (that is, $A$ is nilpotent) then the only eigenvalue of $A$ is zero, consider how the eigenvalues of $A^n$ relate to those of $A$. If $A$ has eigenvalue $\lambda$ with eigenvector $x$, then:
$$Ax=\lambda x$$
$$A^2x=A(Ax)=A(\lambda x)=\lambda Ax=\lambda^2x$$
$$A^3x=A(A^2x)=\lambda^3x$$
and so on. So the eigenvalues of $A^n$ are the eigenvalues of $A$ raised to the $n$-th power. If $A^n$ is the zero matrix, then its only eigenvalue is zero as you have shown; but this means that the only eigenvalue of $A$ is the $n$-th root of zero - which is just zero itself.
