Possible eigenvalues of a matrix satisfying $A*A = 4A$

Question

What is the eigenvalue or what could the eigenvalue be of a matrix $A$ that if multiplied by itself equals $4A$?

Answer 1

Suppose $A$ has an eigenvalue $\lambda$, so there exists $v$ with $Av=\lambda v$. Using the fact that $A^{2}=4A$, we have $$A^{2}v=(4A)v=4(Av)=4(\lambda v)=(4\lambda)v$$ so that $v$ is also an eigenvector of $A^{2}$, with eigenvalue $4\lambda$. But $$A^{2}v=A(Av)=A(\lambda v)=\lambda(Av)=\lambda(\lambda v)=\lambda^{2}v$$ so the eigenvalue of $v$ (with respect to $A^{2}$) is $\lambda^{2}$. Since an eigenvector can only correspond to one eigenvalue, we must have $\lambda^{2}=4\lambda$, so that $\lambda = 0$ or $4$.

Answer 2

The minimal polynomial of $A$ is $x^2-4x=x(x-4)$ or $x$ or $x-4$. So the eigenvalues can be $0,4$.

Answer 3

You are given that $A^2-4A=0$. If $v$ is a nonzero eigenvector of $A$, say with eigenvalue $\lambda$, then $$A^2v=A(Av)=A(\lambda v)=\lambda Av=\lambda^2v,$$ and so $$(A^2-4A)v=A^2v-4Av=\lambda^2v-4\lambda v=(\lambda^2-4\lambda)v.$$ But $A^2-4A=0$, so $\lambda^2-4\lambda=0$. So what are the possible values for $\lambda$?