Eigenvalues of $A$ vs eigenvalues of $2A$

Question

if $A$ is an $n \times n$ matrix with eigenvalues $\lambda_1, \dots , \lambda_n$. Then $2A$ has eigenvalues $2\lambda_1, . . . , 2\lambda_n$.

I don't know how to prove it. Is anybody here have an idea? please help me.

Answer 1

Assume $v$ is an eigenvector that corresponds to $\lambda$ for $A$

Then $(2A)v = 2(Av) = 2\lambda v$ so $v$ is an eigenvector for $2A$ with eigenvalue $2\lambda$