How do you prove the eigenvalues and eigenvectors of matrices are the same?

Question

I wondered if anyone could help me with a couple of proofs.

The question is: Let $A$ be a given $n \times m$ matrix. The collection of scalars $\lambda_i$ and associated $n \times 1$ vectors $q_i$ that solve the equation $Aq=\lambda q$ are known as eigenvalues and eigenvectors of $A$, respectively show that:

i. Suppose $n=m$. Then for any non-singular $n \times n$ square matrix $G$ the eigenvalues of $G^{-1}AG$ are the same as those of $A$.

ii. If $A^{-1}$ exists then it shares the same eigenvectors $q_i$ as $A$ with corresponding eigenvalues $\lambda_i^{-1}$

Thanks in advance!

Answer 1

1) \begin{align*}\det(\lambda I-G^{-1}AG)&=\det(\lambda G^{-1}IG-G^{-1}AG)\\ &=\det(G^{-1}(\lambda I-A)G)\\ &=\det (G^{-1})\cdot \det(\lambda I-A)\cdot \det (G)\\ &=\det (G)^{-1}\cdot \det(\lambda I-A)\cdot \det (G)\\ &=\underbrace{\det (G)^{-1}\cdot \det(G)}_{=1}\cdot \det(\lambda I-A)\\ &=\det(\lambda I-A)\end{align*}

therefore $A$ and $G^{-1}AG$ has the same caracteristic polynomial, and thus the same eigenvalues.

2) $$Au=\lambda u \implies \underbrace{A^{-1}A}_{=I} u=\lambda A^{-1}u\implies u=\lambda A^{-1}u\implies A^{-1}u=\frac{1}{\lambda}u$$

Answer 2

$$Av=\lambda v.$$ Then $$(G^{-1}AG)(G^{-1}v)=G^{-1}Av=G^{-1}\lambda v,$$ so $G^{-1}AG$ has $\lambda$ as an eigenvalue with vector $G^{-1}v$.

2. $Av=\lambda v$, then $v=A^{-1}\lambda v$ and then $=A^{-1} v=\frac{1}{\lambda}v$.