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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!

Clarinetist
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Will Smith
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2 Answers2

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

idm
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  • Thanks, but I thought the characteristic equation was |A-Lambda*I|? Instead of the other way round? Or is it just like that for this equation? Thanks – Will Smith Jan 02 '15 at 13:54
  • It doesn't matter, both are correct. With your formula you'll get something like $(a_1-\lambda)(a_2-\lambda)...(a_k-\lambda)$, and with my formula you'll get $(-1)^k(\lambda-a_1)...(\lambda-a_k)$. But as you can see $$(a_1-\lambda)(a_2-\lambda)...(a_k-\lambda)=(-1)^k(\lambda-a_1)...(\lambda-a_k)$$ – idm Jan 02 '15 at 16:50
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$$Av=\lambda v.$$ 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 $G^{-1}v$.

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