If a linear transformation is similar to another, then they have the same eigenvalues.

Question

I need help getting through this proof

Let $T : R^n \rightarrow R^n$ be a linear transformation. T has an eigenvalue $\lambda$ if there exists some non-zero vector $\vec{x} \in R^n$ such that $T(\vec{x}) = \lambda\vec{x}$. We say that T is similar to another linear transformation $S:R^n \rightarrow R^n$ if there exists an invertible linear transformation $F:R^n\rightarrow R^n$ such that $$ S=F^{-1}\circ T \circ F, $$ which means $S(\vec{x})=(F^{-1} \circ T \circ F)(\vec{x})$ for all $\vec{x} \in R^n$. The following steps will help us prove similar linear transformations have the same eigenvalues.

a) Let $\vec{x}$ be an eigenvector of T with eigenvalue $\lambda$. Show $F^{-1}(\vec{x}) \neq 0$. Hint: Assume $F^{-1}(\vec{x}) = 0$ then apply F to both sides of the equations. Why does this lead to a contradiction?

b) Show $F^{-1}(\vec{x})$ is an eigenvector of S with eigenvalue $\lambda$. (Why is it important that $F^{-1}(\vec{x})\neq 0$?)

c) Show that if $T$ is similar to $S$ then $S$ is similar to $T$.

d) Use the above observations to write a proof of the following statement: If $T$ is similar to $S$ then $T$ and $S$ have the same eigenvalues.

I managed to get the first part (below), but I am having trouble going from there.

Let $x$ be an eigenvector, such that $F^{-1}(\vec{x}) = 0 \rightarrow \vec{x} = 0$, which is a contradiction because eigenvectors are not $0$.

Thanks.

@MhenniBenghorbal I saw that one, but it honestly did not help me; it is not quite the same. — lord_sneed, Dec 08 '13 at 02:04
See this problem. It proves the eigenvalue case and it tells you what is the corresponding eigenvector. — Mhenni Benghorbal, Dec 08 '13 at 02:24

score 0 · Answer 1 · answered Dec 08 '13 at 02:16

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b) $F^{-1}TF\,(F^{-1}x)=F^{-1}Tx=F^{-1}(\lambda x)=\lambda\,(F^{-1}x)$.

c) and d) should be then easy.

answered Dec 08 '13 at 02:16

Berci

90,745

For (c) does "similar" in this case, mean substitute T for S (so wherever there is a T, put a S)? – lord_sneed Dec 08 '13 at 02:24
Yes, that's a way to do it. Specifically, for c), consider $G:=F^{-1}$ then $T=G^{-1}SG$ satisfying the definition of '$S$ similar to $T$'. – Berci Dec 08 '13 at 02:55
so then it becomes: $F^{-1} \circ S \circ F = [F^{-1} \circ S] (x) = F^{-1}(\lambda x) = \lambda F^{-1}(x)$ ? – lord_sneed Dec 08 '13 at 03:08

If a linear transformation is similar to another, then they have the same eigenvalues.

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