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 February 2nd, 2010, 05:26 PM #1 Senior Member   Joined: Nov 2009 Posts: 169 Thanks: 0 prove it's an eigenvalue Soppose that $T,S:R^n-->R^n$ are linear transformations, if $\lambda$ and $\mu$ are eigenvalues for T and S respectively, either prove that $\lambda$$\mu$ is an eigenvalue for ToS or else give a counter-example to show that this is false. Is $\lambda^2$ an eigenvalue for ToT? (Justify!) I am really not familliar with eigenvalue, I know how to find the eigenvalue, but not proving it.
 February 3rd, 2010, 08:50 AM #2 Newbie   Joined: Dec 2007 From: Madrid Posts: 7 Thanks: 0 Re: prove it's an eigenvalue Consider $T,\;S\;:\;\mathbb{R}^2\rightarrow{\mathbb{R}^2}$ defined by: $T\begin{pmatrix}{x}\\{y}\end{pmatrix}=\begin{pmatr ix}{1}&{1}\\{0}=&{1}\end{pmatrix}\begin{pmatrix}{x} \\{y}\end{pmatrix},\;S\begin{pmatrix}{x}\\{y}\end{ pmatrix}=\begin{pmatrix}{2}&{0}\\{1}=&{2}\end{pmatr ix}\begin{pmatrix}{x}\\{y}\end{pmatrix}=$ Then, $(T\circ{S}) \begin{pmatrix}{x}\\{y}\end{pmatrix}= \begin{pmatrix}{1}&{1}\\{0}=&{1}\end{pmatrix}\begin {pmatrix}{2}=&{0}\\{1}=&{2}\end{pmatrix} \begin{pmatrix}{x}\\{y}\end{pmatrix}=\begin{pmatri x}{3}&{2}\\{1}=&{2}\end{pmatrix}\begin{pmatrix}{x}\ \{y}\end{pmatrix}=$ The only eigenvalue of $T$ is $\lambda=1$, the only of $S$ is $\mu=2$. The eigenvalues of $T\circ S$ are $4$ and $1$, so $\lambda \mu$ is not eigenvalue of $T\circ S$. On the other hand, if $\lambda$ is an eigenvalue of $T$ there exists $v\in \mathbb{R}^n$ with $v\neq 0$ such that $T(v)=\lambda v$. Then, $(T\circ T)(v)=T(T(v))=T(\lambda v)=\lambda T(v)=\lambda^2 v$ This proves that $\lambda^2$ is also an eigenvalue of $T\circ T$ Regards. http://ficus.pntic.mec.es/~frej0002/
February 7th, 2010, 11:57 AM   #3
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Re: prove it's an eigenvalue

Quote:
 Originally Posted by pascallimacon Consider $T,\;S\;:\;\mathbb{R}^2\rightarrow{\mathbb{R}^2}$ defined by: $T\begin{pmatrix}{x}\\{y}\end{pmatrix}=\begin{pmatr ix}{1}&{1}\\{0}=&{1}\end{pmatrix}\begin{pmatrix}{x} \\{y}\end{pmatrix},\;S\begin{pmatrix}{x}\\{y}\end{ pmatrix}=\begin{pmatrix}{2}&{0}\\{1}=&{2}\end{pmatr ix}\begin{pmatrix}{x}\\{y}\end{pmatrix}=$ Then, $(T\circ{S}) \begin{pmatrix}{x}\\{y}\end{pmatrix}= \begin{pmatrix}{1}&{1}\\{0}=&{1}\end{pmatrix}\begin {pmatrix}{2}=&{0}\\{1}=&{2}\end{pmatrix} \begin{pmatrix}{x}\\{y}\end{pmatrix}=\begin{pmatri x}{3}&{2}\\{1}=&{2}\end{pmatrix}\begin{pmatrix}{x}\ \{y}\end{pmatrix}=$ The only eigenvalue of $T$ is $\lambda=1$, the only of $S$ is $\mu=2$. The eigenvalues of $T\circ S$ are $4$ and $1$, so $\lambda \mu$ is not eigenvalue of $T\circ S$. On the other hand, if $\lambda$ is an eigenvalue of $T$ there exists $v\in \mathbb{R}^n$ with $v\neq 0$ such that $T(v)=\lambda v$. Then, $(T\circ T)(v)=T(T(v))=T(\lambda v)=\lambda T(v)=\lambda^2 v$ This proves that $\lambda^2$ is also an eigenvalue of $T\circ T$ Regards. tnaks, that is really a great help, now i am clear. http://ficus.pntic.mec.es/~frej0002/

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