Find matrix of linear transformation and show it's diagonalizable

  • #1
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Summary:
Let ##V## be an ##n##-dimensional inner product space, where ##n>0##, and let ##F## be the linear transformation on ##V## defined by ##F(\textbf{u})=\langle \textbf{u},\textbf{c} \rangle \textbf{b}-\langle \textbf{b},\textbf{c} \rangle \textbf{u} ##, where ##\textbf{b},\textbf{c} \in V## and ## \langle \textbf{b},\textbf{c} \rangle \neq 0 ##. Show that ##V## has a basis consisting of eigenvectors of ##F## and find the matrix of ##F## with respect to some such basis.
The strategy here would probably be to find the matrix of ##F##. How would one go about doing that? Since ##V## is finite dimensional, it must have a basis...
 

Answers and Replies

  • #2
tnich
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I think it would be simpler to just figure out what an eigenvector of ##F(.)## looks like. Under what conditions does ##F(\mathbf u) = \lambda \mathbf u## for some ##\lambda## and ##\mathbf u##?
 
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  • #3
WWGD
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This is possible when the arithmetic and geometric multiplicities of each eigenvalue agree.
 

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