Eigenvalues question (1 Viewer)

undalay

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How do you do part (ii) guys ) :

edit: free cookie if u get it
 
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Iruka

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Bx is also an eigenvector, hence it is also in the eigenspace of lambda. What does this tell you about the form of Bx?
 

undalay

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Bx is also an eigenvector, hence it is also in the eigenspace of lambda. What does this tell you about the form of Bx?
Bx = mu x ?

Hence x is an eigen vector?

edit: thank you : )
 
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