I would like to know if there are any rare cases where performing a basis change for describing a transformation on a different basis also changes its eigenvalues.
I know that in general this does not happen, but I would like to know if there are any extreme cases where it does happen.
Are there any extreme instances where a change of basis can change the eigenvalues of the transformed matrix?
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linear-algebra
matrices
linear-transformations
transformation
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4An eigenvalue of a linear transformation $T:V\to W$ is defined as $\lambda$ such that there is nonzero $v\in V$ which satisfies $Tv=\lambda v$; This description is basis-free, so changing basis cannot change eigenvalues – 2017-02-07
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1@user160738: Presumably you meant $T:V\to V$? Unless the range is a subspace of the domain, there cannot be an eigenvector. – 2017-02-07
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1@hardmath Oh, yes; thanks for pointing that out – 2017-02-07
1 Answers
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No. If $T: V \rightarrow V$ is a linear operator where $A$ is the matrix of $T$ with respect to a basis $\mathcal{B}$, and $C$ is the matrix of $T$ with respect to another basis $\mathcal{B}^{'}$, then we necessarily have that $A=PCP^{-1}$ for some invertible matrix P. In other words, $A$ is conjugate (or similar) to $C$. Matrices that are similar always have the exact same eigenvalues.
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0Do you know if something like this could happen in the field of functional analysis(i.e. in infinite dimensional spaces)? – 2017-02-07