babeeb0oL

2021-02-09

Verify if the statement is true: The eigenvalues of a linear transformation T in L(U) are independent of the chosen base of U.

### Answer & Explanation

pattererX

Let M be the matrix representation of T with respect to the basis A and N be the matrix representation w.r.t. the basis B. To convert M to N, recall that you use
$N={P}^{-1}\ast MP$
where P is the change of basis matrix for vectors expressed in the basis B to their expression in the basis A.
Let v be an eigenvector of N with eigenvalue λ. Then
$Nv=\lambda v\therefore {P}^{-1}\ast MPv=\lambda v\therefore M\left[Pv\right]=\lambda \left[Pv\right]$
Hence λ is also an eigenvalue of the matrix M for the eigenvector Pv. As this holds for any arbitrary bases A and B, we see that eigenvalues of one matrix representation are also eigenvalues of any other.

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