Eigenvalues of Matrix vs Eigenvalues of Operator

If you randomly give two linear operators $T$ and $T'$ on $V$, then of course, there is no reason at all they should have same eigenvalues. However, your $T$ and $T'$ are not randomly picked. You first give $T$ and define $A$ as its matrix respect to some basis $\alpha$. Then you choose another basis $\beta$ in $V$ and define $T'$ in such a way that its representation matrix with respect to $\beta$ is $$ [T']_\beta=A. $$ Consequently, your $T$ and $T'$ are related to each other in such a way that $$ [T]_\alpha=[T']_\beta. $$

As a consequence of the following exercise, your $T$ and $T'$ have the same eigenvalues.

Exercise: Show that $T$ and $[T]_\alpha$ have exactly same eigenvalues for any basis $\alpha$ in $V$.


Two matrices $A,B \in M_n(\mathbb{F})$ are called similar if there exists an invertible matrix $P$ such that $P^{-1}AP = B$. Two linear maps $T,S \colon V \rightarrow V$ on a finite dimensional vector space are called similar if there exists an invertible linear transformation $R$ (which is the same thing as an isomorphism) such that $R^{-1} \circ T \circ R = S$. Consider the following statements:

  1. The maps $T,S$ are similar if and only if the matrices $[T]_{\mathcal{B}},[S]_{\mathcal{B}}$ representing the maps with respect to an arbitrary ordered basis $\mathcal{B}$ of $V$ are similar.
  2. Two matrices $A,B \in M_n(\mathbb{F})$ are similar if and only if for any $n$ dimensional vector space $V$, any choice of ordered basis $\mathcal{B}_1$ for $V$ and any operator $T \colon V \rightarrow V$ such that $[T]_{\mathcal{B}_1} = A$ you can find a basis $\mathcal{B}_2$ such that $[T]_{\mathcal{B}_2} = B$. Thus, similar matrices define the same linear operators up to a choice of basis.
  3. Two linear maps $T,S \colon V \rightarrow V$ are similar if and only if they can be represented by the same matrix with respect to two different ordered bases of $V$.

You can check directly that similar linear maps have the same eigenvalues and this will also imply that similar matrices have the same eigenvalues (as the eigenvalues of a matrix $A$ are precisely the eigenvalues of the linear map $T_A \colon \mathbb{F}^n \rightarrow \mathbb{F}^n$ associated to $A$). In your case, the maps $T$ and $T'$ are similar and hence they have the same eigenvalues.

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Linear Algebra

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