Eigenvalues and eigenvectors/Endomorphism/Section

Lemma

Let

$\varphi \colon V\longrightarrow V$

denote an endomorphism on a ${}K$-vector space ${}V$, and let

$f\colon V\longrightarrow W$

denote an isomorphism of ${}K$-vector spaces. Set

${}\psi =f\circ \varphi \circ f^{-1}\,.$

Then the following hold.

1. A vector ${}v\in V$ is an eigenvector of ${}\varphi$ for the eigenvalue ${}a\in K$ if and only if ${}f(v)$ is an eigenvector of ${}\psi$ for the eigenvalue ${}a$.
2. ${}\varphi$ and ${}\psi$ have the same eigenvalues.
3. The mapping ${}f$ induces for every ${}a\in K$ an isomorphism

   $f\colon \operatorname {Eig} _{a}{\left(\varphi \right)}\longrightarrow \operatorname {Eig} _{a}{\left(\psi \right)}.$

Proof

(1). Let ${}v\in V$ be an eigenvector of ${}\varphi$ for the eigenvalue ${}a$. Set

${}w:=f(v)\,.$

Then

${}\psi (w)=(f\circ \varphi \circ f^{-1})(f(v))=(f\circ \varphi )(v)=f(\varphi (v))=f(av)=af(v)=aw\,.$

The reverse statement holds in the same way. (2) and (3) follow directly from (1).

$\Box$

If we have an endomorphism on a finite-dimensional vector space, which is described by the matrix ${}M$ with respect to a given basis, then the eigenvalues and the eigenvectors correspond to each other. Then eigenvector of the matrix is the Coordinate tuple of the corresponding eigenvector with respect to the basis. The eigenvalues do not depend on the chosen basis, but the eigenvectors do.

Corollary

Let ${}\varphi \colon V\rightarrow V$ be an endomorphism on the finite-dimensional ${}K$-vector space ${}V$, and let ${}{\mathfrak {u}}=u_{1},\ldots ,u_{n}$ denote a basis of ${}V$. Let ${}M=M_{\mathfrak {u}}^{\mathfrak {u}}$ be the describing matrix of ${}\varphi$ with respect to the basis. Then ${}v\in V$ is an eigenvector of ${}\varphi$ for the eigenvalue ${}a$ if and only if the coordinate tuple

of ${}v$ with respect to the basis is an eigenvector of ${}M$ for the eigenvalue ${}a$.

Proof

This follows directly from fact  (1), using the diagram

${\begin{matrix}V&{\stackrel {\varphi }{\longrightarrow }}&V&\\\!\!\!\!\!\!\!\!\!\!\!\!\!\psi _{\mathfrak {u}}^{-1}\downarrow &&\downarrow \!\psi _{\mathfrak {u}}^{-1}\!\!\!\!\!&\\K^{n}&{\stackrel {M_{\mathfrak {u}}^{\mathfrak {u}}(\varphi )}{\longrightarrow }}&K^{n}\,.&\!\!\!\!\!\\\end{matrix}}$

$\Box$

Corollary

Let ${}M$ be an ${}n\times n$-matrix over a field ${}K$, and let ${}B$ denote an invertible ${}n\times n$-matrix. Let ${}a\in K$. Then an ${}n$-tuple ${}{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}$ is an eigenvector of ${}M$ for the eigenvalue ${}a$ if and only if

${}{\begin{pmatrix}x'_{1}\\\vdots \\x'_{n}\end{pmatrix}}=B{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\,$

is an eigenvector of the matrix ${}BMB^{-1}$ for the eigenvalue ${}a$. In particular, ${}M$ and ${}BMB^{-1}$

have the same eigenvalues.

Proof

This follows from fact.

$\Box$