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Let $T$ be a linear operator on an $n-$dimensional vector space $V$ with characteristic polynomial $f(t)$. Then

$a)$ A scalar $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is a zero of the polynomial $f(t)$ (i.e., if and only if $f(\lambda)=0$).

$b)$ $T$ has at most $n$ distinct eigenvalues.

Attempt:

$a)$ $\lambda$ is an eigenvalue $\Leftrightarrow$ there is $v\neq \vec{0}$ and $T(v) = \lambda v$ $\Leftrightarrow$ there is $v\neq \vec{0}$, $v \in ker(T − \lambda I)$ $\Leftrightarrow$ $ker(T − \lambda I)\neq \vec{0}$ $\Leftrightarrow$ $ker([T]_\beta −\lambda I) \neq \vec{0}$ for some basis $\beta$ $\Leftrightarrow$ $f(\lambda) = det([T]_\beta − \lambda I) = 0$

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    $\begingroup$ b) follows from a) since $f$ has degree $n$. $\endgroup$ Aug 13, 2021 at 5:07
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    $\begingroup$ Using polynomial long division it can be proved that a polynomial of degree $n$ (with coefficients in a field say) has at most $n$ zeroes. $\endgroup$
    – Mason
    Aug 13, 2021 at 5:42

1 Answer 1

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Suppose that $m$ is a minimal polynomial for $T$, and suppose that $\lambda$ is an eigenvalue of $T$. Then there is a non-zero vector $x$ such that $(T-\lambda I)x=0$. Consequently, $$ (m(T)-m(\lambda)I)x=g(T)(T-\lambda I)x=0, $$ where $g$ is a polynomial. Because $m(T)=0$ and $x\ne 0$, it follows that $m(\lambda)=0$ for any eigenvalue $\lambda$ of $T$.

Conversely, suppose that $m(\lambda)=0$, where $m$ is the minimal polynomial for $T$. By polynomial division, there is a polynomial $g$ of lower order than $m$ such that $$ 0=m(T)=m(T)-m(\lambda)I=(T-\lambda I)g(T). $$ Because $m$ is minimal, it follows that $g(T)\ne 0$. And, from the above, every non-zero vector in the range of $g(T)$ must be an eigenvector of $T$ with eigenvalue $\lambda$. So $\lambda$ is an eigenvalue of $T$.

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