$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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proof
Let $V$ be a finite dimensional vector space and $T\in\mathcal{L}(V)$. Then the roots of $m_T$ are the same as the roots of $P_T$. In particular, every eigenvalue of $T$ is a root of $m_T$.

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Let $V$ be a finite dimensional vector space and $T\in\mathcal{L}(V)$. Then the roots of $m_T$ are the same as the roots of $P_T$. In particular, every eigenvalue of $T$ is a root of $m_T$.

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concepts
If
Only If
proof
FullPage
result
Concepts
If
Only If
proof
Let $V$ be a finite dimensional vector space and $T\in\mathcal{L}(V)$. Then the roots of $m_T$ are the same as the roots of $P_T$. In particular, every eigenvalue of $T$ is a root of $m_T$.

Concepts

Coming soon

If

Coming soon

Only If

Coming soon

Proof

Coming soon
Let $V$ be a finite dimensional vector space and $T\in\mathcal{L}(V)$. Then the roots of $m_T$ are the same as the roots of $P_T$. In particular, every eigenvalue of $T$ is a root of $m_T$.

Concepts

Coming soon

If

Coming soon

Only If

Coming soon

Proof

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concepts
If
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proof