$ \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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Let $V$ be a finite dimensional space over $\mathbb{F}$, where $\mathbb{F}$ is algebraically complete. Then for any linear transformation $T:V\to V$, $\text{det}T$ is the product of the eigenvalues of $T$, and $\text{tr}(T)$ is the sum of the eigenvalues of $T$.

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Let $V$ be a finite dimensional space over $\mathbb{F}$, where $\mathbb{F}$ is algebraically complete. Then for any linear transformation $T:V\to V$, $\text{det}T$ is the product of the eigenvalues of $T$, and $\text{tr}(T)$ is the sum of the eigenvalues of $T$.

Concepts

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Hypothesis

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Proof

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FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof
Let $V$ be a finite dimensional space over $\mathbb{F}$, where $\mathbb{F}$ is algebraically complete. Then for any linear transformation $T:V\to V$, $\text{det}T$ is the product of the eigenvalues of $T$, and $\text{tr}(T)$ is the sum of the eigenvalues of $T$.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
Let $V$ be a finite dimensional space over $\mathbb{F}$, where $\mathbb{F}$ is algebraically complete. Then for any linear transformation $T:V\to V$, $\text{det}T$ is the product of the eigenvalues of $T$, and $\text{tr}(T)$ is the sum of the eigenvalues of $T$.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
FullPage
result
concepts
hypothesis
implications
proof