$ \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 $T:V\to V$ be a linear transformation. Then a scalar $\lambda\in\mathbb{F}$ is am eigenvalue if there exists some nonzero vector $x\in V$ such that $Tx=\lambda x$, where $x$ is the eigenvector.

Concepts

To find the eigenvalues of $T$, let $\lambda$ be a variable and solve for $\text{det}(T-\lambda I)=0$

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Let $T:V\to V$ be a linear transformation. Then a scalar $\lambda\in\mathbb{F}$ is am eigenvalue if there exists some nonzero vector $x\in V$ such that $Tx=\lambda x$, where $x$ is the eigenvector.

Concepts

To find the eigenvalues of $T$, let $\lambda$ be a variable and solve for $\text{det}(T-\lambda I)=0$

Used In

Coming soon

Hypothesis

Coming soon

Results

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FullPage
definition
concepts
used in
hypothesis
results
FullPage
definition
concepts
used in
hypothesis
results
Let $T:V\to V$ be a linear transformation. Then a scalar $\lambda\in\mathbb{F}$ is am eigenvalue if there exists some nonzero vector $x\in V$ such that $Tx=\lambda x$, where $x$ is the eigenvector.

Concepts

To find the eigenvalues of $T$, let $\lambda$ be a variable and solve for $\text{det}(T-\lambda I)=0$

Used In

Coming soon

Hypothesis

Coming soon

Results

Coming soon
Let $T:V\to V$ be a linear transformation. Then a scalar $\lambda\in\mathbb{F}$ is am eigenvalue if there exists some nonzero vector $x\in V$ such that $Tx=\lambda x$, where $x$ is the eigenvector.

Concepts

To find the eigenvalues of $T$, let $\lambda$ be a variable and solve for $\text{det}(T-\lambda I)=0$

Used In

Coming soon

Hypothesis

Coming soon

Results

Coming soon
FullPage
definition
concepts
used in
hypothesis
results