$ \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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Basis Change

Let $B$ and $B$ be a basis for $V$. Then there is an unique invertible matrix $A$ such that $$\begin{align*} A[v]_C=[v]_B \end{align*}$$ and $$\begin{align*} [v]_C=A^{-1}[v]_B. \end{align*}$$ The $j^{\text{th}}$ column of $A$ is $[y_j]_B$, where $C=\{y_1, \dots, y_n\}$

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Basis Change

Let $B$ and $B$ be a basis for $V$. Then there is an unique invertible matrix $A$ such that $$\begin{align*} A[v]_C=[v]_B \end{align*}$$ and $$\begin{align*} [v]_C=A^{-1}[v]_B. \end{align*}$$ The $j^{\text{th}}$ column of $A$ is $[y_j]_B$, where $C=\{y_1, \dots, y_n\}$

Concepts

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Hypothesis

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Results

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Proof

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FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof

Basis Change

Let $B$ and $B$ be a basis for $V$. Then there is an unique invertible matrix $A$ such that $$\begin{align*} A[v]_C=[v]_B \end{align*}$$ and $$\begin{align*} [v]_C=A^{-1}[v]_B. \end{align*}$$ The $j^{\text{th}}$ column of $A$ is $[y_j]_B$, where $C=\{y_1, \dots, y_n\}$

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon

Basis Change

Let $B$ and $B$ be a basis for $V$. Then there is an unique invertible matrix $A$ such that $$\begin{align*} A[v]_C=[v]_B \end{align*}$$ and $$\begin{align*} [v]_C=A^{-1}[v]_B. \end{align*}$$ The $j^{\text{th}}$ column of $A$ is $[y_j]_B$, where $C=\{y_1, \dots, y_n\}$

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
FullPage
result
concepts
hypothesis
implications
proof