$ \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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If $A$ is invertible then it is invertible as a transformation with the inverse $A^{-1}$ if and only if $A$ is bijective.
A is invertible if and only if $\text{row}(A)=\mathbb{F}^n$, which happens if and only if the rows of $A$ form a basis for $\mathbb{F}^n$

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If $A$ is invertible then it is invertible as a transformation with the inverse $A^{-1}$ if and only if $A$ is bijective.
A is invertible if and only if $\text{row}(A)=\mathbb{F}^n$, which happens if and only if the rows of $A$ form a basis for $\mathbb{F}^n$

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result
concepts
If
Only If
proof
FullPage
result
Concepts
If
Only If
proof
If $A$ is invertible then it is invertible as a transformation with the inverse $A^{-1}$ if and only if $A$ is bijective.
A is invertible if and only if $\text{row}(A)=\mathbb{F}^n$, which happens if and only if the rows of $A$ form a basis for $\mathbb{F}^n$

Concepts

Coming soon

If

Coming soon

Only If

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Proof

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If $A$ is invertible then it is invertible as a transformation with the inverse $A^{-1}$ if and only if $A$ is bijective.
A is invertible if and only if $\text{row}(A)=\mathbb{F}^n$, which happens if and only if the rows of $A$ form a basis for $\mathbb{F}^n$

Concepts

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If

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Only If

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Proof

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If
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proof