Invertible - Neutrinos are the coolest particles

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A linear transformation $T:U\to U$ is invertible if there exists a linear transformation $T^{-1}:U\to U$ such that $TT^{-1}=T^{-1}T=I$.
Furthermore if $T:V\to V$, where $V$ is finite dimensional, is left or right invertible, then $T$ is invertible.

Concepts

As usual a function is invertible if and only if its bijective. This means that if $R$ is invertible, then $R$ has no zero rows (if it has zero rows, it would not be subjective).
For infinite matrices, it's possible that it is only left or right invertible, but not the other. For instance, $$\begin{align*} A=\begin{bmatrix} 0 & 0 & 0 & \dots \\ 1 & 0 & 0 & \dots \\ 0 & 1 & 0 & \dots \\ \vdots & \vdots & \vdots &\ddots \end{bmatrix} \end{align*}$$ $A$ is left invertible but it is not right invertible.
If $A$ and $B$ are invertible, then so is $AB$, where $(AB)^{-1}=B^{-1}A^{-1}$. This holds for a an arbitrary number of matrices.
The inverse is calculated using block matrices. Let $R$ be a row reduced echelon matrix that is row equivalent to $A$. Note that $A$ is invertible if and only if $R$ is the identity matrix.
There also exists a matrix $P$ which is the product of elementary matrices such that $R=PA$, as $R$ is row equivalent to $A$. Since $R=I_n$, this implies that $P=A^{-1}$.
We form an augmented matrix, $[A|I_n]$. Multiplying this augmented matrix by $P$ then gives the new augmented matrix $$\begin{align*} P[A|I_n] &= [PA| P]\\ &=[R|P]\\ &=[I_n|P], \end{align*}$$ and so $[A|I_n]$ is row equivalent to $[I_n|P]$ Hence to compute $A^{-1}$, we form the augmented matrix $[A|I_n]$ then row reduce until we get an augmeted matrix of the form $[I_n|A]$.

Used In

Hypothesis

If a matrix $A$ such that $A^k=I_n$ for some $k$, then $A$ is diagonalizable and stuff
If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

Results

If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

A linear transformation $T:U\to U$ is invertible if there exists a linear transformation $T^{-1}:U\to U$ such that $TT^{-1}=T^{-1}T=I$.
Furthermore if $T:V\to V$, where $V$ is finite dimensional, is left or right invertible, then $T$ is invertible.

Concepts

As usual a function is invertible if and only if its bijective. This means that if $R$ is invertible, then $R$ has no zero rows (if it has zero rows, it would not be subjective).
For infinite matrices, it's possible that it is only left or right invertible, but not the other. For instance, $$\begin{align*} A=\begin{bmatrix} 0 & 0 & 0 & \dots \\ 1 & 0 & 0 & \dots \\ 0 & 1 & 0 & \dots \\ \vdots & \vdots & \vdots &\ddots \end{bmatrix} \end{align*}$$ $A$ is left invertible but it is not right invertible.
If $A$ and $B$ are invertible, then so is $AB$, where $(AB)^{-1}=B^{-1}A^{-1}$. This holds for a an arbitrary number of matrices.
The inverse is calculated using block matrices. Let $R$ be a row reduced echelon matrix that is row equivalent to $A$. Note that $A$ is invertible if and only if $R$ is the identity matrix.
There also exists a matrix $P$ which is the product of elementary matrices such that $R=PA$, as $R$ is row equivalent to $A$. Since $R=I_n$, this implies that $P=A^{-1}$.
We form an augmented matrix, $[A|I_n]$. Multiplying this augmented matrix by $P$ then gives the new augmented matrix $$\begin{align*} P[A|I_n] &= [PA| P]\\ &=[R|P]\\ &=[I_n|P], \end{align*}$$ and so $[A|I_n]$ is row equivalent to $[I_n|P]$ Hence to compute $A^{-1}$, we form the augmented matrix $[A|I_n]$ then row reduce until we get an augmeted matrix of the form $[I_n|A]$.

Used In

Hypothesis

If a matrix $A$ such that $A^k=I_n$ for some $k$, then $A$ is diagonalizable and stuff
If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

Results

If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

FullPage

definition

concepts

used in

hypothesis

results

FullPage

definition

concepts

used in

hypothesis

results

A linear transformation $T:U\to U$ is invertible if there exists a linear transformation $T^{-1}:U\to U$ such that $TT^{-1}=T^{-1}T=I$.
Furthermore if $T:V\to V$, where $V$ is finite dimensional, is left or right invertible, then $T$ is invertible.

Concepts

As usual a function is invertible if and only if its bijective. This means that if $R$ is invertible, then $R$ has no zero rows (if it has zero rows, it would not be subjective).
For infinite matrices, it's possible that it is only left or right invertible, but not the other. For instance, $$\begin{align*} A=\begin{bmatrix} 0 & 0 & 0 & \dots \\ 1 & 0 & 0 & \dots \\ 0 & 1 & 0 & \dots \\ \vdots & \vdots & \vdots &\ddots \end{bmatrix} \end{align*}$$ $A$ is left invertible but it is not right invertible.
If $A$ and $B$ are invertible, then so is $AB$, where $(AB)^{-1}=B^{-1}A^{-1}$. This holds for a an arbitrary number of matrices.
The inverse is calculated using block matrices. Let $R$ be a row reduced echelon matrix that is row equivalent to $A$. Note that $A$ is invertible if and only if $R$ is the identity matrix.
There also exists a matrix $P$ which is the product of elementary matrices such that $R=PA$, as $R$ is row equivalent to $A$. Since $R=I_n$, this implies that $P=A^{-1}$.
We form an augmented matrix, $[A|I_n]$. Multiplying this augmented matrix by $P$ then gives the new augmented matrix $$\begin{align*} P[A|I_n] &= [PA| P]\\ &=[R|P]\\ &=[I_n|P], \end{align*}$$ and so $[A|I_n]$ is row equivalent to $[I_n|P]$ Hence to compute $A^{-1}$, we form the augmented matrix $[A|I_n]$ then row reduce until we get an augmeted matrix of the form $[I_n|A]$.

Used In

Hypothesis

If a matrix $A$ such that $A^k=I_n$ for some $k$, then $A$ is diagonalizable and stuff
If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

Results

If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

A linear transformation $T:U\to U$ is invertible if there exists a linear transformation $T^{-1}:U\to U$ such that $TT^{-1}=T^{-1}T=I$.
Furthermore if $T:V\to V$, where $V$ is finite dimensional, is left or right invertible, then $T$ is invertible.

Concepts

As usual a function is invertible if and only if its bijective. This means that if $R$ is invertible, then $R$ has no zero rows (if it has zero rows, it would not be subjective).
For infinite matrices, it's possible that it is only left or right invertible, but not the other. For instance, $$\begin{align*} A=\begin{bmatrix} 0 & 0 & 0 & \dots \\ 1 & 0 & 0 & \dots \\ 0 & 1 & 0 & \dots \\ \vdots & \vdots & \vdots &\ddots \end{bmatrix} \end{align*}$$ $A$ is left invertible but it is not right invertible.
If $A$ and $B$ are invertible, then so is $AB$, where $(AB)^{-1}=B^{-1}A^{-1}$. This holds for a an arbitrary number of matrices.
The inverse is calculated using block matrices. Let $R$ be a row reduced echelon matrix that is row equivalent to $A$. Note that $A$ is invertible if and only if $R$ is the identity matrix.
There also exists a matrix $P$ which is the product of elementary matrices such that $R=PA$, as $R$ is row equivalent to $A$. Since $R=I_n$, this implies that $P=A^{-1}$.
We form an augmented matrix, $[A|I_n]$. Multiplying this augmented matrix by $P$ then gives the new augmented matrix $$\begin{align*} P[A|I_n] &= [PA| P]\\ &=[R|P]\\ &=[I_n|P], \end{align*}$$ and so $[A|I_n]$ is row equivalent to $[I_n|P]$ Hence to compute $A^{-1}$, we form the augmented matrix $[A|I_n]$ then row reduce until we get an augmeted matrix of the form $[I_n|A]$.

Used In

Hypothesis

If a matrix $A$ such that $A^k=I_n$ for some $k$, then $A$ is diagonalizable and stuff
If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

Results

If $f:X\to Y$ is a function, it has an inverse $P^{-1}:Y\to X$ if and only if $f$ is injective

FullPage

definition

concepts

used in

hypothesis

results