notes

InverseTransformation.md (3431B)

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 # Inverse Transformation (and matricies)
 
 Khan U2
 
 **Definition:** The inverse of a transformation is the transformation that undoes the original transformation for the entire domain codomain of the original transformation.
 
 This transformation must be [Bijective](Bijective.md) otherwise there will be issues with mappings either there are outputs without inputs or there are outputs with multiple inputs.
 
 A transformation is invertible if and only if there exists an f^-1 such that f^-1 composed with f is I (identity function).
 
 (in the world of L.T.'s) Assume the standard matrix of f is A and the standard matrix of f^-1 is B. We then know:
 
 AB = I
 
 ## Unique?
 
 Inverse functions are unique (there is only one). 
 
 Let's assume they are not. We then find A(x) = B(x) for all x in R^n. This means A and B are the same function, but they are not. This is a contradiction.
 
 ## Invertible?
 
 To find this we know it needs to be bijective. When solving for RREF if there are instances in the R^m space, where R^m is the codomain, that are not mapped to (found by having a row of zeroes where we can't map to everything based on the combination) then the standard matrix is not invertible as it stands from R^n to R^m.
 
 As such, T is onto iff C(A) = R^m (columns span R^m). **We know this is only true when RREF has a pivot in each row. ([Rank](Rank.md) of the matrix = m)**
 
 For injectivity we test for one-to-one. To find this we need to make sure the rank of the matrix is equal to n where n is the number of columns.
 
 Basically, we need a square matrix that is linearly independent (rows = columns).
 
 As such, the matrix is only invertable if the RREF is the identity matrix in R^n.
 
 Another thing about this; if the determinant is not 0 the matrix is invertible if not then it is not invertible.
 
 ## Solving
 
 #### Intuitive
 
 There are formulas and stuff for solving this, but let's think from an intuitive sense first. To solve for the inverse of some matrix we need to find the matrix that converts from the original matrix to the identity matrix. If we are to create an augmented matrix where the left is the original matrix and the right is the identity matrix we can then perform changes to both sides to convert the left side to the identity matrix. By doing this we will then be left with the right side as the inverse matrix.
 
 That's a lot of text so let's do a simple example of a 2x2 matrix:
 
 $A=\begin{bmatrix}
 2 & 3 &| 1 & 0 \\
 1 & 2 & |0 & 1
 \end{bmatrix}$
 
 $\begin{bmatrix}
 1 & 1 & |1 & -1 \\
 0 & 1 &|-1 & 2
 \end{bmatrix}$
 
 $\begin{bmatrix}
 1 & 0 & |2 & -3 \\
 0 & 1 & |-1 & 2
 \end{bmatrix}$
 
 Solution:
 
 $A^{-1} =\begin{bmatrix}
 2 & -3 \\
 -1 & 2
 \end{bmatrix}$
 
 As can be seen we simply make changes on the left side and update the former identity accordingly to track what matrix would result in such a matrix on the left side starting from the original matrix.
 
 #### Formulaic
 
 The formula is useful, but simply a description of the rules we can apply. For the 2x2 formula we 0 the first column from the second row, zero the second column from the first row, and then normalize both rows.
 
 I was going to show this derivation, but it is trivial and long.
 
 ##### 2x2 Formula:
 
 $A =\begin{bmatrix}
 a & b \\
 c & d
 \end{bmatrix}$
 
 $A^{-1}=
 \frac{1}{ad-bc}
 \begin{bmatrix}
 d & -b \\
 -c & a
 \end{bmatrix}$
 
 The interestinge thing about this is the denominator of the fraction out front is the determinant of the matrix.