commit f3fe72750c97377149734db321b0f166ada60793
parent 9f63a86e5e2238896646752c614fd8efd229d0ca
Author: Andrew <andrewlaack1@gmail.com>
Date: Wed, 7 Aug 2024 22:02:59 -0500
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-:lin-alg:
-# Inverse Transformation (and matricies)
-
-Khan U2
-
-## Notes
-
-**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.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.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.
-
-## Solving
-
-To find the inverse of a matrix we find all of the manipulations done to the original matrix that results in us getting to the identity matrix. These manipulations can then be placed into their own matrix and thus derive the inverse.
-
-#### Example 1:
-
-$A = \begin{bmatrix}
- 2 & 3 \\
- 1 & 2
-\end{bmatrix}$
-
-
