Invertible matrix and Pseudo-inverse

1. Overview

$$AB=BA=I_{n}$$

2. Description

2.1 Pre-requirement

A matrix is invertible if it is

• Square matrix
• Full rank
• Non-zero determinant

2.2 Properties

Let $A$ be a square n by n matrix over a field $K$ (for example the field $R$ of real numbers). The following statements are equivalent for any given matrix are either all true or all false:

• A is invertible, that is, A has an inverse, is nonsingular, or is nondegenerate.
• A is row-equivalent to the n-by-n identity matrix $I_{n}$.
• A is column-equivalent to the n-by-n identity matrix $I_{n}$.
• A has n pivot positions.
• det A ≠ 0. In general, a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring.
• A has full rank; that is, rank A = n.
• The equation Ax = 0 has only the trivial solution x = 0.
• The kernel of A is trivial, that is, it contains only the null vector as an element, ker(A) = {0}.
• Null A = {0}.
• The equation Ax = b has exactly one solution for each b in $K_{n}$.
• The columns of A are linearly independent.
• The columns of A span $K_{n}$.
• Col A = $K_{n}$.
• The columns of A form a basis of $K_{n}$.
• The linear transformation mapping x to Ax is a bijection from Kn to Kn.
• There is an n-by-n matrix B such that AB = $I_{n}$ = BA.
• The transpose $A^{T}$ is an invertible matrix (hence rows of A are linearly independent, span $K^{n}$, and form a basis of $K^{n}$.
• The number 0 is not an eigenvalue of A.
• The matrix A can be expressed as a finite product of elementary matrices.
• The matrix A has a left inverse (that is, there exists a B such that BA = I) or a right inverse (that is, there exists a C such that AC = I), in which case both left and right inverses exist and B = C = $A^{-1}$.
• ($A^{-1})^{-1}$ = $A$;
• $(kA)^{-1}$ = $k^{-1}A^{-1}$ for nonzero scalar k;
• $(Ax)^{+}=x^{+}A^{-1}$ where + denotes the Moore-Penrose inverse and x is a vector;
• $(A^{T})^{-1}=(A^{-1})^{T}$;
• For any invertible n-by-n matrices A and B, $(AB)^{-1}=B^{-1}A^{-1}$. More generally, if $A_{1},\cdots , A_{k}$ are invertible n-by-n matrices, then $(A_{1}A_{2}\cdots A_{k-1}A_{k})^{-1}=A_{k}^{-1}A_{k-1}^{-1}\cdots A_{2}^{-1}A_{1}^{-1}$
• $det\:  A^{-1}=(det\:  A)^{-1}$

2.3 Calculate Inverse Matrix

2.3.1 Example

2.3 Proof of uniqueness

3. Pseudo-inverse

3.1 Properties

• Unlike the true inverses where $AA^{-1}=A^{-1}A=I$, with the pseudoinverse: $AA^{*}\neq A^{*}\neq AI$
• Can compress a rank-deficient matrix down to a size where it has a true inverse(e.g., via PCA), then the project back to the full space
• The Moore-Penrose pseudoinverse is unique, but there are other pseudoinverses

4. Example

4.1 Invertible example

The matrix $A$ is invertible. To check this, one can compute that $det\: A=-\frac{1}{2}$, which is non-zero

4.2 Non-invertible Example(Singular)

The determinant of $B$ is 0, which is a necessary and sufficient condition for a matrix to be non-invertible

5. References

https://en.wikipedia.org/wiki/Invertible_matrix