Geometry Formulary

Inverse of a Matrix

A^{-1} = \frac{1}{det(A)} Adj(A)

Adjugate Matrix

The adjugate of a matrix A is the transpose of the cofactor matrix:
Adj(A) = C^{T}

$(i-j)$-minor (AKA `M_{ij}`)

M_{ij} := Determinant of the matrix B got by removing the $i$-row and the $j$-column from matrix A

Cofactors

C is the matrix of cofactors of a matrix A where all the elements c_{ij} are defined like this:
$ c_{ij} = \left( -1\right)^{i + j}M_{ij} $

Eigenvalues

By starting from the definition of eigenvectors:
A\vec{v} = \lambda\vec{v}

As we can see, the vector \vec{v} was unaffected by this matrix multiplication appart from a scaling factor \lambda, called eigenvalue

By rewriting this formula we get:
$ \left(A - \lambda I\right)\vec{v} = 0 $

This is solved for:
\det(A- \lambda I) = 0

Note

If the determinant is 0, then (A - \lambda I ) is not invertible, so we can't solve the previous equation by using the trivial solution (which can't be taken into account since \vec{v} is not 0 by definition)

Caley-Hamilton

Each square matrix over a commutative ring satisfies its own characteristic equation det(\lambda I - A)

Tip

In other words, once found the characteristic equation, we can substitute the unknown variable \lambda with the matrix itself (known), powered to the correspondent power

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