where mkaz talks tech

Linear Algebra Math

Posted on Aug 1, 1997

Here are some of the definitions and examples used in Linear Algebra and specifically the linear algebra calculators available.

Let A, B, and C represent n x n matrices.

Example of a 3 x 3 Matrix:

1  2  3
1  1  2
0  1  2

A + B = C
The addition of two matrices is straight forward. You just add each matrix position-wise. So the upper-left element of matrix A plus the upper-left element of matrix B is the upper-left element in matrix C. Do the same for all elements.

A x B = C
The multiplication of two matrices is not quite as simple. First we need the matrices to be of proper size. This means matrix A size n x m must be multiplied by a m x p matrix. The resultant matrix will then be n x p. For our case, we are using n x n matrices, so this isn’t a problem.

The equation for multiplying two matrices is : (elementwise)

    [AB]ij = SIGMA [A]ik[B]kj

Where the SIGMA summation goes from k=1…n

A example element from our 3x3 Case. To get the first element in our solution matrix c11

c11 = (a11 * b11) + (a12 *
      b21) + (a13 * b31)

Where aij and bij are from matrices A, B respectively.

The trace of a matrix is simply the summation of its main diagonal.

The transpose of a matrix is switching the rows and columns.

For example:

A = a b c
d e f
g h i
AT = a d g
b e h
c f i

The determinant of a matrix is not quite simple. For a n x n matrix the definition of the determinant is as follows :

det(A) = SIGMA (±)a1j1 a2j2. . .anjn

Where SIGMA is our summation over all permutations j1 j2 … jn of the set S={1, 2, …, n }.

The sign is + or - according to whether the permutation is even or odd.

Example: In our 3x3 case it is a little easier, and boils down to :

det(A) = aei + cdh + bfg - ceg - bdi - afh 

Where are matrix first row is a b c , 2nd row d e f, and 3rd row, g h i

Calculation Technique: For the n x n the calculation of the determinant, by definition, is based upon a factorial number of calculations with respect to the size of the matrix. ie. a 3x3 matrix would have 6 calculations (3!) to make, whereas a 20x20 matrix would have 2.43 x 10^18 calculations (20!).

So instead of brute forcing the calculations, I first do some operations on the matrix, which converts it to a upper triangular matrix, and then calculate the determinant by multiplying down the diagonal, since everything below is 0, this will give the determinant.

The adjoint of A is the transpose of the matrix whose ith, and jth element is the cofactor Aij of the aij element from matrix A.

The cofactor of an element aij from matrix A is :

  aij = (-1)i + j * det (A'), where A' is the matrix obtained from "omitting" the ith and jth rows, of matrix A.

The inverse of A is the matrix which when multiplied to A returns the identity matrix.

Calculation Technique: The inverse was obtained using the Theorem:

Aadj(A) = det(A)In

Which when manipulated gives you:

A-1 = (1 / det(A)) * adj(A)