Linear Algebra Math

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 3×3 Case. To get the first element in our solution matrix c₁₁

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

Where a_ij and b_ij are from matrices A, B respectively.

trace(A)

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

A^T

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

For example:

A = a b c d e f g h i A^T = a d g b e h c f i

det(A)

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 (±)a_1j₁ a_2j₂. . .a_{nj_n}

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

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

Example: In our 3×3 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 3×3 matrix would have 6 calculations (3!) to make, whereas a 20×20 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.

adj(A)

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

The cofactor of an element a_ij from matrix A is :

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

inv(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)I_n

Which when manipulated gives you:

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

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Linear Algebra Math

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