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Linear Algebra/Linear Dependance of Columns

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Let C1, C2, C3, ..., Cn be n columns of m numbers Cn=[a1na2na3n⋮amn].

A linear combination of columns n1C1+n2C2+n3C3+...+nnCn is the column

Cn=[c1c2c3⋮cn].

Where ck=n1ak1+n1ak1+n2ak2+n3ak3+...+nnakn.

Theorem

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If there is a determinant of order n which is A=aij, and there are n columns of n elements such that the ith entry of the jth column is equal to aij, then if one of the columns is a linear combination of the other columns, then the determinant is equal to 0.

Proof

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Suppose that the kth column is a linear combination of the other column,

[a11a12a13…c1a11+c2a12+c3a13+…+cna1n…a1na21a22a23…c1a21+c2a22+c3a23+…+cna2n…a2na31a23a33…c1a31+c2a32+c3a33+…+cna3n…a3n⋮⋮⋮⋮⋮⋮⋮an1an3an3…c1an1+c2an2+c3an3+…+cnann…ann].

Then by the linearity of determinants, the determinant is equal to

c1[a11a12a13…a11…a1na21a22a23…a21…a2na31a23a33…a31…a3n⋮⋮⋮⋮⋮⋮⋮an1an3an3…an1…ann]+c2[a11a12a13…a12…a1na21a22a23…a22…a2na31a23a33…a32…a3n⋮⋮⋮⋮⋮⋮⋮an1an3an3…an2…ann]+c3[a11a12a13…a13…a1na21a22a23…a23…a2na31a23a33…a33…a3n⋮⋮⋮⋮⋮⋮⋮an1an3an3…an3…ann]+…+cn[a11a12a13…a1n…a1na21a22a23…a2n…a2na31a23a33…a3n…a3n⋮⋮⋮⋮⋮⋮⋮an1an3an3…ann…ann].

Since all of those matrices have repeat columns, their determinants are 0, and so their sum is 0.

Rank of a Matrix

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The rank of a matrix is the maximum order of a minor that does not equal 0. The minor of a matrix with the order of the rank of the matrix is called a basis minor of the matrix, and the columns that the minor includes are called the basis columns.