Linear Algebra/Linear Dependence of Columns

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Let C1, C2, C3, ..., Cn be n columns of m numbers Cn=[a1na2na3namn].

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

Cn=[c1c2c3cn].

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

Theorem

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

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

[a11a12a13c1a11+c2a12+c3a13++cna1na1na21a22a23c1a21+c2a22+c3a23++cna2na2na31a23a33c1a31+c2a32+c3a33++cna3na3nan1an3an3c1an1+c2an2+c3an3++cnannann].

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

c1[a11a12a13a11a1na21a22a23a21a2na31a23a33a31a3nan1an3an3an1ann]+c2[a11a12a13a12a1na21a22a23a22a2na31a23a33a32a3nan1an3an3an2ann]+c3[a11a12a13a13a1na21a22a23a23a2na31a23a33a33a3nan1an3an3an3ann]++cn[a11a12a13a1na1na21a22a23a2na2na31a23a33a3na3nan1an3an3annann].

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

Rank of a Matrix

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.

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