Consequence of linear combination in matrix .

Question

If a column of a matrix is linear combination of another column, what are the consequences ?

Several terminology coming into my mind to relate with this such as

• Rank of the matrix ;
• Determinant of the matrix if $0$ , why ?

Answer 1

To answer your question as to why is the determinant zero in this case, you know that the determinant, when viewed as a function of the matrix' columns is an alternating multilinear functional. If you have a matrix, $A\in\mathbb{R}^{n\times n}$, and if its columns are $a_1,...,a_n\in\mathbb{R}^n$, then $$ \det(a_1,...,a_n) $$ is a function that swaps sign every time you exchange two vectors in it. That also means that if you plug in the same vector twice, the determinant will be zero.

If one of your columns is the linear combination of another column(s), it means that your columns are linearly dependent.

If $a_1,...,a_n$ are linearly dependent, then the linear combination $$ 0=\alpha_1a_1+...+\alpha_na_n $$ is possible in such a way that not all $\alpha_i$s are zero. Let us assume that $\alpha_i$ is nonzero. Then $$ a_i=-\frac{\alpha_1}{\alpha_i}a_1-...-\frac{\alpha_n}{\alpha_i}a_n, $$ with obviously $a_i$ being left out of this sum on the rhs. Then, $$ \det(a_1,...,a_i,...,a_n)=\det\left(a_1,...,-\frac{\alpha_1}{\alpha_i}a_1-...-\frac{\alpha_n}{\alpha_i}a_n,...,a_n\right)=\\=-\frac{\alpha_1}{\alpha_i}\det(a_1,...,a_1,...a_n)-...(\mathrm{similiar\ sums\ to\ this})=0, $$ since one vector will always be featured twice in very determinant.