91Ó°ÊÓ

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} + ... + {x_p}{{\bf{v}}_p} = 0\) has a trivial solution, where \({{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_p}\) are the vectors.

For the matrix equation\(A{\bf{x}} = {\bf{b}}\), the equation has at most one solution when\({\bf{b}} = 0\). Thus, the equation\(A{\bf{x}} = 0\)\(\left( {{\bf{x}} = 0} \right)\)has at most one solution, which means the solution is trivial. A trivial solution implies that the columns are linearly independent.

Consider\(A{\bf{x}} = 0\), where\(A = \left[ {\begin{array}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}& \cdots &{{{\bf{a}}_n}}\end{array}} \right]\), and\({\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right]\).

Thus, it is written as shown below:

\(\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}& \cdots &{{{\bf{a}}_n}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{array}} \right] = 0\\{x_1}{{\bf{a}}_1} + {x_2}{{\bf{a}}_2} + ... + {x_n}{{\bf{a}}_n} = 0\end{array}\)

For the equation\({x_1}{{\bf{a}}_1} + {x_2}{{\bf{a}}_2} + ... + {x_n}{{\bf{a}}_n} = 0\), the equation has\({x_1} = {x_2} = ... = {x_n} = 0\). So, the columns of A are linearly independent.

Hence, theequation\(A{\bf{x}} = {\bf{b}}\)has at most one solution, and thecolumns of A are linearly independent.