91Ó°ÊÓ

To prove that the set \(\{e_1, e_2, \ldots, e_n\}\) is linearly independent, we consider a linear combination with coefficients \(c_1, c_2, \ldots, c_n\) such that: \(c_1e_1 + c_2e_2 + \cdots + c_ne_n = \mathbf{0}\). Expanding and rewriting the linear combination as a column-vector equation, we get: \(\begin{pmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}\). This column-vector equation gives us a system of equations where \(c_1 = 0, c_2 = 0, \ldots, c_n = 0\). Since all coefficients are zero, we have proven that the only linear combination of the vectors that equals the zero vector is the trivial linear combination, and thus, the set \(\{e_1, e_2, \ldots, e_n\}\) is linearly independent.