Question

Use Theorem 7 in section 1.7 to explain why the columns of the matrix Aare linearly independent.

\(A = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\2&5&0&0\\3&6&8&0\\4&7&9&{10}\end{aligned}} \right)\)

Short answer

The columns of matrix \(A\) are linearly independent according to theorem 7.

Step-by-step solution

Denote the columns of matrix A

The columns of matrix \(A\) are denoted from right to left by vectors \({{\mathop{\rm v}\nolimits} _1},...,{{\mathop{\rm v}\nolimits} _4}\).

Check if the columns of matrix A are linearly independent

Theorem 7states that an indexed set \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},...,{v_p}} \right\}\) of two or more vectors islinearly dependentif and only if at least one of the vectors in \(S\) is a linear combination of the others. If \(S\) is linearly dependent and \({{\mathop{\rm v}\nolimits} _1} \ne 0\), then some \({{\mathop{\rm v}\nolimits} _j}\) is a linear combination of the preceding vectors \({{\mathop{\rm v}\nolimits} _1},...,{v_{j - 1}}\).

Vector \({{\mathop{\rm v}\nolimits} _1}\) is non-zero; \({{\mathop{\rm v}\nolimits} _2}\) is not a multiple of \({{\mathop{\rm v}\nolimits} _1}\) (since the third entry of \({{\mathop{\rm v}\nolimits} _2}\)is non-zero), and \({{\mathop{\rm v}\nolimits} _3}\) is not a linear combination of \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\) (since the second entry of \({{\mathop{\rm v}\nolimits} _3}\) is non-zero). Furthermore, \({{\mathop{\rm v}\nolimits} _4}\) cannot be a linear combination of \({{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},\) and \({{\mathop{\rm v}\nolimits} _3}\), based on the first entry in the vector. Thus, the columns are linearly independent according to theorem 7.