Let \(A = LU\), where L is an invertible lower triangular matrix and \(U\) is upper triangular. Explain why the first column of \(A\) is a multiple of the first column of \(L\). How is the second column of \(A\) related to the columns of \(L\)?

Here, \(A = LU\) \({{\mathop{\rm col}\nolimits} _1}\left( A \right) = L \cdot {{\mathop{\rm col}\nolimits} _1}\left( U \right)\). \(L \cdot {{\mathop{\rm col}\nolimits} _1}\left( U \right)\) is a linear combination of the columns of L in which all weights, except probably the first, are zero since \({{\mathop{\rm col}\nolimits} _1}\left( U \right)\) contains a zero in every entry. Therefore, \({{\mathop{\rm col}\nolimits} _1}\left( A \right)\) is a multiple of \({{\mathop{\rm col}\nolimits} _1}\left( U \right)\).

Thus, the first column of A is a multiple of the first column of L.

Likewise, \({{\mathop{\rm col}\nolimits} _2}\left( A \right) = L \cdot {{\mathop{\rm col}\nolimits} _2}\left( U \right)\) represents the linear combination of the columns of L using the first two entries in \({{\mathop{\rm col}\nolimits} _2}\left( U \right)\) as weights. It is because other entries in \({{\mathop{\rm col}\nolimits} _2}\left( U \right)\) are zero. It means \({{\mathop{\rm col}\nolimits} _2}\left( A \right)\) is a linear combination of the first two columns of \(L\).

Thus, the second column of A is a linear combination of the first two columns of L.