Question

The first four Laguerre polynomials are \(1,1 - t,2 - 4t + {t^2}\), and \(6 - 18t + 9{t^2} - {t^5}\). Show that these polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

Short answer

It is proved that the first four Laguerre polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

Step-by-step solution

Basis theorem

Theorem 12states that let\(V\)be a p-dimensional vector space;\(p \ge 1\), then anylinearly independent setof exactly \(p\) elements in \(V\) is automatically a basis for \(V\). Any set of exactly \(p\) elements that span \(V\) is automatically a basis for \(V\).

Show that the first four Laguerre polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\)

The columns of the matrix are the coordinate vectors of the Laguerre polynomials corresponding to the standard basis\(\left\{ {1,t,{t^2},{t^3}} \right\}\)of\({{\mathop{\rm P}\nolimits} _3}\). Thus,

\(A = \left[ {\begin{array}{*{20}{c}}1&0&2&6\\0&{ - 1}&{ - 4}&{ - 18}\\0&0&1&9\\0&0&0&{ - 1}\end{array}} \right]\)

There are four pivot columns in the matrix; so its columns are linearly independent. The Laguerre polynomials themselves are linearly independent in\({{\mathop{\rm P}\nolimits} _3}\)because the coordinate vectors of Laguerre polynomials form a linearly independent set. The basis theorem asserts that the Laguerre polynomials form a basis for\({{\mathop{\rm P}\nolimits} _3}\)because there are four Hermite polynomials and\(\dim {{\mathop{\rm P}\nolimits} _3} = 4\).

Thus, it is proved that the first four Laguerre polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).