Question

The first four Hermite polynomials are \(1,2t, - 2 + 4{t^2},\) and \( - 12t + 8{t^3}\). These polynomials arise naturally in the study of certain important differential equations in mathematical physics. Show that the first four Hermite polynomials form a basis of \({{\mathop{\rm P}\nolimits} _3}\).

Short answer

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

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

\(A = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&0\\0&2&0&{ - 12}\\0&0&4&0\\0&0&0&8\end{array}} \right]\)

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