Answer
The columns of the matrix are linearly independent and hence they form the basis for $P_{3}$.
The given Laguerre polynomials form a basis for $P_{3}$.
Work Step by Step
Given the polynomials,
$1,1 - t , 2 - 4 t + t ^ { 2 } , \text { and } 6 - 18 t + 9 t ^ { 2 } - t ^ { 3 }$
We are required to show that these polynomials form a basis of $P_{3}$
Let the given subspace be represented as:
$\begin{pmatrix}1\\1-t\\2-4t+t^{2}\\6-18t+9t^{2}-t^{3}\end{pmatrix}=t^{0}\begin{bmatrix}1\\1\\2\\6\end{bmatrix}+t^{1}\begin{bmatrix}0\\-1\\-4\\-18\end{bmatrix}+t^{2}\begin{bmatrix}0\\0\\1\\9\end{bmatrix}+t^{3}\begin{bmatrix}0\\0\\0\\-1\end{bmatrix}$
Writing in matrix form we have;
Let, $\mathbf{A}=\begin{bmatrix}{ 1 } & { 0 } & { 0 } & { 0 } \\ { 1 } & { -1 } & { 0 } & { 0 }\\2&-4&1&0\\6&-18&9&-1\end{bmatrix}$
Row reducing the augmented matrix to echelon form:
$\mathbf{A}=\begin{bmatrix}{ 1 } & { 0 } & { 0 } & { 0 } \\ { 1 } & { -1 } & { 0 } & { 0 }\\2&-4&1&0\\6&-18&9&-1\end{bmatrix}\sim\begin{bmatrix}{ 1 } & { 0 } & { 0 } & { 0 } \\ { 0 } & { 1 } & { 0 } & { 0 }\\0&0&1&0\\0&0&0&1\end{bmatrix}$
The columns of the matrix are linearly independent and hence they form the basis for $P_{3}$.
Therefore, The given Laguerre polynomials form a basis for $P_{3}$.
