Prove that $1+t^2$, $t+t^2$, $1+2t+t^2$ is a basis for the space of quadratic polynomials $P^{(2)}$.
I have worked it out to the point where I have the following: $(1+t^2)(1, 0, 1)^T +(t+t^2)(1,1,0)^T + (1+2t+t^2)(1,2,1)^t$
Am I on the right track? If so, what should my next step be?
On
You should know, that $\{1,t,t^2\}$ is a basis for the space of quadratic polynomials.
To show that $\{f_1(t)=(1+t^2),f_2(t)=(t+t^2),f_3(t)=(1+2t+t^2)\}$ is also a basis,
you must show the $f_k$ being linearly independent, which is true if the matrix $A$ defined below is invertible, or equivalently $\text{det}A \ne 0$. \begin{equation} A \left( \begin{matrix} 1\\ t\\ t^2 \end{matrix} \right) := \left( \begin{matrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 1 & 2 & 1 \end{matrix} \right) \left( \begin{matrix} 1\\ t\\ t^2 \end{matrix} \right) = \left( \begin{matrix} 1+t^2\\ t+t^2\\ 1+2t+t^2 \end{matrix} \right) \end{equation}
With the rule of Sarrus (see http://en.wikipedia.org/wiki/Rule_of_Sarrus), one gets \begin{equation} \text{det}A = +1+0+0-2-0-1 = -2 \ne 0.\Box \end{equation}
Hint: If you call $f(t)= 1+t^2,\, g(t) = t+t^2,\, h(t) = 1+ 2t+ t^2$, you need to find $3$ values for $t$, let's say $t_1,\, t_2,\, t_3$, such that:
$$\begin{vmatrix} f(t_1) & f(t_2) & f(t_3) \\ g(t_1) & g(t_2) & g(t_3) \\ h(t_1) & h(t_2) & h(t_3) \end{vmatrix}\neq 0 $$
So, you prove that $f(t),\,g(t),\,h(t)$ are linearly independent.
Plus, you know that $\dim P_2[t]=3$.