Basis for the Space of Quadratic Polynomials $P^{(2)}$ -- Homework Help 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? 
 A: 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}
A: 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 $\dim P_2[t]=3$.
