The base $B\subseteq \mathbb{R}[t]_{\leq 2}$ of the space of polynomials of degree less or equal $2$ is given.$$B=\{2t^2+4t+2,t+3,t^2+5t+3\}$$Display the following vector as a linear combination of the base $B$: $8t^2+2\quad \in\mathbb{R}[t]_{\leq 2}$
Do I need to find $\lambda_1,\lambda_2,\lambda_3\in\mathbb{R}$ with $$\lambda_1(2t^2+4t+2)+\lambda_2(t+3)+\lambda_3(t^2+5t+3)=8t^2+2$$or Am I completely mistaken? Can you please give a hint on how to go on?
You're on the right track. To continue, just match the coefficients of the left and right side: $$(2\lambda_1 + \lambda_3) t^2 + (4\lambda_1 + \lambda_2 + 5\lambda_3) t + (2\lambda_1 + 3 \lambda_2 + 3 \lambda_3) = 8t^2 + 0t + 2$$ You can pull three equations in three variables ($\lambda_1, \lambda_2, \lambda_3$) from this one expression. Can you do that?