91影视

Consider the set of all polynomial such that .

The set$\left\{1,t,t_2,...,t_n\right\}$ is linear independent setV of if there exist constant${\mathit{a}}_{\mathbf{i}}\mathbf{鈭�}\mathit{R}$such that${\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}{\mathit{t}}_{\mathbf{1}}\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\mathit{t}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{a}}_{\mathbf{n}}{\mathit{t}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$where ${\mathit{a}}_{\mathbf{0}}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$.

Any polynomial $f\left(t\right)鈭�P_2$is defined as follows.

$f\left(t\right)=b_0+b_{1}t+b_2{t}^2$

Substitute the value 1 fort in the equation$f\left(t\right)=b_0+b_{1}t+b_2{t}^2$as follows.

$$\begin{gathered} f\left(t\right)=b_0+b_{1}t+b_2{t}^2 \\ f\left(1\right)=b_0+b_1+b_2 \end{gathered}$$

As $f\left(1\right)=0$, substitute the value 0 for$f\left(1\right)$in the equation$f\left(1\right)=b_0+b_1+b_2$as follows.

$\begin{array}{l}f\left(1\right)=b_0+b_1+b_2\\ b_0=-\left(b_1+b_2\right)\end{array}$

Substitute the value $-\left(b_1+b_2\right)$for in the equation $f\left(t\right)=b_0+b_{1}t+b_2{t}^2$as follows.

$\begin{array}{l}f\left(t\right)=b_0+b_{i}t+b_2{t}^2\\ f\left(t\right)=-(b_1+b_2)+b_{1}t+b_2{t}^2\\ f\left(t\right)=b_1\left(t-1\right)+b_2\left(t^2-1\right)\end{array}$

From the equation, the set $\left\{(t-1),(t^2-1)\right\}$span$P_2$ if $f\left(1\right)=0$.

Compare the equations $b_1(t-1)+b_2(t^2-1)=0$ both side as follows.

$$\begin{gathered} b_1\left(t-1\right)+b_2\left(t^2-1\right)=0 \\ {b}_1\left(t-1\right)+b_2\left(t^2-1\right)=0\left(t-1\right)+\left(0\right)\left(t^2-1\right) \\ {b}_i=0 \end{gathered}$$

By the definition of linear independence, the subset $\left\{(t-1),(t^2-1)\right\}$is L.I.

Therefore, the set of all polynomial$P^2$ such that$f\left(1\right)=0$ has dimension 2 and spanned by $Span\left\{(t-1),(t^2-1)\right\}$.