Science:Math Exam Resources/Courses/MATH307/April 2006/Question 03 (a)

MATH307 April 2006

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Question 03 (a)
Let $P_{2}=\{\alpha _{0}+\alpha _{1}t+\alpha _{2}t^{2}:\alpha _{j}\in \mathbb {R} (j=0,1,2)\}$ be the set of all polynomials of degree at most 2. Show that $\displaystyle P_{2}$ is a vector space and show that $\displaystyle B=\{1,t,t^{2}\}$ form a basis of $\displaystyle P_{2}$. Express the function $\displaystyle f(t)=1-3t+2t^{2}\in P_{2}$ in a vector form using B as the basis.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If $P_{2}$ is a vector space then the axioms must hold true $\displaystyle \alpha _{0}+\alpha _{1}t+\alpha _{2}t^{2}+\beta _{0}+\beta _{1}t+\beta _{2}t^{2}=(\alpha _{0}+\beta _{0})+(\alpha _{1}+\beta _{1})t+(\alpha _{2}+\beta _{2})t^{2}$ $\displaystyle c(\alpha _{0}+\alpha _{1}t+\alpha _{2}t^{2})=c\alpha _{0}+c\alpha _{1}t+c\alpha _{2}t^{2}$ We can also think of $P_{2}$ to be in $\mathbb {R} ^{3}$ $\alpha _{0}+\alpha _{1}t+\alpha _{2}t^{2}\rightarrow {\begin{bmatrix}\alpha _{0}\\\alpha _{1}\\\alpha _{2}\end{bmatrix}}$ such that commutative and associate properties are also satisfied. To show that B is a basis, we have to show that it spans $P_{2}$ and are linearly independent If we think of $P_{2}$ to be in $\mathbb {R} ^{3}$ then we can immediately identify the basis for $\mathbb {R} ^{3}$ Basis of $\mathbb {R} ^{3}:\left\{{\begin{bmatrix}1\\0\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\0\end{bmatrix}},{\begin{bmatrix}0\\0\\1\end{bmatrix}}\right\}$ We can use this result and build the basis of $P_{2}$ such that $B={\begin{bmatrix}1\\0\\0\end{bmatrix}}+t{\begin{bmatrix}0\\1\\0\end{bmatrix}}+t^{2}{\begin{bmatrix}0\\0\\1\end{bmatrix}}={\begin{bmatrix}1\\t\\t^{2}\end{bmatrix}}$. Therefore $f(t)={\begin{bmatrix}1&-3&2\end{bmatrix}}{\begin{bmatrix}1\\t\\t^{2}\end{bmatrix}}=1-3t+2t^{2}$