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

MATH307 April 2006

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Question 03 (c)
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. Find the transformation matrix A such that $T(f)=Af$ using the B as the basis for $P_{2}$

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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. Given that the basis is $B={\begin{bmatrix}1\\t\\t^{2}\end{bmatrix}}$ If $f(t)=f\cdot B$ Then an arbitrary vector is $f={\begin{bmatrix}\alpha _{0}\\\alpha _{1}\\\alpha _{2}\end{bmatrix}}$ $T(f)=(1+t){\frac {\mathrm {d} u}{\mathrm {d} x}}$ The $(1+t)$ applies to all terms of ${\frac {\mathrm {d} u}{\mathrm {d} x}}$ can write the matrix A to be: $A=(1+t)Df$ where D is the differentiation matrix $D{\begin{bmatrix}\alpha _{0}\\\alpha _{1}\\\alpha _{2}\end{bmatrix}}={\begin{bmatrix}\alpha _{1}\\2\alpha _{2}\\0\end{bmatrix}}$ Therefore $D={\begin{bmatrix}0&1&0\\0&0&2\\0&0&0\end{bmatrix}}$ Now we observe the linear trasnformation $(1+t)$ on ${\begin{bmatrix}\alpha _{1}\\2\alpha _{2}\\0\end{bmatrix}}$ $(1+t)(\alpha _{1}+2\alpha _{2}t)=(\alpha _{1}+(\alpha _{1}+2\alpha _{2})t+2\alpha _{2}t^{2})$ $(1+t){\begin{bmatrix}\alpha _{1}\\2\alpha _{2}\\0\end{bmatrix}}={\begin{bmatrix}\alpha _{1}\\2\alpha _{2}+\alpha _{1}\\2\alpha _{2}\end{bmatrix}}$ therefore $(1+t)$ can be rewritten as ${\begin{bmatrix}1&0&0\\1&1&0\\0&1&0\end{bmatrix}}$ The transformation is therefore $(1+t)D={\begin{bmatrix}1&0&0\\1&1&0\\0&1&0\end{bmatrix}}{\begin{bmatrix}0&1&0\\0&0&2\\0&0&0\end{bmatrix}}={\begin{bmatrix}0&1&0\\0&1&2\\0&2&0\end{bmatrix}}$