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\\[\\left( \\begin{matrix}\\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\end{matrix} \\right.\\] | \n[[0]] | \n[[1]] | \n[[2]] | \n$0$ | \n$0$ | \n\\[\\left) \\begin{matrix}\\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.}\\\\ \\phantom{.} \\\\ \\phantom{.}\\\\ \\phantom{.}\\\\\\end{matrix} \\right.\\] | \n
[[3]] | \n[[4]] | \n[[5]] | \n[[6]] | \n$0$ | \n||
$0$ | \n[[7]] | \n[[8]] | \n[[9]] | \n[[10]] | \n||
$0$ | \n$0$ | \n[[11]] | \n[[12]] | \n[[13]] | \n||
$0$ | \n$0$ | \n$0$ | \n[[14]] | \n[[15]] | \n
Let $P_n$ denote the vector space over the reals of polynomials $p(x)$ of degree $n$ with coefficients in the real numbers.
\nLet the linear map \\[\\phi: P_4 \\rightarrow P_4 \\] be defined by:
\n\\[\\phi(p(x)) = \\simplify[all,!collectnumbers]{{a} * p(x) + ({b} * x + {c}) * p'(x) + (x ^ 2 + {d} * x + {f}) * p''(x)}\\]
\nwhere $p'(x)$ is the first derivative of $p(x)$ and $p''(x)$ the second derivative.
", "tags": ["basis", "checked2015", "linear map", "linear spaces", "MAS2223", "matrix", "matrix given by a basis", "matrix of a linear map", "poynomials", "vector spaces"], "rulesets": {}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "13/02/2013:
\nAdvice to be completed.
", "licence": "Creative Commons Attribution 4.0 International", "description": "Let $P_n$ denote the vector space over the reals of polynomials $p(x)$ of degree $n$ with coefficients in the real numbers.
\nLet the linear map $\\phi: P_4 \\rightarrow P_4$ be defined by:
\n$\\phi(p(x))=ap(x) + (bx + c)p'(x) + (x ^ 2 + dx + f)p''(x)$
\nUsing the standard basis for range and domain find the matrix given by $\\phi$.
"}, "advice": "We have:
\n\\[\\phi(1) =\\simplify[all,!zerofactor,!collectNumbers,!constantsfirst,!noleadingminus]{ {a} * 1 + ({b} * x + {c}) * 0 + (x ^ 2 + {d} * x + {f}) * 0 = {a} = {a} * 1 + 0 * x + 0 * x ^ 2 + 0 * x ^ 3 + 0 * x ^ 4}\\]
\ngives the first column of the matrix.
\n\\[\\phi(x) = \\simplify[all,!zerofactor,!collectNumbers,!constantsfirst,!noleadingminus]{{a} * x + ({b} * x + {c}) * 1 + (x ^ 2 + {d} * x + {f}) * 0 = {c} + {a + b} * x = {c} * 1 + {a + b} * x + 0 * x ^ 2 + 0 * x ^ 3 + 0 * x ^ 4}\\]
\ngives the second column of the matrix.
\n\\[\\phi(x ^ 2) = \\simplify[all,!zerofactor,!collectNumbers,!constantsfirst,!noleadingminus]{{a} * x ^ 2 + ({b} * x + {c}) * 2 * x + (x ^ 2 + {d} * x + {f}) * 2 = {2 * f} + {2 * d + 2 * c} * x + {a + 2 * b + 2} * x ^ 2 = {2 * f} * 1 + {2 * d + 2 * c} * x + {a + 2 * b + 2} * x ^ 2 + 0 * x ^ 3 + 0 * x ^ 4}\\]
\n\n
gives the third column of the matrix.
\nContinuing on in this way for $\\phi(x^3),\\;\\phi(x^4)$ we obtain the matrix for $\\phi$ with respect to the given bases for domain and range.
\n\\[\\begin{pmatrix}\\var{a}&\\var{c}&\\var{2*f}&0&0\\\\0&\\var{a+b}&\\var{2*d+2*c}&\\var{6*f}&0\\\\0&0&\\var{a+2*b+2}&\\var{3*c+6*d}&\\var{12*f}\\\\0&0&0&\\var{a+3*b+6}&\\var{4*c+12*d}\\\\0&0&0&0&\\var{a+4*b+12}\\end{pmatrix}\\]
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}