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Find the matrix $A$ of the linear transformation

$$T(f(t)) = \int_{-6}^4 f(t) \ dt$$

from $P_3$ to $\mathbb{R}$ with respect to the standard bases for $P_3$ and $\mathbb{R}$.

$$A = \left[ \ \ \color{red}{\square} \quad \color{red}{\square} \quad \color{red}{\square} \quad \color{red}{\square} \ \ \right]$$

I'm a little confused about what $P_3$ stands for and where $A$ is coming from.

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Most likely, the vector space of polynomials of degree 3 or less. – David Mitra Mar 28 '12 at 16:46
up vote 3 down vote accepted

$P_{3}$ is the vector space over $\mathbb{R}$ of polynomials with degree at most $3$. This is a vector space, as you can add polynomials together, and multiply them by constants. The standard basis for $P_{3}$ is

$$\mathcal{B} = \{1,t,t^{2},t^{3}\}$$

To find the matrix, apply the transformation $T$ to the basis vectors, to get

$$A = \Big[ \int_{-6}^{4} 1 dt, \int_{-6}^{4} t dt, \int_{-6}^{4} t^{2} dt, \int_{-6}^{4} t^{3} dt \Big]$$

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The notation $P_n$ in this context denotes the vector space of polynomials of degree atmost $n$ with coefficients in $\mathbb R$.

$$P_n:=\{a_0+a_1x+\cdots+a_nx^n \mid a_0,a_1, \cdots, a_n \in \Bbb R\}$$

The matrix $A$ is the matrix of transformation associated with $T:P_n \to \Bbb R$ given by $$T(f(t))=\int_{-6}^4 f(t) \rm dt$$ where $P_n$ and $\Bbb R$ are considered as vector spaces with standard basis.

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