Let $P_n$ denote the vector space of all polynomials in one variable with real coefficients and of degreeless than, or equal to,n, equipped with the standard basis {$1,x,x^2,…,x^n$}. Define $T:P_2→P_3$ by $T(p)(x)=∫_0^xp(t)dt+ p’(x)+p(2)$.
Write down the matrix of this transformation with respect to the standard bases of $P_2$ and $P_3$. And determine the dimension of the kernel of the linear transformation $T$ defined above.

After a calculation I find that the matrix will be $$\begin{bmatrix} 1 & 1 & 0 & 0\\ 3 & 0 & 1/2 & 0\\ 4 & 2 & 0 & 1/3 \end{bmatrix}$$

and $\ker T =\{0\}$. Does my calculation alright.

  • $\begingroup$ Looks good to me. $\endgroup$ – Gerry Myerson Jan 20 '13 at 11:11

$$\begin{align*}T(1)&=\int\limits_0^x1\cdot dt+1'+1=x+1\\T(x)&=\int\limits_0^xt\,dt+1+2=\frac{1}{2}x^2+3\\T(x^2)&=\int\limits_0^xt^2\,dt+2x+4=\frac{1}{3}x^3+2x+4\end{align*}$$

The corresponding matrix is thus


which I think fits your answer but you seem to have interchanged rows and columns.This is right if you apply matrices (and functions) from the right, something almost nobody does in these days. If you people apply matrices and functions from the left then your matrix must be $\,4\times 3\,$


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