# linear algebra kernel image and bases

I know this is a long question but any help would be much appreciated. After the question I've attempted as much as I can and need further help if possible on the rest. Thanks in advance, it's much appreciated.

Let $\mathcal{P}_2(\Bbb R)$ denote the vector space of real polynomial functions of degree less than or equal to two, and let $B := [p_0, p_1, p_2]$ denote the natural ordered basis for $\mathcal{P}_2(\Bbb R)$ (so $p_i(x) = x^i$).

Define $g \in \mathcal{P}_2(\Bbb R)$ by $g(x) = 3x^2 - x + 2$. Write $g$ as a linear combination of the elements of $B$. Compute the coordinate vector $g_B$ of $g$ with respect to $B$.

Define $h_1, h_2, h_3 \in \mathcal{P}_2(\Bbb R)$ by $h_1(x) = x^2+3x-2$, $h_2(x) = x^2-x+1$, and $h_3(x) = x^2-5x-1$. Define $C := [h_1, h_2, h_3]$. Assuming $C$ is an ordered basis for $\mathcal{P}_2(\Bbb R)$, construct the change of coordinate matrix, $A$, which converts $C$-coordinates to $B$-coordinates. Compute $A^{-1}$.

Let $M$ denote the change of coordinate matrix that converts $B$-coordinates to $C$-coordinates. How are $A$ and $M$ related? Explain why the calculations you have performed show that $C$ is a basis for $\mathcal{P}_2(\Bbb R)$. Compute the coordinate vector $g_C$ of $g$ with respect to $C$.

This is what I have so far:

$$A^{-1}= \frac{1}{20} \left(\begin{array}-4&2&6\\8&1&13\\-4&-3&1\end{array}\right).$$

The matrices $A$ and $M$ are inverses of eachother. The fact that $A$ is invertible means that the columns of $A$ form a basis for $\Bbb R^3$, confirming the fact that $B$ is a basis for $\Bbb R^3$. $g_C$ = = $( \frac{2}{5}, \frac{27}{10}, \frac{-1}{10})$

The next part of the question is:

Let $F$ : $\mathcal{P}_2(\Bbb R)$ $\rightarrow$ $\mathcal{P}_3(\Bbb R)$ be the linear transformation determined by: $F(f)(x) = \int ^{3x+1} _{x-1} f(t) dt.$

Compute the dimension of the kernel of $F$. Determine a basis for the image of $F$. Define $A := [p_0, p_1, p_2, p_3]$ and compute $M^A_ B (F)$, the matrix of $F$ with respect to the given ordered bases. Determine the rank of $M^A_ C (F)$.

If anyone could show what to do with this, I would really apreciate it, thank you.

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Well, $M_B^A(F)=\bigl(F(p_0)\: F(p_1)\: F(p_2)\bigr)$, where the $F(p_j)$ are expressed as column vectors with $A$-coordinates. In particular, $$F(p_0)(x)=\int_{x-1}^{3x+1}dt=(3x+1)-(x-1)=2x+2=2p_0+2p_1,$$ $$F(p_1)(x)=\int_{x-1}^{3x+1}tdt=\frac{1}{2}\bigl[(3x+1)^2-(x-1)^2\bigr]=4x^2+4x=4p_1+4p_2,$$ \begin{align}F(p_2)(x)&=\int_{x-1}^{3x+1}t^2dt=\frac{1}{3}\bigl[(3x+1)^3-(x-1)^3\bigr]\\&=\frac{26}{3}x^3+10x^2+2x+\frac{2}{3}=\frac{2}{3}p_0+2p_1+10p_2+\frac{26}{3}p_3.\end{align} Thus, $$M_B^A(F)=\left(\begin{array}{ccc}2&0&2/3\\2&4&2\\0&4&10\\0&0&26/3\end{array}\right).$$ This can be determined to be rank 3 by Gauss-Jordan elimination (for example), so since the dimension of $\mathcal{P}_2(\Bbb R)$ is $3$, then the kernel of the linear transformation $F$ is $0$ by Dimension Theorem. A basis for the image of $F$ would be the columns of $M_B^A(F)$.
Everything else looks good to me, except I think you mean "confirming the fact that $C$ is a basis for $\Bbb R^3$", instead of what you put.
thank you so much, this is exactly what i needed. thanks so much. yes i did mean $C$ is a basis for $R^3$. thank you – user34742 Jul 14 '12 at 13:24