# 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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If this is homework, please add the tag. –  Siminore Jul 11 '12 at 14:31
I have reformatted your question to make clearer which portions are part of the problem and which are your reasoning and answers. Please make sure I've not made any errors or altered your meaning. –  Cameron Buie Jul 11 '12 at 14:46
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