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I am having trouble with this problem. I have to find the matrix representation of a linear transformation. The example in my book got me my answer below but I do not feel that it is right/sufficient. Can someone explain matrix representation of a linear transformation?

Given $P_2(x)$ and $P_3(x)$ and the linear transformation: $L:P_2(x)\rightarrow P_3(x)$ defined by $L(p(x)) = \displaystyle \int p(x)dx$. Find the matrix representation $A$ of the linear transformation $L$. Then find the rank of $A$ and the null space of $A$.

Here is what I have: $$A = \begin{bmatrix}0&1&0\\ 0&0&2\\ 0&0&0\end{bmatrix}$$

$R(A)$ = 2

$N(A)$ = 1

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Do $P_2$ and $P_3$ have the same dimension? If not, then why is this matrix square? You should say what $P_2$ and $P_3$ are, also. –  Dylan Moreland Dec 5 '11 at 20:10
    
The indefinite integral is indefinite. What exactly is this mapping that takes polynomials of degree $\le 2$ to polynomials of degree $\le 3$? Integral from where to where? –  André Nicolas Dec 5 '11 at 20:11
    
But indefinite integral is only defined up to a constant. How is it fixed in your definition ? –  Sasha Dec 5 '11 at 20:12
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They do. A matrix would need to have $4$ rows. Your matrix comes from a problem about the differentiation operator. Idea can be copied, matrix can't. –  André Nicolas Dec 5 '11 at 20:15
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That matrix $A$ looks more like the matrix of the derivation. –  Andrea Mori Dec 5 '11 at 20:19
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2 Answers

up vote 0 down vote accepted

choose a basis for the polynomial spaces, say $\{1,x,x^2\}$ and $\{1,x,x^2,x^3\}$. then integration $\int_0^xp(t)dt$ takes the basis for $P_2$ to $x,x^2/2,x^3/3$. in terms of vectors $$ (1,0,0)\mapsto(0,1,0,0), (0,1,0)\mapsto(0,0,1/2,0), (0,0,1)\mapsto(0,0,0,1/3) $$ so you get the matrix (wrt these bases) $$ \left( \begin{array}{ccc} 0&0&0\\ 1&0&0\\ 0&1/2&0\\ 0&0&1/3\\ \end{array} \right) $$

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This is when you "normalize" $L$ imposing that $L(P)(0)=0$. –  Andrea Mori Dec 5 '11 at 20:21
    
But where did you get the integration from 0 to x? Where does this come from? –  jmendegan Dec 5 '11 at 20:24
    
@jmendegan i imposed it for the question to make sense –  yoyo Dec 5 '11 at 20:27
    
$L(0) = 0$ is forced upon you by linearity, isn't it? So this should be fine. –  Dylan Moreland Dec 5 '11 at 20:30
    
Can you explain the null space of this matrix? –  jmendegan Dec 6 '11 at 2:18
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Adding to @yoyo answer.

Mechanical way is:

$x \mapsto y$
0: $(1,0,0)\mapsto(0,1,0,0)$
1: $(0,1,0)\mapsto(0,0,1/2,0)$
2: $(0,0,1)\mapsto(0,0,0,1/3)$
.
.
.
n: . . .

trannsformation required = $\displaystyle \sum\limits_0^n y^Tx$

Every $y^Tx$ will give you $4 \times 3$ matrix and you add them up.

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