3

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

Sasha
  • 70,631
jmendegan
  • 231
  • 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 05 '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 05 '11 at 20:11
  • But indefinite integral is only defined up to a constant. How is it fixed in your definition ? – Sasha Dec 05 '11 at 20:12
  • I thought $P_2$ had dim = 3 and $P_3$ had dim = 4????? That is what my notes say. – jmendegan Dec 05 '11 at 20:12
  • 1
    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 05 '11 at 20:15
  • $P_3(x)$ is all polynomials with degree 3 or less and $P_2(x)$ is all polynomials with degree 2 or less. – jmendegan Dec 05 '11 at 20:16
  • 1
    That matrix $A$ looks more like the matrix of the derivation. – Andrea Mori Dec 05 '11 at 20:19
  • @jmendegan The following link may be useful. http://math.stackexchange.com/questions/64907/correspondences-between-linear-transformations-and-matrices/64911#64911 –  Dec 05 '11 at 22:11

2 Answers2

3

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 $$ (1,0,0)\mapsto(0,1,0,0), \quad(0,1,0)\mapsto(0,0,1/2,0), \quad (0,0,1)\mapsto(0,0,0,1/3), $$ so you get the matrix (w.r.t. these bases) $$ \left( \begin{array}{ccc} 0&0&0\\ 1&0&0\\ 0&1/2&0\\ 0&0&1/3\\ \end{array} \right). $$

yoyo
  • 9,943
0

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.

Pratik Deoghare
  • 13,573