I need to find the matrix corresponding to the linear map $f:V_3 \rightarrow V_3$, where $V_3$ is the vector space of all polynomials of degree less than or equal to 3, $$f(p(X))=p(X)-p'(X)$$,

with regards to the bases $ B=\{1,X,X^2,X^3\}$ and $C=\{1, X-1, X^2-X,X^3-X^2\}$.

My attempt:

$f(p(X)) = \Sigma_{j=0}^3 \alpha_j X^j + \Sigma_{j=1}^3 j\alpha_jX^{j-1} = (\alpha_0+\alpha_1)1 + (\alpha_1+2\alpha_2)X+(\alpha_2+3\alpha_3)X^2+\alpha_3X^3$

Now I need to come up with the values for $f(1),f(X),f(X^2),f(X^3)$ and express those values in terms of the basis C.

So e.g. for $f(1)=(\alpha_0+\alpha_1)1$ which expressed in terms of basis C is: $(\alpha_0+\alpha_1)c_0$

$f(X)=(\alpha_1+2\alpha_2)X$ which in terms of C is: $(\alpha_1+2\alpha_2)c_1+(\alpha_1+2\alpha_2)c_0$

Continuing in this way I get the matrix:

$\left( \begin{array}{cc} \alpha_0+\alpha_1 & \alpha_1+2\alpha_2 & \alpha_2+3\alpha_3 & \alpha_3\\ 0 & \alpha_1+2\alpha_2 & \alpha_2+3\alpha_3 & \alpha_3 \\ 0 & 0 & \alpha_2+3\alpha_3 & \alpha_3 \\ 0 & 0 & 0 & \alpha_3 \end{array} \right)$

Now I'm not sure whether this is correct? I guess if I would want to test this, I would need to plug in some random polynomial of degree less than or equal to see and see if the result equals $p(X)-p'(X)$, which it does not if I try to, so the matrix is probably incorrect.

One more question. Does it even make sense to use bases other than the standard basis?

  • $\begingroup$ It sounds correct. Of course it makes sense to use other bases than the standard basis. They're just non-standard bases! $\endgroup$ Jan 18 '14 at 12:14
  • $\begingroup$ Related. $\endgroup$
    – Git Gud
    Jan 18 '14 at 12:18
  • $\begingroup$ Thanks for your answer. If I would want to test the correctness of a matrix, how would I go about doing that? If I'd use a polynomial such as $4x^2+3x+2$, I should get $8x+3$ as a result after plugging it in the matrix. But if I plug it in I get a column vector with 4 elements, don't I? $\endgroup$ Jan 18 '14 at 12:21

One thing looks wrong to me off the bat: with respect to both bases, the first column should represent the image of $1$. That is $1 - (1)' = 1-0=1$. So the first column should just have a 1 in the top entry and nothing else.

(If you're not familiar, the matrix of a linear transformation $T$ with respect to an ordered basis $v_1, \dotsc, v_4$ is $T(v_j) = \sum a_{ij}v_i$. So the first column should be such that $T(v_1)=\sum a_{i1}v_i$.

Also, in general, the matrix you end up with should not be filled with $\alpha$'s...it should just be numbers from the field (in this case, real numbers).

  • $\begingroup$ Thanks for the answer. One thing I don't understand is, how you would get numbers from the field inside the matrix if the matrix is for general polynomials of degree $\leq3$? $\endgroup$ Jan 18 '14 at 12:36
  • $\begingroup$ @eager2learn They are the numbers $a_{ij}$ such that $T(v_j)=\sum a_{ij}v_i$. You know that given a basis for $V_3$, each element in $V_3$ has a unique expression with respect to that basis. That is, it has a unique $4$-tuple of coefficients for the basis elements. Those are your real numbers. For example, with respect to the basis $C$, the element $2x^2-2x+1$ has the unique expression $(1)\cdot 1 + (0)\cdot (X-1) + (2)\cdot (X^2-X) + (0)\cdot (X^3-X^2)$. So your real numbers are $(1,0,2,0)$. $\endgroup$
    – Eric Auld
    Jan 18 '14 at 12:45

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