2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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
0
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.