-1
$\begingroup$

I'm asked to find the matrix that represents the endomorphism $g$ with respect to the basis that I found. I'm told that the endomorphism is defined by

$$\Delta : S_3 \to S_3, \ \ \ \Delta (f)(x)=f''(x)$$

and $S_3$ is the vector space spanned by the basis $$\left\{ \frac{1}{\sqrt 2} , \cos x, \cos 2x, \cos 3x\right\}$$ How should I approach this exactly? I know the orthogonal basis but nothing else. Thanks!

$\endgroup$
  • $\begingroup$ ^And what is your basis? $\endgroup$ – ASKASK Jan 8 '18 at 8:55
  • $\begingroup$ Once you've answered the above two questions, the columns of the matrix representation are the images of the basis (column) vectors by the linear map. $\endgroup$ – Arthur Jan 8 '18 at 8:56
  • $\begingroup$ That's not a basis for the vector space of all functions from $\Bbb R$ to $\Bbb R$. It's way too small. $\endgroup$ – Arthur Jan 8 '18 at 8:59
  • $\begingroup$ @Arthur Yes sorry, I have $S_{N}$ that is generated by the basis $(1, \cos (x), \cos (2x), \cos(3x), ..., \cos(Nx))$ but we're only looking at the subspace $S_{3}$ $\endgroup$ – Niktaneous Jan 8 '18 at 9:04
1
$\begingroup$

Since:

  • $\Delta\left(\frac1{\sqrt2}\right)=0$;
  • $\Delta\bigl(\cos(x)\bigr)=-\cos(x)$;
  • $\Delta\bigl(\cos(2x)\bigr)=-4\cos(2x)$;
  • $\Delta\bigl(\cos(3x)\bigr)=-9\cos(3x)$,

the matrix that you are interested in is $\begin{bmatrix}0&0&0&0\\0&-1&0&0\\0&0&-4&0\\0&0&0&-9\end{bmatrix}$.

$\endgroup$
  • $\begingroup$ Oh, it's that simple? Thank you so much! It's greatly appreciated $\endgroup$ – Niktaneous Jan 8 '18 at 9:30

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.