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I have am studying Linear Algebra, and one of our homework assignments is this:

Find a matrix representation for the linear operator $F = \frac d{dx}$ with the orthonormal basis

$$Y_0 = \frac 3{\sqrt{5}} \\ Y_1 = \sqrt{\frac 83}x \\ Y_2 = \frac 52(5x^2-1)$$

We are also supposed to use the scalar product:

$$\langle F,G\rangle = \int_{-1}^1 F^*G\,dx$$

But I am not sure how to go about this problem. What do I call $F$ and what do I call $G$. Please make the explanations clear, I really want to learn how to work with this stuff. Thanks!

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MoparGuy
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  • I don't know what that integral is supposed to mean, but it doesn't really matter. The matrix representation of an operator has nothing to do with an inner product. You just need to find $FY_0$, $FY_1$, and $FY_2$ in terms of $Y_0$, $Y_1$, and $Y_2$. Alternatively you can find the matrix in the standard basis and then do a change of basis. –  Oct 09 '16 at 19:23
  • Hi Bye World, the (-1,1) is the limits on the integral, and the * is supposed to mean "F star". Sorry about that. Could you explain to me the process for finding the matrix for a standard basis then doing the change of basis? The last part of my question requires me to use the matrix to calculate the derivative of 4x^3 - 4. Thank you for your help. – MoparGuy Oct 09 '16 at 19:36
  • OK. I've fixed the equation in your question. In the future, try to learn some MathJax so you can do it yourself. As for finding the matrix -- you'll find $FY_0$, $FY_1$, and $FY_2$ in terms of the basis ${1,x,x^2}$. Then your standard matrix will be the coeff's of those vectors as the columns. The change of basis matrices will be $\pmatrix{Y_0 & Y_1 & Y_2}$ (again, coeff's as columns) and the inverse of this matrix. Then just multiply $\Lambda^{-1} M\Lambda$. –  Oct 09 '16 at 19:38
  • For more info on change of basis, you can read my answers to these questions: 1, 2. –  Oct 09 '16 at 19:44
  • Hi Bye World, Thank you for your help. I have bookmarked the MathJax link so I can begin learning it. Can you explain to me how the integral comes into play with this at all? It seems like it is not even needed then, right? Thanks. – MoparGuy Oct 09 '16 at 19:51
  • It's not needed. –  Oct 09 '16 at 19:52

1 Answers1

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Apply the linear operator to the basis vectors, these will be the column of your matrix $$ (y_0)'=0\\ (y_1)'=\sqrt{\frac{8}{3}}=\frac{\sqrt{40}}{3^{3/2}}y_0\\ (y_2)''=25x=25\sqrt{\frac{3}{8}}y_1 $$ yielding $$ \begin{bmatrix}0&\frac{\sqrt{40}}{3^{3/2}}&0\\0&0&25\sqrt{\frac{3}{8}}\\0&0&0 \end{bmatrix} $$

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