Lauren1234 0 Posted January 18 Let A={ex,sin(x),excos(x),sin(x),cos(x)} and let V be the subspace of C(R) equal to span(A). Define T:V→V,f↦df/dx. How do I prove that T is a linear transformation? (I can do this with numbers but the trig is throwing me). 0 Share this post Link to post Share on other sites

studiot 1945 Posted January 18 I think you have missed out an e^{x} before the second sinx. Looking then at your transformation, and noting that the domain and codomain of, are the same, (so the transformation is closed), you can surely test against the linear space axioms by forming the derivatives and sums and showing there are inverses etc. 0 Share this post Link to post Share on other sites

taeto 92 Posted January 19 (edited) 23 hours ago, Lauren1234 said: Define T:V→V,f↦df/dx. How do I prove that T is a linear transformation? The assignment is wrong. It is like asking how to prove that \(f: [0,1]\to [0,1],\, x\mapsto 2x\) defines a linear function \(f\), when \([0,1]\) means the closed interval of real numbers between \(0\) and \(1.\) Edited January 19 by taeto 0 Share this post Link to post Share on other sites

studiot 1945 Posted January 19 1 hour ago, taeto said: The assignment is wrong. It is like asking how to prove that f:[0,1]→[0,1],x↦2x defines a linear function f , when [0,1] means the closed interval of real numbers between 0 and 1. I think perhaps there have been some transciption errors in the OP? 0 Share this post Link to post Share on other sites

taeto 92 Posted January 19 (edited) 14 minutes ago, studiot said: I think perhaps there have been some transciption errors in the OP? Something is wrong, I may not have pinpointed it exactly. The notation is suspect; \(A\) is a set of what kind of elements, and then what is span(\(A\))? It makes sense if we read the OP as \(A = \{\exp,\exp \cos, \exp \sin, \cos, \sin \}\), and taking span(\(A\)) to be the space of all linear combinations with real coefficients of those five functions. Then the only thing to prove is that the image of the span under \(T\) is contained in the span. Hence my original comment. Edited January 19 by taeto 0 Share this post Link to post Share on other sites

Lauren1234 0 Posted January 19 Sorry it should be Let A={e^xsin(x),e^xcos(x),sin(x),cos(x)} 0 Share this post Link to post Share on other sites

taeto 92 Posted January 19 4 hours ago, Lauren1234 said: Sorry it should be Let A={e^xsin(x),e^xcos(x),sin(x),cos(x)} That makes sense. The meaning is that \(A\) is a set of four functions. The role of \(x\) remains a little ambiguous. If we tread lightly, we can surmise that \(x\) is a coordinate, and its range is the reals. Do you see what is the span(\(A\))\(=V\)? Is it true that if a function \(f\) is any one of the four functions in \(A,\) then \(f'\) belongs to \(V?\) And what if \(f\) is any function in \(V?\) 0 Share this post Link to post Share on other sites

joigus 107 Posted May 23 Linearity is one thing. That's inherited from linearity of the derivative operator. Closure is another thing. That's what Studiot and Taeto are talking about, I think. Because e^{x}, e^{x}sin x, e^{x}cos(x), sin(x), cos(x) are meromorphic (analytic in C ==> analytic in all R) I see no problem with the domains. Closure can be assured by inspection. The most involved cases are e^{x}sin(x), e^{x}cos(x). But, \[\frac{d}{dx}e^{x}\sin x=e^{x}\sin x+e^{x}\cos x\in\textrm{span}\left(A\right)\] \[\frac{d}{dx}e^{x}\cos x=e^{x}\cos x-e^{x}\sin x\in\textrm{span}\left(A\right)\] That would be my answer. 0 Share this post Link to post Share on other sites