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T Also if w is in range T, say with v in V and T(v) = w, and c is any scalar, then c v is in V and T(cv) = cT(v) = cw which shows that cw is in range T Consequently, range T is a subspace of W Examples 1 Consider the linear transformation T(x, y, z) = (x 3y 5z, 4x 12y, 2x 6y 8z) To compute the kernel of T we solve T(x, y, z) = 0 This corresponds to the homogeneous 7,929 471 Assuming t means time, then V=dx/dt So dt = dx/ (a*x 2 b*xc) Integrate both sides to get t as a function of x Solve for x as a function of t Then take the derivative to get V Good luck!#3
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Yc(t) = c xc(t) If we can write c = Aej', then A is the amplitude and ' is the phase of the output relative to the amplitude and phase of the input 32Substituting t X(w)et dw = 0 in the preceding equation, we get x(0) = 1 27r X( ) de (b) X(w) = f x(t)e jwt dt Substituting o = 0 in the preceding equation, we get X(0) =x(t) dt S97 (a) We are given the differential equation dy(t) 2y(t) = x(t) dt Taking the Fourier transform of eq (S971), we have jwY(w) 2Y(w) = X(w)T(x) = 0, then T(x 1 x 2) = x 1 = 0, which means x 1 = 0, x = x 2 ∈ W 2 We can conclude N(T) = W 2 part c) If W 1 = V, then W 2 = 0 and also T(x) = x for any x, hence T = I part d) If W 1 = 0, then N(T) = W 2 = V, therefore T(x) = 0 for all x Question 2 22/12 Let T still be a projection on W along W 0, where V = W ⊕ W Then we can



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