Suppose T M22 rightarrow P2 is a linear transformation defin

Suppose T: M_2,2 rightarrow P_2 is a linear transformation defined by T([a b c d]) = (a - b) x + (b - c)x^2, find

Solution

(a) T(x + y) = T(x) + T(y) (b) T(ax) = aT(x) . Letting a = 0 and -1 shows T(0) = 0 and 216 LINEAR TRANSFORMATIONS AND MATRICES T(-x) = -T(x) . We also see that T(x - y) = T(x + (-y)) = T(x) + T(-y) = T(x) - T(y) . It should also be clear that by induction we have, for any finite sum, T(Íaixi) = ÍT(aáxá) = ÍaáT(xá) for any vectors xá V and scalars aá F.

Proof Let {uè, . . . , uñ} be a basis for U and suppose that Ker T = {0}. Then for any x U we have T(x) = T(Íxáuá) = ÍxáT(uá) for some set of scalars xá, and therefore {T(uá)} spans Im T. If ÍcáT(uá) = 0, then 0 = ÍcáT(uá) = ÍT(cáuá) = T(Ícáuá) which implies that Ícáuá = 0 (since Ker T = {0}). But the uá are linearly independent so that we must have cá = 0 for every i, and hence {T(uá)} is linearly independent. Since nul T = dim(Ker T) = 0 and r(T) = dim(Im T) = n = dim U, we see that r(T) + nul T = dim U. Now suppose that Ker T {0}, and let {wè, . . . , wÉ} be a basis for Ker T. By Theorem 2.10, we may extend this to a basis {wè, . . . , wñ} for U. Since T(wá) = 0 for each i = 1, . . . , k it follows that the vectors T(wk+1), . . . , T(wñ) span Im T. If cj T (wj) = 0 j=k+1 n ! for some set of scalars cá, then 0 = cj T (wj) = j=k+1 n ! T (cj wj) = T ( cj wj) j=k+1 n ! j=k+1 n ! so that Íj ˆ= k+1céwé Ker T. This means that cj wj = aj wj j=1 k ! j=k+1 n ! for some set of scalars aá. But this is just ajwj j=1 k ! \" cjwj = 0 j=k+1 n ! and hence aè = ~ ~ ~ = aÉ = ck+1 = ~ ~ ~ = cn = 0 since the wé are linearly independent. Therefore T(wk+1 ), . . . , T(wñ) are linearly independent and thus form a basis for Im T. We have therefore shown that dim U = k + (n - k) = dim(Ker T) + dim(Im T) = nul T + r(T)

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