×)-Daniel Norman. Dan w \\eChegg Study | Guide x 122-Fin-F 17.p &d2lSessionVal;
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×)-Daniel Norman. Dan w eChegg Study | Guide x 122-Fin-F 17.p &d2lSessionVal; -3 /173273-MathematicsAssistanceCentre/Mocks/Mock-Ma122-Fin-F17.pdf? š. l8 markal let L be a linecur napping with the following matrix representation, which can be red 10 0 - to row-echelon formn as show: I4112 1 0o 10 0 -6 1 4 -61 4 o 0 1 -1 (a) State the domain and codomain for the linear mapping L Domain: :Codomin (b) Give a basis for the range of L (c) Give a basis for the rowspace of [L. (d) Determine 'he nullity(IL) justifying your answer. U EPICExplanation / Answer
(a). [L] has 3 rows and 4 columns. Further, the column space of [L] is the subspace of the codomain which is spanned by the columns of [L]. It is apparent from the RREF of [L] that the 4th column of [L] is a linear combination of its first 3 columns , and the first 3 columns of [L] are linearly independent. Hence the codomain of L is R3. The domain of the linear transformation L is the vector space on which the it acts i.e. the domain of L is a subspace of R4. The RREF of [L]T is
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Hence the domain of L is the subspace of R4 spanned by { (1,0,0,0)T,(0,1,0,0)T,(0,0,1,0)T}.
(b). The range of L is same as col[L] so that a basis for range (L) is {(1,0,0)T,(0,1,0)T,(0,0,1)T}.
(c ). It is apparent from the RREF of [L]Tthat a basis for the row space of [ L] is { (1,0,0,0)T, (0,1,0,0)T, (0,0,1,0)T}.
(d).The nullspace of [L] is the set of solutions to the equation [L] X = 0. If X = (x,y,z,w)T, then this equation is equivalent to x -2w/3 = 0 or, x = 2w/3, y-5w/6 = 0 or, y = 5w/6 and z-w = 0 or, z= w. Then X = (2w/3,5w/6,w,w)T = w/6(4,5,6,6)T. Hence { (4,5,6,6)T} is a basis for the nullspace of [L].
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